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Precalculus : Fundamental Trigonometric Identities

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Example Questions

Example Question #1522 : Pre Calculus

Evaluate the following.

$2sin\frac{\pi}{12}cos\frac{\pi}{12}$

Possible Answers:

$\frac{\sqrt3}{2}$

$\frac{1}{2}$

$-\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

Here we use the double angle identity for sine, which is

$sin(2\alpha) = 2sin\alpha cos\alpha$ .

We can rewrite the originial expression as $sin\left(\frac{\pi}{6}\right)$ using the double angle identity.

From here we can calculate that

$sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ .

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Example Question #1523 : Pre Calculus

Evaluate the following expression.

$2cos^2112.5^{\circ} - 1$

Possible Answers:

$\frac{1}{2}$

$\frac{\sqrt2}{2}$

$-\frac{\sqrt2}{2}$

$\frac{\sqrt3}{2}$

Correct answer:

$-\frac{\sqrt2}{2}$

Explanation:

One of the double angle formuals for cosine is

$cos2\alpha = 2cos^2\alpha - 1$ .

We can use this double angle formula for cosine to rewrite the expression given as the $cos(225^{\circ})$ because $2\cdot 112.5 = 225$ and $\alpha = 112.5$ .

We can then calculate that

$cos(225^{\circ}) = -\frac{\sqrt2}{2}$ .

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Example Question #161 : Trigonometric Functions

Evaluate the following.

$cos^2165^{\circ} - sin^2165^{\circ}$

Possible Answers:

$\frac{\sqrt3}{2}$

$-\frac{\sqrt3}{2}$

$-\frac{\sqrt2}{2}$

$\frac{\sqrt2}{2}$

Correct answer:

$\frac{\sqrt3}{2}$

Explanation:

Here we can use another double angle formula for cosine,

$cos2\alpha = cos^2\alpha - sin^2\alpha$ .

Here $\alpha = 165^{\circ}$ , and so we can use the double angle formula for cosine to rewrite the expression as

$cos(2\alpha) = cos(2\cdot 165^{\circ}) = cos(330^{\circ})$ .

From here we just recognize that

$cos(330^{\circ}) = \frac{\sqrt3}{2}$ .

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Example Question #51 : Fundamental Trigonometric Identities

Evaluate the following expression.

$1- 2sin^2135^{\circ}$

Possible Answers:

$\frac{-1}{2}$

$\frac{1}{2}$

Correct answer:

Explanation:

Here we can use yet another double angle formula for cosine:

$cos(2\alpha) = 1 - 2sin^2\alpha$ .

First, realize that $\alpha = 135^{\circ}$ .

Next, plug this in to the double angle formula to find that

$1 - 2sin^2(135^{\circ}) = cos(2\cdot 135^{\circ}) = cos(270^{\circ})$ .

Here we recognize that $cos(270^{\circ}) = 0$

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Example Question #162 : Trigonometric Functions

Simplify the following. Leave your answer in terms of a trigonometric function.

$sin120^{\circ}cos17^{\circ}+sin17^{\circ}cos120^{\circ}$

Possible Answers:

$cos103^{\circ}$

$cos137^{\circ}$

$sin127^{\circ}$

$sin103^{\circ}$

$sin137^{\circ}$

Correct answer:

$sin137^{\circ}$

Explanation:

This is a quick test of being able to recall the angle sum formula for sine.

Since,

$sin(u\pm v) = sinucov \pm sinvcosu$ , and here

$u = 120^{\circ} and\: v = 17^{\circ}$ , we can rewrite the expression as

$sin(120+17) = sin137^{\circ}$ .

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Example Question #51 : Fundamental Trigonometric Identities

Which of the following is equivalent to the expression:

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

Possible Answers:

cos^2(x)sin^2(x)

tan^2(x)

cot^2(x)

Correct answer:

cos^2(x)sin^2(x)

Explanation:

Which of the following is equivalent to the following expression?

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

Recall our Pythagorean trig identity:

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

It can be rearranged to look just like our numerator:

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

So go ahead and change our original expression to:

$\frac{cos^2(x)}{csc^2(x)}$

Then recall the definition of cosecant:

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

So our original expression can be rewritten as:

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

So our answer is:

cos^2(x)sin^2(x)

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Example Question #52 : Fundamental Trigonometric Identities

Which of the following trigonometric identities is INCORRECT?

Possible Answers:

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

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

$cot (x) = \frac{cos (x)}{sin (x)}$

$sec (x) = \frac{1}{cos (x)}$

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

Correct answer:

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

Explanation:

Cosine and sine are not reciprocal functions.

$sin (x) = \frac{1}{csc (x)}$ and $cos (x) = \frac{1}{sec (x)}$

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Example Question #172 : Trigonometric Functions

Using the trigonometric identities prove whether the following is valid:

$\frac{\csc^{2}A}{\sec^{2}A} = \cot^{2}A$

Possible Answers:

False

Uncertain

True

Only in the range of: $0 < A < 2\pi$

Only in the range of: $0< A < \frac{\pi}{2}$

Correct answer:

True

Explanation:

We begin with the left-hand side of the equation and utilize basic trigonometric identities, beginning with converting the inverse functions to their corresponding base functions:

$\frac{\frac{1}{\sin^{2}A}}{\frac{1}{\cos^{2}A}} = \cot^{2}A$

Next, we rewrite the fractional division in order to simplify the equation:

$\frac{1}{\sin^{2}A}\div\frac{1}{\cos^{2}A} = \cot^{2}A$

In fractional division we multiply by the reciprocal as follows:

$\frac{1}{\sin^{2}A}\times \cos^{2}A = \cot^{2}A$

If we reduce the fraction using basic identities we see that the equivalence is proven:

$\frac{\cos^{2}A}{\sin^{2}A} = \cot^{2}A$

$\cot^{2}A = \cot^{2}A$

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Example Question #53 : Fundamental Trigonometric Identities

Which of the following identities is incorrect?

Possible Answers:

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

$\mathrm{cos}(-x)=-\mathrm{cos}(x)$

$\small \csc(-x)=-\csc(x)$

$\small \sec(-x)=\sec(x)$

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

Correct answer:

$\mathrm{cos}(-x)=-\mathrm{cos}(x)$

Explanation:

The true identity is $\small \cos(-x)=\cos(x)$ because cosine is an even function.

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Example Question #54 : Fundamental Trigonometric Identities

State $\tan x$ in terms of sine and cosine.

Possible Answers:

$\sin x \cos x$

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

$1+\sin x$

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

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

Correct answer:

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

Explanation:

The definition of tangent is sine divided by cosine.

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

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Precalculus : Fundamental Trigonometric Identities

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1522 : Pre Calculus

Example Question #1523 : Pre Calculus

Example Question #161 : Trigonometric Functions

Example Question #51 : Fundamental Trigonometric Identities

Example Question #162 : Trigonometric Functions

Example Question #51 : Fundamental Trigonometric Identities

Example Question #52 : Fundamental Trigonometric Identities

Example Question #172 : Trigonometric Functions

Example Question #53 : Fundamental Trigonometric Identities

Example Question #54 : Fundamental Trigonometric Identities

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