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Precalculus : Fundamental Trigonometric Identities

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Example Questions

Example Question #175 : Trigonometric Functions

$\sin(-x)\cos(-x)\tan(-x)$

Possible Answers:

$\sin^2 x$

$\tan^2 x$

$-\sin x\cos x$

$-\sin^2 x$

$-\tan^2 x$

Correct answer:

$\sin^2 x$

Explanation:

Using these basic identities:

$\sin(-x)=-\sin x$

$\cos(-x)=\cos(x)$

$\tan(-x)=-\tan x$

we find the original expression to be

$(-\sin x)(\cos x)(-\tan x)$

which simplifies to

$\sin x\cos x\tan x$ .

Further simplifying:

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

The cosines cancel, giving us

$\sin ^2 x$

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Example Question #176 : Trigonometric Functions

Which of the following is the best answer for $cos^2(\theta)$ ?

Possible Answers:

$-cot^2(\theta)$

$-sin^2(\theta)$

$-sec^2(\theta)$

$tan^2(\theta)-1$

$1-sin^2(\theta)$

Correct answer:

$1-sin^2(\theta)$

Explanation:

Write the Pythagorean identity.

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

Substract $sin^2(\theta)$ from both sides.

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

The other answers are incorrect.

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Example Question #176 : Trigonometric Functions

Express $\frac{1}{\sec x \csc x \tan x}$ in terms of only sines and cosines.

Possible Answers:

$\sin^2x$

$\cos^2 x$

$\sin x \cos^2 x$

$\cos x \sin^2 x$

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

Correct answer:

$\cos^2 x$

Explanation:

The correct answer is $\cos^2 x$ . Begin by substituting $\frac{1}{\sec x}=\cos x$ , $\frac{1}{\csc x}=\sin x$ , and $\frac{1}{\tan x}=\frac{\cos x}{\sin x}$ . This gives us:

$\frac{1}{\sec x\csc x\tan x}=\frac{\cos x\sin x\cos x}{\sin x}=\cos^2 x$ .

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Example Question #177 : Trigonometric Functions

Express $\sec x\tan x+\frac{\cot x}{\csc x}$ in terms of only sines and cosines.

Possible Answers:

$\sin x +1$

$\cos x +1$

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

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

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

Correct answer:

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

Explanation:

To solve this problem, use the identities $\sec x =\frac{1}{\cos x}$ , $\tan x=\frac{\sin x}{\cos x}$ , $\cot x =\frac{\cos x}{\sin x}$ , and $\frac{1}{\csc x}=\sin x$ . Then we get

$\sec x\tan x+\frac{\cot x}{\csc x}$

$=\frac{1}{\cos x}\cdot \frac{\sin x}{\cos x}+\sin x\cdot \frac{\cos x}{\sin x}$

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

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Precalculus : Fundamental Trigonometric Identities

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #175 : Trigonometric Functions

Example Question #176 : Trigonometric Functions

Example Question #176 : Trigonometric Functions

Example Question #177 : Trigonometric Functions

All Precalculus Resources

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