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Trigonometry : Trigonometry

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Example Questions

Example Question #71 : Trigonometry

Simplify using identities. Leave no fractions in your answer.

$\sec \theta(cos\theta\sin\theta) - \csc\theta\tan\theta$

Possible Answers:

$\cot\theta$

$\sin\theta - \sec\theta$

$\cos\theta - \sec\theta$

$\sin\theta - \cos\theta$

$\cos\theta-\sin\theta$

Correct answer:

$\sin\theta - \sec\theta$

Explanation:

The easiest first step is to simplify our inverse identities:

$\sec \theta(cos\theta\sin\theta) - \csc\theta\tan\theta = \bigg(\frac{1}{cos\theta}\bigg)cos\theta\sin\theta-\bigg(\frac{1}{\sin\theta}\bigg)\bigg(\frac{sin\theta}{\cos\theta}\bigg)$

Cross cancelling, we end up with

$\bigg(\frac{1}{cos\theta}\bigg)cos\theta\sin\theta-\bigg(\frac{1}{\sin\theta}\bigg)\bigg(\frac{\sin\theta}{\cos\theta}\bigg)=\sin\theta-\frac{1}{\cos\theta}$

Finally, eliminate the fraction:

$\sin\theta - \frac{1}{\cos\theta}= \sin\theta-\sec\theta$

Thus,

$\sec \theta(cos\theta\sin\theta) - \csc\theta\tan\theta = \sin\theta - \sec\theta$

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Example Question #71 : Trigonometry

Simplify each expression below. Your answer should have (at most) one trigonometric function and no fractions.

1. $\frac{cot(x)}{tan(x)}$

Possible Answers:

cot^2(x)

tan^2(x)

cos^2(x)

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

Correct answer:

cot^2(x)

Explanation:

Using the quotient identities for trig functions, you can rewrite,

$cot(x)=\frac{cos(x)}{sin(x)}$

and

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

Then the fraction becomes

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

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Example Question #71 : Trigonometric Identities

Simplify each expression below. Your answer should have (at most) one trigonometric function and no fractions.

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

Possible Answers:

-cos^4(x)

cos^4(x)

-cos^2(x)(1-sin^2(x))

1-sin^2(x)

Correct answer:

-cos^4(x)

Explanation:

Use the Pythagorean Identities:

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

and

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

Thus the expression becomes,

(-cos^2(x))(cos^2(x))=-cos^4(x) .

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Example Question #4 : Identities Of Inverse Operations

Simplify each expression below. Your answer should have (at most) one trigonometric function and no fractions.

(cot(x) - csc(x))(cot(x)+csc(x))

Possible Answers:

sin^2(x)

cos^2(x)

Correct answer:

Explanation:

Use the distributive property (FOIL method) to simplify the expression.

(cot(x) - csc(x))(cot(x)+csc(x))

=cot^2(x)-cot(x)csc(x)+cot(x)csc(x)-csc^2(x)

=cot^2(x)-csc^2(x)

Using Pythagorean Identities:

=cot^2(x)-csc^2(x)=-1 .

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Example Question #5 : Identities Of Inverse Operations

Simplify each expression below. Your answer should have (at most) one trigonometric function and no fractions.

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

Possible Answers:

2-sin(x)

-sin(x)

sin(x)

2+sin(x)

Correct answer:

-sin(x)

Explanation:

First, simplify the first term in the expression to 1 because of the Pythagorean Identity.

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

Then, simplify the second term to

$\frac{(1 - sin^2x)}{(1 - sinx)}$ .

This reduces to

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

The expression is now,

1 - (1+sinx) .

Distribute the negative and get,

-sinx .

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Example Question #72 : Trigonometric Identities

Solve each question over the interval $0 < x < 2\pi$

tanx = cotx

Possible Answers:

$0, 2\pi$

$\frac{\pi}{4}, \frac{5\pi}{4}$

$\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$

$\frac{\pi}{2}, \frac{3\pi}{2}, 2\pi$

Correct answer:

$\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$

Explanation:

Divide both sides by cotx to get tan^2x = 1 .

Take the square root of both sides to get that tanx = 1 and tanx = -1 .

The angles for which this is true (this is taking the arctan) are every angle when sinx = cosx and sinx = -cosx .

These angles are all the multiples of $\frac{\pi}{4}$ .

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Example Question #7 : Identities Of Inverse Operations

tan(x) can be stated as all of the following except...

Possible Answers:

tan(-x)

$cot(x)^{-1}$

$cot(\frac{\pi}{2} - x)$

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

-tan(-x)

Correct answer:

tan(-x)

Explanation:

Let's look at these individually:

$\frac{sin(x)}{cos(x)}$ is true by definition, as is $cot(x)^{-1}$ .

$cot(\frac{\pi}{2} - x)$ is also true because of a co-function identity.

This leaves two - and we can tell which of these does not work using the fact that tan(-x) = -tan(x) , which means that tan(-x) is our answer.

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Example Question #83 : Trigonometry

Rt_triangle_letters

In this figure, if angle $c=90^\circ$ , side X=12 , and side Z=8 , what is the measure of angle ?

Possible Answers:

$48.2^\circ$

$\frac{2}{3}^\circ$

$41.8^\circ$

Undefined

$90^\circ$

Correct answer:

$41.8^\circ$

Explanation:

Since $c=90^\circ$ , we know we are working with a right triangle.

That means that $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$ .

In this problem, that would be:

$\sin(a)=\frac{\text{Z}}{\text{X}}$

Plug in our given values:

$\sin(a)=\frac{8}{12}$

$\sin(a)=\frac{2}{3}$

$a=\sin^{-1}(\frac{2}{3})$

$a=41.8^\circ$

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Example Question #72 : Trigonometry

If $sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$ , give the value of $cot\ x$ .

Possible Answers:

$cot\ x=1\ or\ cot\ x=2$

$cot\ x=2$

$cot\ x=1\ or\ cot\ x=-2$

$cot\ x=-1\ or\ cot\ x=2$

$cot\ x=1$

Correct answer:

$cot\ x=1\ or\ cot\ x=2$

Explanation:

$sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$

$\Rightarrow sin^2x+(cos^2x+cos^2x)+3(sin\ x)(cos\ x)=3$

Use the identity sin^2x+cos^2x=1 .

$\Rightarrow 1+cos^2x+3(sin\ x)(cos\ x)=3\Rightarrow cos^2x+3(sin\ x)(cos\ x)=3-1=2$

Now we should divide both sides by sin^2x :

$\Rightarrow \frac{cos^2x}{sin^2x}+\frac{3(sin\ x)(cos\ x)}{sin^2x}=\frac{2}{sin^2x}$

We can use the identity $\frac{1}{sin^2x}=1+cot^2x$ .

$\Rightarrow cot^2x+3cot\ x=2(1+cot^2x)$

$\Rightarrow cot^2x+3cot\ x=2+2cot^2x$

$\Rightarrow 2cot^2x-cot^2x-3cot\ x+2=0$

$\Rightarrow cot^2x-3cot\ x+2=0$

$\Rightarrow (cot\ x-1)(cot\ x-2)=0$

$cot\ x-1=0\Rightarrow cot\ x=1$

$cot\ x-2=0\Rightarrow cot\ x=2$

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Example Question #1 : Trigonometric Functions And Graphs

If $sin\ x+\frac{1}{sin\ x}=2$ , find the value of sin^5x+sin^7x .

Possible Answers:

Correct answer:

Explanation:

$sin\ x+\frac{1}{sin\ x}=2\Rightarrow \frac{sin^2x+1}{sin\ x}=2$

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

$\Rightarrow sin^2x-2sin\ x+1$

$\Rightarrow (sinx-1)^2=0$

$\Rightarrow sin\ x-1=0$

$\Rightarrow sin\ x=1$

Therefore, sin^5x+sin^7x=1^5+1^7=1+1=2 .

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #71 : Trigonometry

Example Question #71 : Trigonometry

Example Question #71 : Trigonometric Identities

Example Question #4 : Identities Of Inverse Operations

Example Question #5 : Identities Of Inverse Operations

Example Question #72 : Trigonometric Identities

Example Question #7 : Identities Of Inverse Operations

Example Question #83 : Trigonometry

Example Question #72 : Trigonometry

Example Question #1 : Trigonometric Functions And Graphs

All Trigonometry Resources

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