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Precalculus : Pre-Calculus

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Example Questions

Example Question #153 : Trigonometric Functions

Find the exact answer of: $cos^2(15^{\circ})$

Possible Answers:

$\frac{\sqrt3}{3}+3$

$\frac{\sqrt3}{4}-\frac{1}{2}$

$\frac{\sqrt3}{4}+\frac{1}{2}$

$-\frac{\sqrt3}{4}-\frac{1}{2}$

$\frac{1}{2}-\frac{\sqrt3}{4}$

Correct answer:

$\frac{\sqrt3}{4}+\frac{1}{2}$

Explanation:

The product and sum formula can be used to solve this question.

Write the formula for cosine identity.

$cos(x)cos(y)= \frac{1}{2}[cos(x+y)+cos(x-y)]$

Split up cos^2(15) into two separate cosine expressions.

cos^2(15)=cos(15)cos(15)

Substitute the 2 known angles into the formula and simplify.

$cos(15)cos(15)=\frac{1}{2}[cos(15+15)+cos(15-15)]$

$cos(15)cos(15)=\frac{1}{2}[cos(30)+cos(0)]$

$cos(15)cos(15)=\frac{1}{2}[\frac{\sqrt3}{2}+1]=\frac{\sqrt3}{4}+\frac{1}{2}$

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Example Question #1521 : Pre Calculus

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

$\sqrt\frac{1+cos46^{\circ}}{2}$

Possible Answers:

$sin46^{\circ}$

$cos11.5^{\circ}$

$cos23^{\circ}$

$cos46^{\circ}$

$sin11.5^{\circ}$

Correct answer:

$cos23^{\circ}$

Explanation:

This is a simple exercise to recognize the half angle formula for cosine.

The half angle formula for cosine is

$cos(\frac{1}{2}\alpha) = \sqrt\frac{1+cos\alpha}{2}$ .

In the expression given $\alpha = 46^{\circ}$ .

With this in mind we can rewrite the expression as the $cos\left(\frac{46}{2}\right)$ , or, after dividing by two, $cos23^{\circ}$

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Example Question #45 : Fundamental Trigonometric Identities

Solve the following over the domain to $2\pi$ .

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

Possible Answers:

$\frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{12}, \frac{17\pi}{12}$

$\frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}$

$\frac{\pi}{6}, \frac{5\pi}{6}$

$\frac{\pi}{12}, \frac{5\pi}{12}$

$\frac{13\pi}{12}, \frac{17\pi}{12}$

Correct answer:

$\frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}$

Explanation:

Here we can rewrite the left side of the equation as sin2x because of the double angle formula for sin, which is 2sinxcosx = sin2x .

Now our equation is

$sin2x = \frac{1}{2}$ ,

and in order to get solve for we take the $sin^{-1}$ of both sides. Just divide by two from there to find .

The only thing to keep in mind here is that the period of the function is half of what it normally is, which is why we have to solve for and then add $\pi$ to each answer.

$\frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}$

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Example Question #1522 : Pre Calculus

Solve over the domain to $2\pi$ .

$cos(x)cos\frac{\pi}{4} + sin(x)sin\frac{\pi}4{} = -1$

Possible Answers:

$\pi$

$\frac{\pi}{4}$

$\frac{\pi}{2}$

$\frac{5\pi}{4}$

$\frac{3\pi}{2}$

Correct answer:

$\frac{5\pi}{4}$

Explanation:

We can rewrite the left side of the equation using the angle difference formula for cosine

( cos(u - v) = cosucosv + sinusinv) as

$cos\left(x - \frac{\pi}{4}\right)$ .

From here we just take the $cos^{-1}$ of both sides and then add $\frac{\pi}{4}$ to get $x = \frac{5\pi}{4}$ .

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Example Question #41 : Fundamental Trigonometric Identities

Which expression is equivalent to

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

Possible Answers:

$\sin ( \theta + 1 )$

$\cos ( 3 \theta + 2 )$

$\cos ( -\theta + 2 )$

$\sin (3 \theta - 2 )$

$\sin (3 \theta + 2 )$

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 #42 : Fundamental Trigonometric Identities

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 #161 : Trigonometric Functions

Evaluate the following.

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

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

$\frac{\sqrt3}{2}$

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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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #153 : Trigonometric Functions

Example Question #1521 : Pre Calculus

Example Question #45 : Fundamental Trigonometric Identities

Example Question #1522 : Pre Calculus

Example Question #41 : Fundamental Trigonometric Identities

Example Question #1522 : Pre Calculus

Example Question #1523 : Pre Calculus

Example Question #42 : Fundamental Trigonometric Identities

Example Question #161 : Trigonometric Functions

Example Question #1523 : Pre Calculus

All Precalculus Resources

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