Simplify expressions using trigonometric identities - Pre-Calculus

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Question

Which of the following statements is false?

Answer

Of the six trigonometric functions, four are odd, meaning . These four are:

sin x

tan x

cot x

csc x

That leaves two functions which are even, which means that . These are:

cos x

sec x

Of the aforementioned, only is incorrect, since secant is an even function, which implies that

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Question

Find the exact value

$sin\left[2tan^{-1}\left(\frac{3}{4}\right)\right]$ .

Answer

By the double angle formula

${\right }sin(2x)=2sin(x)cos(x)$

$sin\left[2tan^{-1}\left(\frac{3}{4}\right)\right]=2sin\left[tan^{-1}\left(\frac{3}{4}\right)\right]cos\left[tan^{-1}\left(\frac{3}{4}\right)\right]$

$=2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)=\frac{24}{25}$

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Question

Find the exact value

$sin\left[2tan^{-1}\left(\frac{8}{15}\right)\right]$ .

Answer

By the double angle formula

$sin\left[2tan^{-1}\left(\frac{8}{15}\right)\right]=2sin\left[tan^{-1}\left(\frac{8}{15}\right)\right]cos\left[tan^{-1}\left(\frac{8}{15}\right)\right]$

$=2\left(\frac{8}{17}\right)\left(\frac{15}{17}\right)=\frac{240}{289}$

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Question

Find the exact value

$cos\left[2tan^{-1}\left(\frac{5}{12}\right)\right]$ .

Answer

By the double-angula formula for cosine

For this problem

$cos\left[2tan^{-1}\left(\frac{5}{12}\right)\right]=cos^2\left[tan^{-1}\left(\frac{5}{12}\right)\right]-sin^2\left[tan^{-1}\left(\frac{5}{12}\right)\right]$

$=\left(\frac{12}{13}\right)^2-\left(\frac{5}{13}\right)^2=\frac{119}{169}$

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Question

Find the exact value

$sin\left[2sin^{-1}\left(\frac{12}{37}\right)\right]$ .

Answer

By the double-angle formula for the sine function

we have

$x=sin^{-1}\left( \frac{12}{37}\right)$

thus the double angle formula becomes,

$sin\left[2sin^{-1}\left(\frac{12}{37}\right)\right]=2sin\left[sin^{-1}\left(12/37\right)\right]cos\left[sin^{-1}\left(12/37\right)\right]$

$=2\left(\frac{12}{37}\right)\left(\frac{35}{37}\right)=\frac{840}{1369}$

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Question

Which of the following is equivalent to $f(x)=\sin(x)?$

Answer

When trying to identify equivalent equations that use trigonometric functions it is important to recall the general formula and understand how the terms affect the translations.

The general formula for sine is as follows.

$f(x)=a\sin(bx-c)+d$ where is the amplitude, is used to find the period of the function $\frac{2\pi}{|b|}$ , represents the phase shift $\frac{c}{b}$ , and is the vertical shift.

This is also true for,

$f(x)=a\cos(bx-c)+d$ .

Looking at the possible answer choices lets first focus on the ones containing sine.

$f(x)=\sin (x)+\frac{\pi}{2}$ has a vertical shift of $\frac{\pi}{2}$ therefore it is not an equivalent function as it is moving the original function up.

$f(x)=\sin\left(x-\frac{\pi}{2}\right)$ has a phase shift of $\frac{\pi}{2}$ therefore it is not an equivalent function as it is moving the original function to the right.

Now lets shift our focus to the answer choices that contain cosine.

$f(x)=\cos(x)-\frac{\pi}{2}$ has a vertical shift down of $\frac{\pi}{2}$ units. This will create a graph that has a range that is below the -axis. It is important to remember that has a range of . Therefore this cosine function is not an equivalent equation.

$f(x) = \cos\left(x-\frac{\pi}{2}\right)$ has a phase shift to the right $\frac{\pi}{2}$ units. Plugging in some values we see that,

$f(0)=\cos\left(0-\frac{\pi}{2}\right)=0$ ,

$f\left(\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}-\frac{\pi}{2}\right)=\cos(0)=1$ .

Now, looking back at our original function and plugging in those same values of and we get,

,

$f\left(\frac{\pi}{2}\right)=sin\left(\frac{\pi}{2}\right)=1$ .

Since the function values are the same for each of the input values, we can conclude that $f(x) = \cos\left(x-\frac{\pi}{2}\right)$ is equivalent to .

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Question

If $\theta=X$ , which of the following best represents $sin(2\theta)$ ?

Answer

The expression $sin(2\theta)$ is a double angle identity that can also be rewritten as:

$2sin(\theta) cos(\theta)$

Replace the value of theta for $\theta=X$ .

The correct answer is:

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Question

Compute

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

Answer

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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Question

Compute

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

Answer

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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Question

Using the half-angle identities, which of the following answers best resembles ?

Answer

Write the half angle identity for sine.

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

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

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

Substitute this value into the formula.

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

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Question

Let and two reals. Given that:

What is the value of:

?

Answer

We have:

and :

(1)-(2) gives:

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

This gives:

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Question

Let , , and be real numbers. Given that:

What is the value of in function of ?

Answer

We note first, using trigonometric identities that:

This gives:

Since,

We have :

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Question

Using the fact that,

.

What is the result of the following sum:

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

Answer

We can write the above sum as :

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

From the given fact, we have :

$\vdots$

and we have : .

This gives :

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

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Question

Compute in function of .

Answer

Using trigonometric identities we have :

and we know that:

This gives us :

Hence:

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Question

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 ?

Answer

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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Question

Using the fact that , what is the result of the following sum:

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

Answer

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 :

$\vdots$

This gives us :

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

Therefore we have:

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

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Question

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

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

Answer

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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Question

Given that :

and,

Compute :

in function of .

Answer

We have using the given result:

This gives us:

Hence :

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Question

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

Answer

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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Question

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

Answer

Using trigonometric identities we know that:

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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