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Precalculus : Fundamental Trigonometric Identities

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Example Questions

Example Question #1499 : Pre Calculus

Find the exact value

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

Possible Answers:

$\frac{6}{5}$

$\frac{24}{25}$

$\frac{9}{8}$

$\frac{12}{25}$

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 #1500 : Pre Calculus

Find the exact value

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

Possible Answers:

$\frac{240}{289}$

$\frac{16}{289}$

$\frac{16}{17}$

$\frac{120}{289}$

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 #131 : Trigonometric Functions

Find the exact value

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

Possible Answers:

$\frac{7}{13}$

$\frac{119}{169}$

$-\frac{119}{169}$

$\frac{50}{169}$

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 #132 : Trigonometric Functions

Find the exact value

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

Possible Answers:

$\frac{840}{37}$

$\frac{420}{1369}$

$\frac{840}{1369}$

$\frac{24}{37}$

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 #133 : Trigonometric Functions

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

Possible Answers:

sinXcosX

1-cos^2X

2sinXcosX

cos^2X-sin^2X

sin^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 #134 : Trigonometric Functions

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

Possible Answers:

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

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

$6 \cos ^ 2 x - 1$

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

$1 - 6 \sin ^ 2 x$

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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Example Question #141 : Trigonometric Functions

Compute

$\small 2\cos^2 \frac{\pi}{8}$

Possible Answers:

$\small 0$

$\small \pi-2$

$\small \frac{\sqrt{2}}{2}+1$

$\small \frac{\sqrt{3}}{2}-1$

Correct answer:

$\small \frac{\sqrt{2}}{2}+1$

Explanation:

A useful trigonometric identity to remember for this problem is

$\small 2\cos^2 \theta-1=\cos2\theta$

or equivalently,

$\small \small 2\cos^2 \theta=1+\cos2\theta$

If we substitute $\small \theta$ for $\small \frac{\pi}{8}$ , we get

$\small \small \small 2\cos^2 \frac{\pi}{8}=1+\cos2\frac{\pi}{8}=1+\cos\frac{\pi}{4}=1+\frac{\sqrt{2}}{2}$

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Example Question #142 : Trigonometric Functions

Compute

$\small \cos\left(\frac{\pi}{8}\right)\sin\left(\frac{\pi}{8}\right)$

Possible Answers:

$\small \small \frac{\sqrt{6}}{9}$

$\small \frac{\sqrt{3}}{4}$

$\small \frac{1+\pi}{\sqrt{5}}$

$\small \small \frac{\sqrt{2}}{4}$

Correct answer:

$\small \small \frac{\sqrt{2}}{4}$

Explanation:

A useful trigonometric identity to remember is

$\small 2\cos x \sin x = \sin2x$

If we plug in $\small \pi/8$ into this equation, we get

$\small \small 2\cos \left(\frac{\pi}{8}\right)\sin \left(\frac{\pi}{8}\right) = \sin2\left(\frac{\pi}{8}\right)=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$

We can divide the equation by 2 to get

$\small \cos \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2}}{4}$

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Example Question #143 : Trigonometric Functions

Using the half-angle identities, which of the following answers best resembles sin(30) ?

Possible Answers:

$\pm\sqrt{\frac{1+cos(15)}{2}}$

$\pm\sqrt{\frac{1-cos(60)}{2}}$

$\pm\sqrt{\frac{1-cos(15)}{2}}$

$\pm\sqrt{\frac{1+sin(30)}{2}}$

$\pm\sqrt{\frac{1+sin(60)}{1-sin(60)}}$

Correct answer:

$\pm\sqrt{\frac{1-cos(60)}{2}}$

Explanation:

Write the half angle identity for sine.

$sin\left(\frac{A}{2}\right)= \pm\sqrt{\frac{1-cosA}{2}}$

Since we are given sin(30) , the angle is equal to $\frac{A}{2}$ . Set these two angles equal to each other and solve for .

$\frac{A}{2}=30$

A=60

Substitute this value into the formula.

$sin\left(\frac{60}{2}\right)= \pm\sqrt{\frac{1-cos(60)}{2}}$

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Example Question #144 : Trigonometric Functions

Let and two reals. Given that:

$$sin(a+b)=sin(a)cos(b)+cos(a)sin(b)$$

$$sin(a-b)=sin(a)cos(b)-cos(a)sin(b)$$

What is the value of:

$sin^2(a+b)-sin^2(a-b)$ ?

Possible Answers:

sin(2a)cos(2b)

sin(a)sin(b)

cos(2a)sin(2b)

sin(2a)sin(2b)

2sin(2a)sin(2b)

Correct answer:

sin(2a)sin(2b)

Explanation:

We have:

sin^2(a+b)=(sin(a)cos(b)+cos(a)sin(b))^2$$

=sin^2(a)cos^2(b)+cos^2(a)sin^2(b)+2sin(a)cos(b)cos(a)sin(b)(1)$$

and :

sin^2(a-b)=(sin(a)cos(b)-cos(a)sin(b))^2$$

=sin^2(a)cos^2(b)+cos^2(a)sin^2(b)-2sin(a)cos(b)cos(a)sin(b)(2)$$

(1)-(2) gives:

4sin(a)cos(b)cos(a)sin(b)=[2sin(a)cos(a)] [2sin(b)cos(b)]$$

Knowing from the above formula that:( take a=b in the formula above)

sin(2a)=2sin(a)cos(a) and sin(2b)=2sin(b)cos(b)$

This gives:

sin^2(a+b)-sin^2(a-b)=sin(2a)sin(2b)$

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Precalculus : Fundamental Trigonometric Identities

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1499 : Pre Calculus

Example Question #1500 : Pre Calculus

Example Question #131 : Trigonometric Functions

Example Question #132 : Trigonometric Functions

Example Question #133 : Trigonometric Functions

Example Question #134 : Trigonometric Functions

Example Question #141 : Trigonometric Functions

Example Question #142 : Trigonometric Functions

Example Question #143 : Trigonometric Functions

Example Question #144 : Trigonometric Functions

All Precalculus Resources

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