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Precalculus : Trigonometric Functions

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Example Questions

Example Question #1492 : Pre Calculus

Simplify.

$1-\tan x(\csc x)$

Possible Answers:

None of these answers are correct.

$\sin^2x$

$\tan^2 x$

$1-\sec x$

Correct answer:

$1-\sec x$

Explanation:

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

$\frac{1}{\sin x}=\csc x$

Given these identities...

$1-\tan x(\csc x)$

$1- \frac{\sin x}{\cos x}(\frac{1}{\sin x})$

$1-\frac{1}{\cos x}$

$1- \sec x$

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Example Question #1493 : Pre Calculus

Simplify $\frac{\sin (x)}{1-\cos (x)}-\cot(x)$ completely.

Possible Answers:

$-\sec (x)$

$\csc (x)$

$-\tan (x)$

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

$1-\tan(x)$

Correct answer:

$\csc (x)$

Explanation:

First simplify the fraction

$\frac{\sin (x)}{1-\cos (x)}$

by multiplying it by its conjugate

$\frac{1+\cos (x)}{1+\cos (x)}$ .

After doing so, continue simplying:

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

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Example Question #1494 : Pre Calculus

Fully simplify.

Simplify:

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

Possible Answers:

$\sin ^2 x$

None of these answers are correct.

$\sin x + \cos x$

$\cos^2 x$

Correct answer:

$\sin ^2 x$

Explanation:

$\sin^2 x + \cos^2 x = 1$

$1+ \cot^2x= \csc^2 x$

Given the above identities:

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

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Example Question #1495 : Pre Calculus

Simplify:

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

Possible Answers:

None of these answers are correct.

$1-2\cos^2x$

$1+2\sin^2x$

Correct answer:

Explanation:

$\sin^2x+\cos^2x=1$

and

$\sin^2x=1-\cos^2x$

Therefore...

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

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

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

=1+1=2

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Example Question #131 : Trigonometric Functions

Find the exact value

$sin\left[2tan^{-1}\left(\frac{3}{4}\right)\right]$ .

Possible Answers:

$\frac{12}{25}$

$\frac{6}{5}$

$\frac{24}{25}$

$\frac{9}{8}$

Correct answer:

$\frac{24}{25}$

Explanation:

By the double angle formula

${\right }sin(2x)=2sin(x)cos(x)$

$sin\left[2tan^{-1}\left(\frac{3}{4}\right)\right]=2sin\left[tan^{-1}\left(\frac{3}{4}\right)\right]cos\left[tan^{-1}\left(\frac{3}{4}\right)\right]$

$=2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)=\frac{24}{25}$

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Example Question #132 : Trigonometric Functions

Find the exact value

$sin\left[2tan^{-1}\left(\frac{8}{15}\right)\right]$ .

Possible Answers:

$\frac{16}{17}$

$\frac{120}{289}$

$\frac{16}{289}$

$\frac{240}{289}$

Correct answer:

$\frac{240}{289}$

Explanation:

By the double angle formula

{}sin(2x)=2sin(x)cos(x)

$sin\left[2tan^{-1}\left(\frac{8}{15}\right)\right]=2sin\left[tan^{-1}\left(\frac{8}{15}\right)\right]cos\left[tan^{-1}\left(\frac{8}{15}\right)\right]$

$=2\left(\frac{8}{17}\right)\left(\frac{15}{17}\right)=\frac{240}{289}$

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Example Question #22 : Fundamental Trigonometric Identities

Find the exact value

$cos\left[2tan^{-1}\left(\frac{5}{12}\right)\right]$ .

Possible Answers:

$-\frac{119}{169}$

$\frac{50}{169}$

$\frac{7}{13}$

$\frac{119}{169}$

Correct answer:

$\frac{119}{169}$

Explanation:

By the double-angula formula for cosine

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

For this problem

$cos\left[2tan^{-1}\left(\frac{5}{12}\right)\right]=cos^2\left[tan^{-1}\left(\frac{5}{12}\right)\right]-sin^2\left[tan^{-1}\left(\frac{5}{12}\right)\right]$

$=\left(\frac{12}{13}\right)^2-\left(\frac{5}{13}\right)^2=\frac{119}{169}$

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Example Question #23 : Fundamental Trigonometric Identities

Find the exact value

$sin\left[2sin^{-1}\left(\frac{12}{37}\right)\right]$ .

Possible Answers:

$\frac{840}{1369}$

$\frac{840}{37}$

$\frac{24}{37}$

$\frac{420}{1369}$

Correct answer:

$\frac{840}{1369}$

Explanation:

By the double-angle formula for the sine function

sin(2x)=2sin(x)cos(x)

we have

$x=sin^{-1}\left( \frac{12}{37}\right)$

thus the double angle formula becomes,

$sin\left[2sin^{-1}\left(\frac{12}{37}\right)\right]=2sin\left[sin^{-1}\left(12/37\right)\right]cos\left[sin^{-1}\left(12/37\right)\right]$

$=2\left(\frac{12}{37}\right)\left(\frac{35}{37}\right)=\frac{840}{1369}$

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Example Question #132 : Trigonometric Functions

If $\theta=X$ , which of the following best represents $sin(2\theta)$ ?

Possible Answers:

2sinXcosX

sinXcosX

cos^2X-sin^2X

sin^2X

1-cos^2X

Correct answer:

2sinXcosX

Explanation:

The expression $sin(2\theta)$ is a double angle identity that can also be rewritten as:

$2sin(\theta) cos(\theta)$

Replace the value of theta for $\theta=X$ .

The correct answer is: 2sin(X) cos(X)

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Example Question #25 : Fundamental Trigonometric Identities

Which expression is equivalent to $\cos(6x)$ ?

Possible Answers:

$1 - 6 \sin ^ 2 x$

$3 \cos ^ 2 (2x ) - 1$

$\cos ^ 2 x - \sin ^ 2 x$

$6 \cos ^ 2 x - 1$

$1- 2 \sin^2 (3x)$

Correct answer:

$1- 2 \sin^2 (3x)$

Explanation:

The relevant trigonometric identity is:

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

In this case, "u" is since 2*3x = 6x .

The only one that actually follows this is $1- 2 \sin^2 (3x)$

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Precalculus : Trigonometric Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1492 : Pre Calculus

Example Question #1493 : Pre Calculus

Example Question #1494 : Pre Calculus

Example Question #1495 : Pre Calculus

Example Question #131 : Trigonometric Functions

Example Question #132 : Trigonometric Functions

Example Question #22 : Fundamental Trigonometric Identities

Example Question #23 : Fundamental Trigonometric Identities

Example Question #132 : Trigonometric Functions

Example Question #25 : Fundamental Trigonometric Identities

All Precalculus Resources

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