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Precalculus : Trigonometric Functions

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Example Questions

Example Question #141 : Trigonometric Functions

Compute

$\small 2\cos^2 \frac{\pi}{8}$

Possible Answers:

$\small 0$

$\small \frac{\sqrt{2}}{2}+1$

$\small \frac{\sqrt{3}}{2}-1$

$\small \pi-2$

Correct answer:

$\small \frac{\sqrt{2}}{2}+1$

Explanation:

A useful trigonometric identity to remember for this problem is

$\small 2\cos^2 \theta-1=\cos2\theta$

or equivalently,

$\small \small 2\cos^2 \theta=1+\cos2\theta$

If we substitute $\small \theta$ for $\small \frac{\pi}{8}$ , we get

$\small \small \small 2\cos^2 \frac{\pi}{8}=1+\cos2\frac{\pi}{8}=1+\cos\frac{\pi}{4}=1+\frac{\sqrt{2}}{2}$

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Example Question #141 : Trigonometric Functions

Compute

$\small \cos\left(\frac{\pi}{8}\right)\sin\left(\frac{\pi}{8}\right)$

Possible Answers:

$\small \small \frac{\sqrt{6}}{9}$

$\small \frac{1+\pi}{\sqrt{5}}$

$\small \frac{\sqrt{3}}{4}$

$\small \small \frac{\sqrt{2}}{4}$

Correct answer:

$\small \small \frac{\sqrt{2}}{4}$

Explanation:

A useful trigonometric identity to remember is

$\small 2\cos x \sin x = \sin2x$

If we plug in $\small \pi/8$ into this equation, we get

$\small \small 2\cos \left(\frac{\pi}{8}\right)\sin \left(\frac{\pi}{8}\right) = \sin2\left(\frac{\pi}{8}\right)=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$

We can divide the equation by 2 to get

$\small \cos \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2}}{4}$

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Example Question #142 : Trigonometric Functions

Using the half-angle identities, which of the following answers best resembles sin(30) ?

Possible Answers:

$\pm\sqrt{\frac{1+cos(15)}{2}}$

$\pm\sqrt{\frac{1+sin(30)}{2}}$

$\pm\sqrt{\frac{1-cos(15)}{2}}$

$\pm\sqrt{\frac{1+sin(60)}{1-sin(60)}}$

$\pm\sqrt{\frac{1-cos(60)}{2}}$

Correct answer:

$\pm\sqrt{\frac{1-cos(60)}{2}}$

Explanation:

Write the half angle identity for sine.

$sin\left(\frac{A}{2}\right)= \pm\sqrt{\frac{1-cosA}{2}}$

Since we are given sin(30) , the angle is equal to $\frac{A}{2}$ . Set these two angles equal to each other and solve for .

$\frac{A}{2}=30$

A=60

Substitute this value into the formula.

$sin\left(\frac{60}{2}\right)= \pm\sqrt{\frac{1-cos(60)}{2}}$

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Example Question #26 : Fundamental Trigonometric Identities

Let and two reals. Given that:

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

$$sin(a-b)=sin(a)cos(b)-cos(a)sin(b)$$

What is the value of:

$sin^2(a+b)-sin^2(a-b)$ ?

Possible Answers:

sin(2a)sin(2b)

sin(a)sin(b)

sin(2a)cos(2b)

2sin(2a)sin(2b)

cos(2a)sin(2b)

Correct answer:

sin(2a)sin(2b)

Explanation:

We have:

sin^2(a+b)=(sin(a)cos(b)+cos(a)sin(b))^2$$

=sin^2(a)cos^2(b)+cos^2(a)sin^2(b)+2sin(a)cos(b)cos(a)sin(b)(1)$$

and :

sin^2(a-b)=(sin(a)cos(b)-cos(a)sin(b))^2$$

=sin^2(a)cos^2(b)+cos^2(a)sin^2(b)-2sin(a)cos(b)cos(a)sin(b)(2)$$

(1)-(2) gives:

4sin(a)cos(b)cos(a)sin(b)=[2sin(a)cos(a)] [2sin(b)cos(b)]$$

Knowing from the above formula that:( take a=b in the formula above)

sin(2a)=2sin(a)cos(a) and sin(2b)=2sin(b)cos(b)$

This gives:

sin^2(a+b)-sin^2(a-b)=sin(2a)sin(2b)$

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Example Question #144 : Trigonometric Functions

Let , , and be real numbers. Given that:

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

sin(a-b)=sin(a)cos(b)-cos(a)sin(b)

What is the value of sin(3a) in function of sin(a) ?

Possible Answers:

3sin(a)-4sin^3(a)

3sin(a)-4sin^4(a)

4sin(a)-3sin^3(a)

6sin(a)-4sin^3(a)

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

Correct answer:

3sin(a)-4sin^3(a)

Explanation:

We note first, using trigonometric identities that:

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

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

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

This gives:

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

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

We have :

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

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Example Question #146 : Trigonometric Functions

Using the fact that,

sin(x+180)=-sin(x) .

What is the result of the following sum:

$sin(1)+sin(2)+sin(3)\cdots +sin(359)$

Possible Answers:

sin(1)

cos(1)

Correct answer:

Explanation:

We can write the above sum as :

$sin(1)+sin(181)+sin(2)+sin(182)\cdots$

+ sin(179)+sin(379) + sin(180)

From the given fact, we have :

sin(1)+sin(181)=sin(1)+sin(1+180)=sin(1)-sin(1)=0

sin(2)+sin(182)=sin(2)+sin(2+180)=sin(2)-sin(2)=0

$\vdots$

sin(179)+sin(379)=sin(179)+sin(179+180)=sin(179)-sin(179)=0

and we have : sin(180)=0 .

This gives :

$sin(1)+sin(2)+sin(3)\cdots +sin(359)=0$

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Example Question #145 : Trigonometric Functions

Compute cos(4x) in function of sin(x) .

Possible Answers:

1-8sin^2(x)-8sin^4(x)

1+8sin^2(x)+8sin^4(x)

1-8sin^2(x)+8sin^4(x)

1-6sin^2(x)+8sin^4(x)

1-8sin^2(x)+6sin^4(x)

Correct answer:

1-8sin^2(x)+8sin^4(x)

Explanation:

Using trigonometric identities we have :

cos(4x)=1-2sin^2(2x) and we know that:

sin(2x)=2sin(x)cox(x)

This gives us :

cos(4x)=1-2(2sin(x)cox(x))^2=1-8sin^2(x)cos^2(x)

= 1-8sin^2(x)(1-sin^2(x)= 1-8sin^2(x)+8sin^4(x)

Hence:

cos(4x)=1-8sin^2(x)+8sin^4(x)

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Example Question #146 : Trigonometric Functions

Given that :

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

Let,

$tan\left(\frac{x}{8}\right)=s$

What is $tan\left(\frac{x}{4}\right)$ in function of ?

Possible Answers:

$\frac{4s}{1-4s^2}$

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

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

$\frac{4s}{1-s^2}$

$\frac{s}{1-4s^2}$

Correct answer:

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

Explanation:

We will use the given formula :

We have in this case:

$tan\left(\frac{x}{4}\right)=\frac{2tan(\frac{x}{8})}{1-tan^2(\frac{x}{8})}$

Since we know that :

$tan\left(\frac{x}{8}\right)=s$

This gives :

$\frac{2tan(\frac{x}{8})}{1-tan^2(\frac{x}{8})}=\frac{2s}{1-s^2}$

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Example Question #1511 : Pre Calculus

Using the fact that cos(x+180)=-cos(x) , what is the result of the following sum:

$cos(1)+cos(2)+cos(3)\cdots +cos(359)$

Possible Answers:

Correct answer:

Explanation:

We can write the above sum as :

$cos(1)+cos(181)+cos(2)+cos(182)\cdots + cos(179)+cos(379) + cos(180)$

From the given fact, we have :

cos(1)+cos(181)=cos(1)+cos(1+180)=cos(1)-cos(1)=0

cos(2)+cos(182)=cos(2)+cos(2+180)=cos(2)-cos(2)=0

$\vdots$

cos(179)+cos(379)=cos(179)+cos(179+180)=cos(179)-cos(179)=0

This gives us :

$cos(1)+cos(181)+cos(2)+cos(182)\cdots + cos(179)+cos(379) + cos(180)$

=0+cos(180)=-1

Therefore we have:

$cos(1)+cos(2)+cos(3)\cdots +cos(359)=-1$

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Example Question #147 : Trigonometric Functions

Let $\alpha, \beta$ be real numbers. If $tan(2x)=\alpha$ and $tan(x)=\beta$

What is the value of $\alpha$ in function of $\beta$ ?

Possible Answers:

$\alpha=\frac{2\beta}{1-\beta^2}$

$\alpha=\frac{2\beta^2}{1-\beta^2}$

$\alpha=\frac{2\beta}{1-2\beta^2}$

$\alpha=\frac{2\beta}{2-\beta^2}$

$\alpha=\frac{2\beta}{3-2\beta^2}$

Correct answer:

$\alpha=\frac{2\beta}{1-\beta^2}$

Explanation:

Using trigonometric identities we know that :

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

This gives :

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

We also know that $\alpha =tan(2x)$

This gives :

$\alpha=\frac{2\beta}{1-\beta^2}$

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Precalculus : Trigonometric Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #141 : Trigonometric Functions

Example Question #141 : Trigonometric Functions

Example Question #142 : Trigonometric Functions

Example Question #26 : Fundamental Trigonometric Identities

Example Question #144 : Trigonometric Functions

Example Question #146 : Trigonometric Functions

Example Question #145 : Trigonometric Functions

Example Question #146 : Trigonometric Functions

Example Question #1511 : Pre Calculus

Example Question #147 : Trigonometric Functions

All Precalculus Resources

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