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Trigonometry : Trigonometric Functions

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Example Questions

Example Question #1 : Simplifying Trigonometric Functions

If $cos\ x=\frac{1}{3}$ and $sin\ x\neq 0$ , give the value of $\frac{sin\ x+sin\ 2x}{sin\ x-sin\ 2x}$ .

Possible Answers:

Correct answer:

Explanation:

Based on the double angle formula we have, $sin\ 2x=2(sin\ x)(cos\ x)$ .

$\frac{sin\ x+sin\ 2x}{sin\ x-sin\ 2x}=\frac{sin\ x+2(sin\ x)(cos\ x)}{sin\ x-2(sin\ x)(cos\ x)}$

$=\frac{sinx(1+2cos\ x )}{sinx(1-2cos\ x )}=\frac{1+2cos\ x}{1-2cos\ x}$

$=\frac{1+2(\frac{1}{3})}{1-2(\frac{1}{3})}=\frac{\frac{3+2}{3}}{\frac{3-2}{3}}=5$

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Example Question #3 : Simplifying Trigonometric Functions

If $x=\frac{\pi}{3}$ , give the value of $\frac{sin\ x}{sin\ x+cos\ 2x}$ .

Possible Answers:

$\frac{3-\sqrt{3}}{2}$

$\frac{3+\sqrt{3}}{2}$

$\frac{3+\sqrt{2}}{2}$

$\frac{3-\sqrt{2}}{2}$

Correct answer:

$\frac{3+\sqrt{3}}{2}$

Explanation:

$x=\frac{\pi}{3}\Rightarrow sin\ x=sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$

Now we can write:

$x=\frac{\pi}{3}\Rightarrow 2x=2(\frac{\pi}{3})=\frac{2\pi}{3}\Rightarrow cos\ 2x=cos(\frac{2\pi}{3})$

$=cos(\pi-\frac{\pi}{3})$

$=-cos (\frac{\pi}{3})$

$=-\frac{1}{2}$

Now we can substitute the values:

$\frac{sin\ x}{sin\ x+cos\ 2x}=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}-\frac{1}{2}}=\frac{\sqrt{3}}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\times \frac{{\sqrt{3}+1}}{{\sqrt{3}+1}}=\frac{3+\sqrt{3}}{2}$

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Example Question #11 : Simplifying Trigonometric Functions

If $cot\ 34^{\circ}=1.5$ , give the value of $\frac{2sin\ 326^{\circ}+3sin\ 56^{\circ}}{cos\ 304^{\circ}}$ .

Possible Answers:

1.5

0.5

2.5

Correct answer:

2.5

Explanation:

$326^{\circ}=360^{\circ}-34^{\circ}\Rightarrow sin\ 326^{\circ}=sin(360^{\circ}-34^{\circ})\Rightarrow sin\ 326^{\circ}=-sin\ 34^{\circ}$

$56^{\circ}=90^{\circ}-34^{\circ}\Rightarrow sin\ 56^{\circ}=sin(90^{\circ}-34^{\circ})\Rightarrow sin\ 56^{\circ}=cos\ 34^{\circ}$

$304^{\circ}=270^{\circ}+34^{\circ}\Rightarrow cos\ 304^{\circ}=cos(270^{\circ}+34^{\circ})\Rightarrow cos\ 304^{\circ}=sin\ 34^{\circ}$

Now we can simplify the expression as follows:

$\frac{2sin\ 326^{\circ}+3sin\ 56^{\circ}}{cos\ 304^{\circ}}=\frac{-2sin\ 34^{\circ}+3cos\ 34^{\circ}}{sin\ 34^{\circ}}$

$=-2+3cot\ 34^{\circ}=-2+3\times 1.5=-2+4.5=2.5$

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Example Question #31 : Trigonometric Functions

If $sin\ x\times cos\ x=\frac{1}{2}$ , what is the value of $(sin\ x+cos\ x)^2$ ?

Possible Answers:

2.5

1.5

Correct answer:

Explanation:

$(sin\ x+cos\ x)^2=sin^2x+cos^2x+2sin\x cos\ x$

$=1+2(sin\ x)(cos\ x)=1+2(\frac{1}{2})=1+1=2$

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Example Question #33 : Trigonometric Functions And Graphs

If $sin^2x-cos^2x=\frac{1}{2}$ , give the value of cos^4x-sin^4x .

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

Correct answer:

$-\frac{1}{2}$

Explanation:

cos^4x-sin^4x = (cos^2x-sin^2x)(cos^2x+sin^2x)

We know that cos^2x+sin^2x=1 .

cos^4x-sin^4x = (cos^2x-sin^2x)(cos^2x+sin^2x)

$=(cos^2x-sin^2x)\times 1=-(sin^2x-cos^2x)=-\frac{1}{2}$

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Example Question #31 : Trigonometric Functions And Graphs

Simplify:

A=(cos^4x-sin^4x)(1+tan^2x)+tan^2x

Possible Answers:

1.5

0.5

2.5

Correct answer:

Explanation:

A=(cos^4x-sin^4x)(1+tan^2x)+tan^2x

= (cos^2x-sin^2x)(cos^2x+sin^2x)(1+tan^2x)+tan^2x

We know that cos^2x+sin^2x=1 .

Then we can write:

A= (cos^2x-sin^2x)(1+tan^2x)+tan^2x

$=(cos^2x-sin^2x)(1+\frac{sin^2x}{cos^2x})+tan^2x$

$=(cos^2x-sin^2x)(\frac{cos^2x+sin^2x}{cos^2x})+tan^2x$

$\Rightarrow A=\frac{(cos^2x-sin^2x)}{cos^2x}+tan^2x$

$=\frac{cos^2x}{cos^2x}-\frac{sin^2x}{cos^2x}+tan^2x$

=1-tan^2x+tan^2x

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Example Question #33 : Trigonometric Functions

Which of the following is equal to $\sin x\tan x$ ?

Possible Answers:

$\cos x \cot x$

$\frac{\cos^2x}{\sin x}$

$\sec x \csc x$

$\frac{\sin^2x}{\cos x}$

$\sin x \csc x$

Correct answer:

$\frac{\sin^2x}{\cos x}$

Explanation:

Break $\tan x$ apart: $\tan x =\frac{\sin x }{\cos x}$ .

This means that $\sin x\tan x=\sin x *\frac{\sin x}{\cos x}$ or $\frac{\sin^2 x}{\cos x}$

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Example Question #12 : Simplifying Trigonometric Functions

Simplify: $\frac{sin(\theta)}{tan(\theta)}-\frac{csc(\theta)}{sec(\theta)}$

Possible Answers:

$cos(\theta)+cot(\theta)$

$sec(\theta)-tan(\theta)$

$cos(\theta)-sin(\theta)cos(\theta)$

$cos(\theta)cot(\theta)-cot(\theta)$

$cos(\theta)-cot(\theta)$

Correct answer:

$cos(\theta)-cot(\theta)$

Explanation:

Rewrite $\frac{sin(\theta)}{tan(\theta)}-\frac{csc(\theta)}{sec(\theta)}$ in terms of sines and cosines.

$\frac{sin(\theta)}{tan(\theta)}-\frac{csc(\theta)}{sec(\theta)} = \frac{sin(\theta)}{\frac{sin(\theta)}{cos(\theta)}}-\frac{\frac{1}{sin(\theta)}}{\frac{1}{cos(\theta)}}$

Simplify the complex fractions.

$\frac{sin(\theta)}{\frac{sin(\theta)}{cos(\theta)}}-\frac{\frac{1}{sin(\theta)}}{\frac{1}{cos(\theta)}}=sin(\theta)\cdot \frac{cos(\theta)}{sin(\theta)}-\frac{1}{sin(\theta)}\cdot \frac{cos(\theta)}{1}$

$=cos(\theta)-\frac{cos(\theta)}{sin(\theta)}= cos(\theta)-cot(\theta)$

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Example Question #113 : Trigonometry

Simplify the following expression:

$sec^2xcot^2x \ - sec^2x + tan^2x$

Possible Answers:

$cos \ 2x$

cos^2x

tan^2x

cot^2x

Correct answer:

cot^2x

Explanation:

We will first invoke the appropriate ratio for cotangent, and then use pythagorean identities to simplify the expression:

$sec^2xcot^2x \ - sec^2x + tan^2x$

$= \frac{1}{cos^2x} * \frac{cos^2}{sin^2x} - (sec^2 x - tan^2x){}$

$= \frac{1}{sin^2x} - (sec^2x - tan^2x){}$

$= csc^2x - 1{}$

since

$sec^2x - tan^2x = 1; \ csc\ x = \frac{1}{sin \ x}{}$

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Example Question #34 : Trigonometric Functions

Simplify the trigonometric expression.

$\frac{\tan(\theta)\csc(\theta)}{\sec(\theta)}$

Possible Answers:

$\tan(\theta)$

$\cos(\theta)$

$\sec\theta$

Correct answer:

Explanation:

Using basic trigonometric identities, we can simplify the problem to

$\frac{\frac{\sin(\theta)}{\cos(\theta)}*\frac{1}{\sin(\theta)}}{\frac{1}{\cos(\theta)}}$ .

We can cancel the sine in the numerator and the one over cosine cancels on top and bottom, leaving us with 1.

$\frac{\frac{\sin(\theta)}{\cos(\theta)}*\frac{1}{\sin(\theta)}}{\frac{1}{\cos(\theta)}}=\frac{\frac{1}{\cos(\theta)}}{\frac{1}{\cos(\theta)}}=\frac{1}{\cos(\theta)}\cdot \frac{\cos(\theta)}{1}=1$

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Trigonometry : Trigonometric Functions

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Simplifying Trigonometric Functions

Example Question #3 : Simplifying Trigonometric Functions

Example Question #11 : Simplifying Trigonometric Functions

Example Question #31 : Trigonometric Functions

Example Question #33 : Trigonometric Functions And Graphs

Example Question #31 : Trigonometric Functions And Graphs

Example Question #33 : Trigonometric Functions

Example Question #12 : Simplifying Trigonometric Functions

Example Question #113 : Trigonometry

Example Question #34 : Trigonometric Functions

All Trigonometry Resources

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