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All Common Core: High School - Functions Resources

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Flashcards: Use Special Triangles to Determine Trigonometric Functions: CCSS.Math.Content.HSF-TF.A.3

Calculate sine, cosine, and tangent of the following angle.

$\theta=\frac{2\pi}{6}$

$\\\sin\left(\frac{2\pi}{6} \right )=\frac{\sqrt{3}}{2} \\ \\\cos\left(\frac{2\pi}{6} \right )=\frac{1}{2} \\ \\ \tan\left( \frac{2\pi}{6}\right )=\sqrt{3}$

$\\\sin\left(\frac{2\pi}{6} \right )=\frac{\sqrt{3}}{2} \\ \\\cos\left(\frac{2\pi}{6} \right )=\frac{1}{2} \\ \\ \tan\left( \frac{2\pi}{6}\right )=\frac{1}{2}$

$\\\sin\left(\frac{2\pi}{6} \right )=\frac{1}{2} \\ \\\cos\left(\frac{2\pi}{6} \right )=\frac{\sqrt{3}}{2} \\ \\ \tan\left( \frac{2\pi}{6}\right )=\frac{1}{\sqrt{3}}$

$\\\sin\left(\frac{2\pi}{6} \right )=\frac{\sqrt{3}}{2} \\ \\\cos\left(\frac{2\pi}{6} \right )=\frac{1}{2} \\ \\ \tan\left( \frac{2\pi}{6}\right )=\frac{\sqrt{3}}{2}$

$\\\sin\left(\frac{2\pi}{6} \right )=\frac{1}{2} \\ \\\cos\left(\frac{2\pi}{6} \right )=\frac{\sqrt{3}}{2} \\ \\ \tan\left( \frac{2\pi}{6}\right )=\sqrt{3}$

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Flashcards: Use Special Triangles to Determine Trigonometric Functions: CCSS.Math.Content.HSF-TF.A.3

Calculate sine, cosine, and tangent of the following angle.

$\theta=\frac{2\pi}{6}$

Your answer:

$\\\sin\left(\frac{2\pi}{6} \right )=\frac{\sqrt{3}}{2} \\ \\\cos\left(\frac{2\pi}{6} \right )=\frac{1}{2} \\ \\ \tan\left( \frac{2\pi}{6}\right )=\frac{1}{2}$

Your answer:

$\\\sin\left(\frac{2\pi}{6} \right )=\frac{1}{2} \\ \\\cos\left(\frac{2\pi}{6} \right )=\frac{\sqrt{3}}{2} \\ \\ \tan\left( \frac{2\pi}{6}\right )=\frac{1}{\sqrt{3}}$

Your answer:

$\\\sin\left(\frac{2\pi}{6} \right )=\frac{\sqrt{3}}{2} \\ \\\cos\left(\frac{2\pi}{6} \right )=\frac{1}{2} \\ \\ \tan\left( \frac{2\pi}{6}\right )=\frac{\sqrt{3}}{2}$

Your answer:

$\\\sin\left(\frac{2\pi}{6} \right )=\frac{1}{2} \\ \\\cos\left(\frac{2\pi}{6} \right )=\frac{\sqrt{3}}{2} \\ \\ \tan\left( \frac{2\pi}{6}\right )=\sqrt{3}$

Correct answer:

$\\\sin\left(\frac{2\pi}{6} \right )=\frac{\sqrt{3}}{2} \\ \\\cos\left(\frac{2\pi}{6} \right )=\frac{1}{2} \\ \\ \tan\left( \frac{2\pi}{6}\right )=\sqrt{3}$

Explanation:

This question tests one's ability to understand the connection between special triangles (30-60-90 and 45-45-90), trigonometric functions related to them (sine, cosine, tangent), and the corresponding degree/radian on the unit circle. It relies on the understanding that the measure of an angle can be described using the properties of the unit circle and arc length. Furthermore, questions involving concepts dealing with trigonometry use the foundation of corresponding identities with angles that come from the unit circle. Recall that the unit circle is a circle that has a radius of one and is centered at the origin. Each (x,y) pair that lies on the circle can be found by creating a right triangle that has a base on the -axis and a hypotenuse that is from the center of the circle to the desired point on the edge of the circle. The height of the triangle is the vertical line that connects the point on the circle's edge to the -axis.

For the purpose of Common Core Standards, understanding "use special triangles to determine geometrically the values of sine, cosine, tangent for $\frac{\pi}{3}, \frac{\pi}{4},\frac{\pi}{6}$ " , falls within the Cluster A of "extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF.TF.A).

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.

Possible approaches to solving this problem:

I. Plot the angle on the unit circle to find the $(x,y)=(cos(\theta),sin(\theta))$ .

II. Identify the special triangle that corresponds to the angle.

III. Using memorization of the unit circle.

For this particular question let's us approach II.

Step 1: Simplify the angle if possible.

$\theta=\frac{2\pi}{6}=\frac{2\cdot 1\pi}{2\cdot 3}=\frac{\pi}{3}$

Step 2: Identify the special right triangle that corresponds to the angle.

Let's convert the angle to degrees to more easily see the special right triangle that it creates.

$\\ \frac{\pi}{3}\times \frac{180^\circ}{\pi}\\ \\=\frac{3\cdot 60}{3}\\\\=60^\circ$

Therefore the special triangle that is created with $\theta$ is 30-60-90 degree triangle. The angle which is made at the origin will be 60 degrees and the length of the segment from the origin to the edge of the unit circle is one by definition.

Step 3: Calculate the side lengths of the special triangle.

From here, recall that the special 30-60-90 degree triangle has the following side identities.

$\\\textup{short side}= \frac{1}{2}\times\textup{hypotenuse} \\\textup{long side}=\textup{short side}\times \sqrt{3}$

Substituting in one for the hypotenuse, the short and long leg of the triangle can be found.

$\\\textup{short side}=\frac{1}{2}\times 1=\frac{1}{2} \\ \\ \textup{long side}=\frac{1}{2}\times\sqrt{3}=\frac{\sqrt{3}}{2}$

It is important to remember that the short side of the triangle is on the -axis and adjacent to $\theta$ . The long side is the vertical line that is parallel to the -axis and opposite of $\theta$ .

Step 4: Calculate sine, cosine, and tangent.

Recall the following trigonometric identities.

$\\ \sin(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}} \\ \\ \\ \cos(\theta)=\frac{\textup{adjacent}}{\textup{hypotenuse}} \\ \\ \\ \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

From Step 2,

$\\ \textup{opposite}=\frac{\sqrt{3}}{2} \\ \\ \textup{adjacent}=\frac{1}{2} \\ \\ \textup{hypotenuse}=1$

Substituting the values found in Step 2 into the above identities is as follows.

$\\ \sin(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}=\frac{\frac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3}}{2} \\ \\ \\ \cos(\theta)=\frac{\textup{adjacent}}{\textup{hypotenuse}}=\frac{\frac{1}{2}}{1}=\frac{1}{2} \\ \\ \\ \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2}\times \frac{2}{1}=\sqrt{3}$

Flashcards: Use Special Triangles to Determine Trigonometric Functions: CCSS.Math.Content.HSF-TF.A.3

Calculate sine of the following angle.

$\theta=\frac{2\pi}{8}$

$\sin\left(\frac{2\pi}{8} \right )={\sqrt{2}}$

$\sin\left(\frac{2\pi}{8} \right )={1}$

$\sin\left(\frac{2\pi}{8} \right )=\frac{\sqrt{3}}{2}$

$\sin\left(\frac{2\pi}{8} \right )=\frac{1}{\sqrt{2}}$

$\sin\left(\frac{2\pi}{8} \right )=\frac{1}{2}$

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Flashcards: Use Special Triangles to Determine Trigonometric Functions: CCSS.Math.Content.HSF-TF.A.3

Calculate sine of the following angle.

$\theta=\frac{2\pi}{8}$

Your answer:

$\sin\left(\frac{2\pi}{8} \right )={\sqrt{2}}$

Your answer:

$\sin\left(\frac{2\pi}{8} \right )={1}$

Your answer:

$\sin\left(\frac{2\pi}{8} \right )=\frac{\sqrt{3}}{2}$

Your answer:

$\sin\left(\frac{2\pi}{8} \right )=\frac{1}{2}$

Correct answer:

$\sin\left(\frac{2\pi}{8} \right )=\frac{1}{\sqrt{2}}$

Explanation:

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.

Possible approaches to solving this problem:

I. Plot the angle on the unit circle to find the $(x,y)=(cos(\theta),sin(\theta))$ .

II. Identify the special triangle that corresponds to the angle.

III. Using memorization of the unit circle.

For this particular question let's us approach II.

Step 1: Simplify the angle if possible.

$\theta=\frac{2\pi}{8}=\frac{2\cdot 1\pi}{2\cdot 4}=\frac{\pi}{4}$

Step 2: Identify the special right triangle that corresponds to the angle.

Let's convert the angle to degrees to more easily see the special right triangle that it creates.

$\\ \frac{\pi}{4}\times \frac{180^\circ}{\pi}\\ \\=\frac{4\cdot 45}{4}\\\\=45^\circ$

Therefore the special triangle that is created with $\theta$ is 45-45-90 degree triangle. The angle which is made at the origin will be 45 degrees and the length of the segment from the origin to the edge of the unit circle is one by definition.

Step 3: Calculate the side lengths of the special triangle.

From here, recall that the special 45-45-90 degree triangle has the following side identities.

$\\ \text{side lengths}: \\\text{opposite side}=\text{adjacent side}=1 \\ \text{hypotenuse}=\sqrt{2}$

Step 4: Calculate sine.

Recall the following trigonometric identities.

$\\ \sin(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}$

$\\ \sin(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}=\frac{1}{\sqrt{2}}$

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All Common Core: High School - Functions Resources

Flashcards: Use Special Triangles to Determine Trigonometric Functions: CCSS.Math.Content.HSF-TF.A.3

Flashcards: Use Special Triangles to Determine Trigonometric Functions: CCSS.Math.Content.HSF-TF.A.3

Flashcards: Use Special Triangles to Determine Trigonometric Functions: CCSS.Math.Content.HSF-TF.A.3

Flashcards: Use Special Triangles to Determine Trigonometric Functions: CCSS.Math.Content.HSF-TF.A.3

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