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Precalculus : Trigonometric Functions

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Example Questions

Example Question #151 : Trigonometric Functions

Given that :

cos(a+b)=cos(a)cos(b)-sin(a)sin(b) and,

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

Compute :

cos(3a) in function of cos(a) .

Possible Answers:

4cos^3(a)+3cos(a)

-4cos^3(a)-3cos(a)

4cos^3(a)-3cos(a)

3cos^3(a)-4cos(a)

3cos^2(a)-4cos(a)

Correct answer:

4cos^3(a)-3cos(a)

Explanation:

We have using the given result:

cos(3a)=cos(2a)cos(a)-sin(2a)sin(a)

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

sin(2a)=2sin(a)cos(a)

This gives us:

cos(3a)=(2cos^2(a)-1)cos(a)-2cos(a)sin^2(a)

=2cos^3(a)-cos(a)-2cos(a)+2cos^3(a)=4cos^3(a)-3cos(a)

Hence :

cos(3a)=4cos^3(a)-3cos(a)

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Example Question #1512 : Pre Calculus

Let be an integer and a real number. Compute $cos(x+n\pi)$ as a function of cos(x) .

Possible Answers:

- cos(x)

-(-1)^n cos(x)

$(-1)^{n+1} cos(x)$

(-1)^n cos(x)

$(-1)^{n-1}cos(x)$

Correct answer:

(-1)^n cos(x)

Explanation:

Using trigonometric identities we have :

$cos(x+n\pi)=cos(x)cos(n\pi)-sin(x)sin(n\pi)$

We know that :

$sin(n\pi)=0$ and $cos(n\pi)=(-1)^n$

This gives :

$cos(x+n\pi)=(-1)^ncos(x)$

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Example Question #151 : Trigonometric Functions

Compute $sin\left(\frac{\pi}{2}+n\pi\right)$ .

Possible Answers:

(-1)^n

sin(1)

Correct answer:

(-1)^n

Explanation:

Using trigonometric identities we know that:

sin(x+y)=sin(x)cos(y)+cos(x)sin(y)

Letting $x=\frac{\pi}{2}$ and $y=n\pi$ in the above expression we have:

$sin\left(\frac{\pi}{2}+n\pi\right)=sin\left(\frac{\pi}{2}\right)cos(n\pi)+sin(n\pi)cos\left(\frac{\pi}{2}\right)$

We also know that:

$cos(n\pi)=(-1)^n$ and $sin(n\pi)=0$ .

This gives:

$sin\left(\frac{\pi}{2}+n\pi\right)=(-1)^n$

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Example Question #153 : Trigonometric Functions

Given that:

tan(x)=s , what is the value of $tan\left(x+\frac{\pi}{4}\right)$ in function of ?

Possible Answers:

$\frac{1+2s}{1-2s}$

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

$\frac{1+s}{1-s}$

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

$\frac{2+s}{2-s}$

Correct answer:

$\frac{1+s}{1-s}$

Explanation:

We know by definition that:

$tan\left(x+\frac{\pi}{4}\right)=\frac{sin(x+\frac{\pi}{4})}{cos(x+\frac{\pi}{4})}$

We also have by trigonometric identities:

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

$cos\left(x+\frac{\pi}{4}\right)=cos(x)cos\left(\frac{\pi}{4}\right)-sin\left(\frac{\pi}{4}\right)sin(x)=\frac{\sqrt{2}}{2}(cos(x)-sin(x)) $$$

Thus :

$tan\left(x+\frac{\pi}{4}\right)=\frac{cos(x)+sin(x)}{cos(x)-sin(x)}$

Now we have:

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

This gives us:

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

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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:

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

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

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

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

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

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:

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

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

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

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

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

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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Precalculus : Trigonometric Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #151 : Trigonometric Functions

Example Question #1512 : Pre Calculus

Example Question #151 : Trigonometric Functions

Example Question #153 : Trigonometric Functions

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

All Precalculus Resources

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