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Precalculus : Fundamental Trigonometric Identities

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Example Questions

Example Question #121 : Trigonometric Functions

Which of the following is equal to:

-sec(x)csc(-x)cot(-x)cos(-x)

Possible Answers:

-csc(x)cot(x)

cot(x)csc(x)

sec(x)csc(x)

-tan(-x)sec(x)csc(x)

Correct answer:

-csc(x)cot(x)

Explanation:

Recall that $sec(x)=\frac {1}{cos(x)}$ , and that cos(-x)=cos(x)

Therefore:

$sec(x)\cdot cos(-x)=sec(x)\cdot cos(x)=1$

Since that term is eliminated, we have left:

-csc(-x)cot(-x)

Recall that

csc(-x)=-csc(x)

cot(-x)=-cot(x)

Therefore:

$-csc(-x)cot(-x)=-(-csc(x)\cdot-cot(x))=-csc(x)cot(x)$

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Example Question #11 : Fundamental Trigonometric Identities

Compute

$\tan\left(\frac{\pi}{12}\right)$ .

Possible Answers:

$\small \frac{2+\sqrt{3}}{2}$

$\small 2+\sqrt{3}$

$\sqrt3$

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

$\small 2-\sqrt{3}$

Correct answer:

$\small 2-\sqrt{3}$

Explanation:

We can use the following trigonometric identity to help us in the calculation:

$\small \tan(X)=\frac{1-\cos2X}{\sin2X}$

We plug in $\small X=\frac{\pi}{12}$ to get

$\tan\left(\frac{\pi}{12}\right)=\frac{1-\cos \pi/6}{\sin\pi/6}=\frac{1-\frac{\sqrt{3}}{2}}{1/2}$

$\small =2-\sqrt{3}$ .

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Example Question #12 : Fundamental Trigonometric Identities

Simplify

$\cot\left(\frac{\pi}{8}\right)$ .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

We can use the trigonometric identity,

$\small \tan(x)=\frac{1-\cos2x}{\sin2x}$

along with the fact that

$\small \cot(x)=\frac{1}{\tan(x)}$

to compute $\small \cot(\pi/8)$ .

We have

$\small \small \cot(\pi/8)=\frac{1}{\tan \pi/8}=\frac{\sin \pi/4}{1-\cos\pi/4}$

$\small =\frac{\sqrt{2}/2}{1-\sqrt{2}/2}=\frac{\sqrt{2}}{2-\sqrt{2}}$

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Example Question #13 : Fundamental Trigonometric Identities

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

Possible Answers:

$f(x)=\sin (x)+\frac{\pi}{2}$

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

$f(x)=\cos(x)-\frac{\pi}{2}$

$f(x)=\tan(x-\pi)$

$f(x)=\sin\left(x-\frac{\pi}{2}\right)$

Correct answer:

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

Explanation:

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 f(x)=sin(x) has a range of (-1,1) . 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 x=0 and x=1 we get,

f(0)=sin(0)=0 ,

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

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 f(x)=sin(x) .

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Example Question #14 : Fundamental Trigonometric Identities

Suppose:

sin A + cos A=x

sin A-cosA=y

What must be the value of sin^4A-cos^4A ?

Possible Answers:

x^2y^2

x+y

x-y

x^2+y^2

Correct answer:

Explanation:

First, factor sin^4A-cos^4A into their simplified form.

sin^4A-cos^4A=(sin^2A+cos^2A)(sin^2A-cos^2A)

The identity sin^2A+cos^2A equals to 1.

Factor sin^2A-cos^2A .

sin^2A-cos^2A=(sinA+cosA)(sinA-cosA)

Since:

sin A + cos A=x

sin A-cosA=y

Substitute the values of the simplified equation.

(sinA+cosA)(sinA-cosA)=xy

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Example Question #15 : Fundamental Trigonometric Identities

Find the exact value of each expression below without the aid of a calculator.

$sin165^{\circ}$

Possible Answers:

$\pm\sqrt\frac{\left(1-\frac{1}{2}\right)}{2}$

$\pm\sqrt\frac{\left(1+\frac{\sqrt(3)}{2}\right)}{2}$

$\pm\sqrt\frac{\left(1-\frac{\sqrt(3)}{2}\right)}{2}$

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

$\pm1$

Correct answer:

$\pm\sqrt\frac{\left(1-\frac{\sqrt(3)}{2}\right)}{2}$

Explanation:

In order to find the exact value of $\sin165^\circ$ we can use the half angle formula for sin, which is

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

This way we can plug in a value for alpha for which we know the exact value. $165^\circ$ is equal to $330^\circ$ divided by two, and so we can plug in $330^\circ$ for the alpha above.

The cosine of $330^\circ$ is $\frac{\sqrt3}{2}$ .

Therefore our final answer becomes,

$\pm\sqrt\frac{\left(1-\frac{\sqrt(3)}{2}\right)}{2}$ .

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Example Question #13 : Fundamental Trigonometric Identities

Simplify.

$1-\tan x(\csc x)$

Possible Answers:

None of these answers are correct.

$\sin^2x$

$\tan^2 x$

$1-\sec x$

Correct answer:

$1-\sec x$

Explanation:

$\frac{\sin x}{\cos x}=\tan x$

$\frac{1}{\sin x}=\csc x$

Given these identities...

$1-\tan x(\csc x)$

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

$1-\frac{1}{\cos x}$

$1- \sec x$

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Example Question #15 : Fundamental Trigonometric Identities

Simplify $\frac{\sin (x)}{1-\cos (x)}-\cot(x)$ completely.

Possible Answers:

$\csc (x)$

$1-\tan(x)$

$\cos (x)-\sin(x)$

$-\sec (x)$

$-\tan (x)$

Correct answer:

$\csc (x)$

Explanation:

First simplify the fraction

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

by multiplying it by its conjugate

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

After doing so, continue simplying:

$\\ \frac{\sin(x)(1+\cos(x))}{(1-\cos(x))(1+\cos (x))}- \cot(x)\\ \\=\frac{\sin(x)(1+\cos(x))}{1-\cos^2(x)}-\cot(x)\\ \\=\frac{\sin(x)(1+\cos(x))}{\sin^2(x)}-\cot(x)\\ \\=\frac{1+\cos(x)}{\sin(x)}-\frac{\cos(x)}{\sin(x)}\\ \\=\frac{1}{\sin(x)}\\ \\=\csc (x)$

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Example Question #1494 : Pre Calculus

Fully simplify.

Simplify:

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

Possible Answers:

$\sin ^2 x$

None of these answers are correct.

$\sin x + \cos x$

$\cos^2 x$

Correct answer:

$\sin ^2 x$

Explanation:

$\sin^2 x + \cos^2 x = 1$

$1+ \cot^2x= \csc^2 x$

Given the above identities:

$\frac{(\sin ^2 x+ \cos^2 x)}{(1+ \cot^2 x)}=\frac{1}{\csc^2 x}=\sin^2x$

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Example Question #1495 : Pre Calculus

Simplify:

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

Possible Answers:

None of these answers are correct.

$1-2\cos^2x$

$1+2\sin^2x$

Correct answer:

Explanation:

$\sin^2x+\cos^2x=1$

and

$\sin^2x=1-\cos^2x$

Therefore...

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

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

$=1+(\sin ^2x+\cos ^2x)$

=1+1=2

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Precalculus : Fundamental Trigonometric Identities

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #121 : Trigonometric Functions

Example Question #11 : Fundamental Trigonometric Identities

Example Question #12 : Fundamental Trigonometric Identities

Example Question #13 : Fundamental Trigonometric Identities

Example Question #14 : Fundamental Trigonometric Identities

Example Question #15 : Fundamental Trigonometric Identities

Example Question #13 : Fundamental Trigonometric Identities

Example Question #15 : Fundamental Trigonometric Identities

Example Question #1494 : Pre Calculus

Example Question #1495 : Pre Calculus

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