Prove Pythagorean Identity: CCSS.Math.Content.HSF-TF.C.8 - Common Core: High School - Functions

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Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{12}{13}$

find

$\sin(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{12}{13}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{12}{13}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{13^2}=1$

$\sin^2(\theta)+\frac{144}{169}=1$

$\sin^2(\theta)+\frac{144}{169}-\frac{144}{169}=1-\frac{144}{169}$

$\sin^2(\theta)=\frac{169-144}{169}$

$\sin^2(\theta)=\frac{25}{169}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{25}{169}}$

$\sin(\theta)=\pm\frac{5}{13}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{5}{13}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{12}{13}$

find

$\tan(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{12}{13}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{12}{13}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{13^2}=1$

$\sin^2(\theta)+\frac{144}{169}=1$

$\sin^2(\theta)+\frac{144}{169}-\frac{144}{169}=1-\frac{144}{169}$

$\sin^2(\theta)=\frac{169-144}{169}$

$\sin^2(\theta)=\frac{25}{169}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{25}{169}}$

$\sin(\theta)=\pm\frac{5}{13}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{5}{13}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{5}{13}}{\frac{12}{13}}$

$\tan(\theta)=-\frac{5}{13}\times\frac{13}{12}$

$\tan(\theta)=-\frac{5}{12}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{8}{17}$

find

$\sin(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{8}{17}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{8}{17}\right)^2=1$

$\sin^2(\theta)+\frac{8^2}{17^2}=1$

$\sin^2(\theta)+\frac{64}{289}=1$

$\sin^2(\theta)+\frac{64}{289}-\frac{64}{289}=1-\frac{64}{289}$

$\sin^2(\theta)=\frac{289-64}{289}$

$\sin^2(\theta)=\frac{225}{289}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{225}{289}}$

$\sin(\theta)=\pm\frac{15}{17}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{15}{17}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{8}{17}$

find

$\tan(\theta)$

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{8}{17}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{8}{17}\right)^2=1$

$\sin^2(\theta)+\frac{8^2}{17^2}=1$

$\sin^2(\theta)+\frac{64}{289}=1$

$\sin^2(\theta)+\frac{64}{289}-\frac{64}{289}=1-\frac{64}{289}$

$\sin^2(\theta)=\frac{289-64}{289}$

$\sin^2(\theta)=\frac{225}{289}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{225}{289}}$

$\sin(\theta)=\pm\frac{15}{17}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{15}{17}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{15}{17}}{\frac{8}{17}}$

$\tan(\theta)=-\frac{15}{17}\times\frac{17}{8}$

$\tan(\theta)=-\frac{15}{8}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{12}{37}$

find

$\sin(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{12}{37}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{12}{37}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{37^2}=1$

$\sin^2(\theta)+\frac{144}{1369}=1$

$\sin^2(\theta)+\frac{144}{1369}-\frac{144}{1369}=1-\frac{144}{1369}$

$\sin^2(\theta)=\frac{1369-144}{1369}$

$\sin^2(\theta)=\frac{1225}{1369}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{1225}{1369}}$

$\sin(\theta)=\pm\frac{35}{37}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{35}{37}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{12}{37}$

find

$\tan(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{12}{37}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{12}{37}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{37^2}=1$

$\sin^2(\theta)+\frac{144}{1369}=1$

$\sin^2(\theta)+\frac{144}{1369}-\frac{144}{1369}=1-\frac{144}{1369}$

$\sin^2(\theta)=\frac{1369-144}{1369}$

$\sin^2(\theta)=\frac{1225}{1369}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{1225}{1369}}$

$\sin(\theta)=\pm\frac{35}{37}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{35}{37}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{35}{37}}{\frac{12}{37}}$

$\tan(\theta)=-\frac{35}{37}\times\frac{37}{12}$

$\tan(\theta)=-\frac{35}{12}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{20}{29}$

find

$\sin(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{20}{29}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{20}{29}\right)^2=1$

$\sin^2(\theta)+\frac{20^2}{29^2}=1$

$\sin^2(\theta)+\frac{400}{841}=1$

$\sin^2(\theta)+\frac{400}{841}-\frac{400}{841}=1-\frac{400}{841}$

$\sin^2(\theta)=\frac{841-144}{841}$

$\sin^2(\theta)=\frac{441}{841}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{441}{841}}$

$\sin(\theta)=\pm\frac{21}{29}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{21}{29}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{20}{29}$

find

$\tan(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{20}{29}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{20}{29}\right)^2=1$

$\sin^2(\theta)+\frac{20^2}{29^2}=1$

$\sin^2(\theta)+\frac{400}{841}=1$

$\sin^2(\theta)+\frac{400}{841}-\frac{400}{841}=1-\frac{400}{841}$

$\sin^2(\theta)=\frac{841-144}{841}$

$\sin^2(\theta)=\frac{441}{841}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{441}{841}}$

$\sin(\theta)=\pm\frac{21}{29}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{21}{29}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{21}{29}}{\frac{20}{29}}$

$\tan(\theta)=-\frac{21}{29}\times\frac{29}{20}$

$\tan(\theta)=-\frac{21}{20}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{4}{5}$

find

$\sin(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{4}{5}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{4}{5}\right)^2=1$

$\sin^2(\theta)+\frac{4^2}{5^2}=1$

$\sin^2(\theta)+\frac{16}{25}=1$

$\sin^2(\theta)+\frac{16}{25}-\frac{16}{25}=1-\frac{16}{25}$

$\sin^2(\theta)=\frac{25-16}{25}$

$\sin^2(\theta)=\frac{9}{25}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{9}{25}}$

$\sin(\theta)=\pm\frac{3}{5}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{3}{5}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{4}{5}$

find

$\tan(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{4}{5}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{4}{5}\right)^2=1$

$\sin^2(\theta)+\frac{4^2}{5^2}=1$

$\sin^2(\theta)+\frac{16}{25}=1$

$\sin^2(\theta)+\frac{16}{25}-\frac{16}{25}=1-\frac{16}{25}$

$\sin^2(\theta)=\frac{25-16}{25}$

$\sin^2(\theta)=\frac{9}{25}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{9}{25}}$

$\sin(\theta)=\pm\frac{3}{5}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{3}{5}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{3}{5}}{\frac{4}{5}}$

$\tan(\theta)=-\frac{3}{5}\times\frac{5}{4}$

$\tan(\theta)=-\frac{3}{4}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{11}{61}$

find

$\sin(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{11}{61}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{11}{61}\right)^2=1$

$\sin^2(\theta)+\frac{11^2}{61^2}=1$

$\sin^2(\theta)+\frac{121}{3721}=1$

$\sin^2(\theta)+\frac{121}{3721}-\frac{121}{3721}=1-\frac{121}{3721}$

$\sin^2(\theta)=\frac{3721-121}{3721}$

$\sin^2(\theta)=\frac{3600}{3721}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{3600}{3721}}$

$\sin(\theta)=\pm\frac{60}{61}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{60}{61}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{11}{61}$

find

$\tan(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{11}{61}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{11}{61}\right)^2=1$

$\sin^2(\theta)+\frac{11^2}{61^2}=1$

$\sin^2(\theta)+\frac{121}{3721}=1$

$\sin^2(\theta)+\frac{121}{3721}-\frac{121}{3721}=1-\frac{121}{3721}$

$\sin^2(\theta)=\frac{3721-121}{3721}$

$\sin^2(\theta)=\frac{3600}{3721}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{3600}{3721}}$

$\sin(\theta)=\pm\frac{60}{61}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{60}{61}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{60}{61}}{\frac{11}{61}}$

$\tan(\theta)=-\frac{60}{61}\times\frac{61}{11}$

$\tan(\theta)=-\frac{60}{11}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{12}{13}$

find

$\sin(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{12}{13}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{12}{13}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{13^2}=1$

$\sin^2(\theta)+\frac{144}{169}=1$

$\sin^2(\theta)+\frac{144}{169}-\frac{144}{169}=1-\frac{144}{169}$

$\sin^2(\theta)=\frac{169-144}{169}$

$\sin^2(\theta)=\frac{25}{169}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{25}{169}}$

$\sin(\theta)=\pm\frac{5}{13}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{5}{13}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{12}{13}$

find

$\tan(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{12}{13}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{12}{13}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{13^2}=1$

$\sin^2(\theta)+\frac{144}{169}=1$

$\sin^2(\theta)+\frac{144}{169}-\frac{144}{169}=1-\frac{144}{169}$

$\sin^2(\theta)=\frac{169-144}{169}$

$\sin^2(\theta)=\frac{25}{169}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{25}{169}}$

$\sin(\theta)=\pm\frac{5}{13}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{5}{13}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{5}{13}}{\frac{12}{13}}$

$\tan(\theta)=-\frac{5}{13}\times\frac{13}{12}$

$\tan(\theta)=-\frac{5}{12}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{8}{17}$

find

$\sin(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{8}{17}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{8}{17}\right)^2=1$

$\sin^2(\theta)+\frac{8^2}{17^2}=1$

$\sin^2(\theta)+\frac{64}{289}=1$

$\sin^2(\theta)+\frac{64}{289}-\frac{64}{289}=1-\frac{64}{289}$

$\sin^2(\theta)=\frac{289-64}{289}$

$\sin^2(\theta)=\frac{225}{289}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{225}{289}}$

$\sin(\theta)=\pm\frac{15}{17}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{15}{17}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{8}{17}$

find

$\tan(\theta)$

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{8}{17}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{8}{17}\right)^2=1$

$\sin^2(\theta)+\frac{8^2}{17^2}=1$

$\sin^2(\theta)+\frac{64}{289}=1$

$\sin^2(\theta)+\frac{64}{289}-\frac{64}{289}=1-\frac{64}{289}$

$\sin^2(\theta)=\frac{289-64}{289}$

$\sin^2(\theta)=\frac{225}{289}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{225}{289}}$

$\sin(\theta)=\pm\frac{15}{17}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{15}{17}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{15}{17}}{\frac{8}{17}}$

$\tan(\theta)=-\frac{15}{17}\times\frac{17}{8}$

$\tan(\theta)=-\frac{15}{8}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{12}{37}$

find

$\sin(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{12}{37}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{12}{37}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{37^2}=1$

$\sin^2(\theta)+\frac{144}{1369}=1$

$\sin^2(\theta)+\frac{144}{1369}-\frac{144}{1369}=1-\frac{144}{1369}$

$\sin^2(\theta)=\frac{1369-144}{1369}$

$\sin^2(\theta)=\frac{1225}{1369}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{1225}{1369}}$

$\sin(\theta)=\pm\frac{35}{37}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{35}{37}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{12}{37}$

find

$\tan(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{12}{37}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{12}{37}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{37^2}=1$

$\sin^2(\theta)+\frac{144}{1369}=1$

$\sin^2(\theta)+\frac{144}{1369}-\frac{144}{1369}=1-\frac{144}{1369}$

$\sin^2(\theta)=\frac{1369-144}{1369}$

$\sin^2(\theta)=\frac{1225}{1369}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{1225}{1369}}$

$\sin(\theta)=\pm\frac{35}{37}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{35}{37}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{35}{37}}{\frac{12}{37}}$

$\tan(\theta)=-\frac{35}{37}\times\frac{37}{12}$

$\tan(\theta)=-\frac{35}{12}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{20}{29}$

find

$\sin(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{20}{29}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{20}{29}\right)^2=1$

$\sin^2(\theta)+\frac{20^2}{29^2}=1$

$\sin^2(\theta)+\frac{400}{841}=1$

$\sin^2(\theta)+\frac{400}{841}-\frac{400}{841}=1-\frac{400}{841}$

$\sin^2(\theta)=\frac{841-144}{841}$

$\sin^2(\theta)=\frac{441}{841}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{441}{841}}$

$\sin(\theta)=\pm\frac{21}{29}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{21}{29}$

Compare your answer with the correct one above

Question

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{20}{29}$

find

$\tan(\theta)$ ?

Answer

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{20}{29}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{20}{29}\right)^2=1$

$\sin^2(\theta)+\frac{20^2}{29^2}=1$

$\sin^2(\theta)+\frac{400}{841}=1$

$\sin^2(\theta)+\frac{400}{841}-\frac{400}{841}=1-\frac{400}{841}$

$\sin^2(\theta)=\frac{841-144}{841}$

$\sin^2(\theta)=\frac{441}{841}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{441}{841}}$

$\sin(\theta)=\pm\frac{21}{29}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{21}{29}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{21}{29}}{\frac{20}{29}}$

$\tan(\theta)=-\frac{21}{29}\times\frac{29}{20}$

$\tan(\theta)=-\frac{21}{20}$

Compare your answer with the correct one above

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