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Precalculus : Fundamental Trigonometric Identities

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Example Questions

Example Question #41 : Fundamental Trigonometric Identities

Given that:

tan(x)=s

Compute

tan(2x) in function of .

Possible Answers:

$\frac{s(1-s^2)}{(1-s^2-2s)(1-s^2+2s)}$

$\frac{4s^2(1-s^2)}{(1-s^2-2s)(1-s^2+2s)}$

$\frac{s(1-s^2)}{(1-3s^2)(1+s^2)}$

$\frac{4s(1-s^2)}{(1+s^2-2s)(1-s^2+2s)}$

$\frac{4s-4s^3}{1-6s^2+s^4}$

Correct answer:

$\frac{4s-4s^3}{1-6s^2+s^4}$

Explanation:

We will use the following formulas for calculating the tangent:

$tan(2x)=\frac{2tan(x)}{1-tan^2(x)}$

$tan(4x)=\frac{2tan(2x)}{1-tan^2(2x)}$

We have then:

$tan(4x)=\frac{2tan(2x)}{1-tan^2(2x)}= 2\frac{\frac{2tan(x)}{1-tan^2(x)}}{1-\left(\left(\frac{2tan(x)}{1-tan^2(x)}\right)^2\right)}$

This gives us :

$tan(4x)=\frac{tan(2x)}{1-tan^2(2x)}= \frac{\frac{4tan(x)}{1-tan^2(x)}}{1-(\frac{2tan(x)}{1-tan^2(x)})^2}$

Simplifying a bit we get :

$\frac{4tan(x)}{(1-tan^2(x))} \cdot \frac{(1-tan^2(x))^2}{((1-tan^2(x))^2-4tan^2(x))}$

Writing

((1-tan^2(x))^2-4tan^2(x))=1-6tan^2x+tan^4x

We now have :

$tan(4x)=\frac{4tan(x)(1-tan^2x)}{1-6tan^2x+tan^4x}\rightarrow \frac{4tan(x)-4tan^3(x)}{1-6tan^2x+tan^4x}$

Using the fact that tan(x)=s , we have finally:

$tan(4x)= \frac{4s-4s^3}{1-6s^2+s^4}$

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Example Question #41 : Fundamental Trigonometric Identities

Which of the following expressions best represent cos(x)sin(x) ?

Possible Answers:

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

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

$\frac{sin(2x)}{2}$

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

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

Correct answer:

$\frac{sin(2x)}{2}$

Explanation:

Write the trigonometric product and sum identity for cos(A)sin(B) .

$cos(A)sin(B)=\frac{1}{2}[sin(A+B)-sin(A-B)]$

For cos(x)sin(x) , replace A,B with and simplify the expression.

$\frac{1}{2}[sin(x+x)-sin(x-x)]=\frac{1}{2}[sin(2x)-sin(0)]$

$\frac{sin(2x)}{2}$

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Example Question #42 : Fundamental Trigonometric Identities

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

Possible Answers:

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

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

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

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

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

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 #1521 : Pre Calculus

Which expression is equivalent to

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

Possible Answers:

$\sin (3 \theta - 2 )$

$\cos ( 3 \theta + 2 )$

$\cos ( -\theta + 2 )$

$\sin (3 \theta + 2 )$

$\sin ( \theta + 1 )$

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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Precalculus : Fundamental Trigonometric Identities

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #41 : Fundamental Trigonometric Identities

Example Question #41 : Fundamental Trigonometric Identities

Example Question #42 : Fundamental Trigonometric Identities

Example Question #1521 : Pre Calculus

Example Question #45 : Fundamental Trigonometric Identities

Example Question #1522 : Pre Calculus

Example Question #1521 : Pre Calculus

Example Question #1522 : Pre Calculus

Example Question #1523 : Pre Calculus

Example Question #1524 : Pre Calculus

All Precalculus Resources

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