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Precalculus : Trigonometric Functions

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Example Questions

Example Question #1523 : Pre Calculus

Which expression is equivalent to

$\sin(\theta + 1 ) \cos (2 \theta - 3 ) + \cos (\theta + 1 ) \sin (2 \theta - 3 )$ ?

Possible Answers:

$\sin (3 \theta - 2 )$

$\sin ( \theta + 1 )$

$\sin (3 \theta + 2 )$

$\cos ( 3 \theta + 2 )$

$\cos ( -\theta + 2 )$

Correct answer:

$\sin (3 \theta - 2 )$

Explanation:

The relevant trigonometric identity is

$\sin (u + v ) = \sin u \cdot \cos v + \cos u \cdot \sin v$

In this case, "u" is $\theta + 1$ and "v" is $2 \theta - 3$ .

Our answer is

$\sin ( \theta + 1 + 2 \theta - 3 ) = \sin ( 3 \theta - 2 )$ .

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Example Question #1522 : Pre Calculus

Evaluate the following.

$cos15^{\circ}sin15^{\circ}+cos15^{\circ}sin15^{\circ}$

Possible Answers:

$\frac{\sqrt2}{2}$

$\frac{1}{2}$

$\frac{-1}{2}$

$\frac{\sqrt3}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

We can use the angle sum formula for sine here.

If we recall that,

sin(u + v) = sinucosv + sinvcosu ,

we can see that the equation presented is equal to

$sin(15^{\circ}+15^{\circ})$ because $u = 15^{\circ} and\: v=15^{\circ}$ .

We can simplify this to $sin(30^{\circ})$ , which is simply $\frac{1}{2}$ .

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Example Question #1523 : Pre Calculus

Evaluate the following.

$cos15^{\circ}cos300^{\circ} - sin15^{\circ}sin300^{\circ}$

Possible Answers:

$\frac{\sqrt2}{2}$

$-\frac{1}2}$

$\frac{1}{2}$

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

Correct answer:

$\frac{\sqrt2}{2}$

Explanation:

The angle sum formula for cosine is,

$cos(u\pm v) = cosucosv \mp sinusinv$ .

First, we see that $u = 15^{\circ} and \: v = 300^{\circ}$ . We can then rewrite the expression as,

$cos(15^{\circ}+300^{\circ}) = cos(315^{\circ})$ .

All that is left to do is to recall the unit circle to evaluate,

$cos(315^\circ)=\frac{\sqrt2}{2}$ .

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Example Question #1524 : Pre Calculus

Evaluate the following.

$\frac{tan150^{\circ} - tan90^{\circ}}{1 + tan150^{\circ} tan90^{\circ}}$

Possible Answers:

$-\sqrt3$

$\sqrt3$

$\frac{\sqrt3}{3}$

Correct answer:

$\sqrt3$

Explanation:

This one is another angle sum/difference problem, except it is using the trickier tangent identity.

The angle sum formula for tangent is

$tan(u\pm v) = \frac{tanu \pm tanv}{1 \mp tanutanv}$ .

We can see that $u = 150^{\circ} and\: v = 90^{\circ}$ .

We can then rewrite the expression as $tan(150-90) = tan(60^{\circ})$ , which is $\sqrt3$ .

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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 #53 : Fundamental Trigonometric Identities

Evaluate the following.

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

Possible Answers:

$\frac{\sqrt2}{2}$

$\frac{\sqrt3}{2}$

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

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

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 #161 : Trigonometric Functions

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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Precalculus : Trigonometric Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1523 : Pre Calculus

Example Question #1522 : Pre Calculus

Example Question #1523 : Pre Calculus

Example Question #1524 : Pre Calculus

Example Question #1522 : Pre Calculus

Example Question #1523 : Pre Calculus

Example Question #53 : Fundamental Trigonometric Identities

Example Question #161 : Trigonometric Functions

Example Question #162 : Trigonometric Functions

Example Question #51 : Fundamental Trigonometric Identities

All Precalculus Resources

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