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Trigonometry : Identities of Inverse Operations

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Example Question #1 : Identities Of Inverse Operations

Simplify using identities. Leave no fractions in your answer.

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

Possible Answers:

$\cos\theta-\sin\theta$ $\displaystyle \cos\theta-\sin\theta$

$\cot\theta$ $\displaystyle \cot\theta$

$\sin\theta - \cos\theta$ $\displaystyle \sin\theta - \cos\theta$

$\sin\theta - \sec\theta$ $\displaystyle \sin\theta - \sec\theta$

$\cos\theta - \sec\theta$ $\displaystyle \cos\theta - \sec\theta$

Correct answer:

$\sin\theta - \sec\theta$ $\displaystyle \sin\theta - \sec\theta$

Explanation:

The easiest first step is to simplify our inverse identities:

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

Cross cancelling, we end up with

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

Finally, eliminate the fraction:

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

Thus,

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

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Example Question #1 : Identities Of Inverse Operations

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

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

Possible Answers:

$\displaystyle cos^2(x)$

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

$\displaystyle cot^2(x)$

$\displaystyle tan^2(x)$

Correct answer:

$\displaystyle cot^2(x)$

Explanation:

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

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

and

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

Then the fraction becomes

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

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Example Question #3 : Identities Of Inverse Operations

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

$\displaystyle (sin^2(x)-1)(1-sin^2(x))$

Possible Answers:

$\displaystyle 1-sin^2(x)$

$\displaystyle -cos^4(x)$

$\displaystyle cos^4(x)$

$\displaystyle -cos^2(x)(1-sin^2(x))$

Correct answer:

$\displaystyle -cos^4(x)$

Explanation:

Use the Pythagorean Identities:

$\displaystyle sin^2(x)-1=cos^2(x)$

and

$\displaystyle sin^2(x)-1=-cos^2(x)$

Thus the expression becomes,

$\displaystyle (-cos^2(x))(cos^2(x))=-cos^4(x)$.

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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.

$\displaystyle (cot(x) - csc(x))(cot(x)+csc(x))$

Possible Answers:

$\displaystyle cos^2(x)$

$\displaystyle sin^2(x)$

$\displaystyle 1$

$\displaystyle -1$

Correct answer:

$\displaystyle -1$

Explanation:

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

$\displaystyle (cot(x) - csc(x))(cot(x)+csc(x))$

$\displaystyle =cot^2(x)-cot(x)csc(x)+cot(x)csc(x)-csc^2(x)$

$\displaystyle =cot^2(x)-csc^2(x)$

Using Pythagorean Identities:

$\displaystyle =cot^2(x)-csc^2(x)=-1$.

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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.

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

Possible Answers:

$\displaystyle 2+sin(x)$

$\displaystyle 2-sin(x)$

-sin(x) $\displaystyle -sin(x)$

sin(x) $\displaystyle sin(x)$

Correct answer:

-sin(x) $\displaystyle -sin(x)$

Explanation:

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

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

Then, simplify the second term to

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

This reduces to

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

The expression is now,

$\displaystyle 1 - (1+sinx)$.

Distribute the negative and get,

-sinx $\displaystyle -sinx$.

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Example Question #6 : Identities Of Inverse Operations

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

$\displaystyle tanx = cotx$

Possible Answers:

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

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

$0, 2\pi$ $\displaystyle 0, 2\pi$

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

Correct answer:

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

Explanation:

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

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

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

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

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Example Question #71 : Trigonometry

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

Possible Answers:

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

$\displaystyle -tan(-x)$

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

tan(-x) $\displaystyle tan(-x)$

$cot(x)^{-1}$ $\displaystyle cot(x)^{-1}$

Correct answer:

tan(-x) $\displaystyle tan(-x)$

Explanation:

Let's look at these individually:

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

$cot(\frac{\pi}{2} - x)$ $\displaystyle cot(\frac{\pi}{2} - x)$ 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 $\displaystyle tan(-x) = -tan(x)$, which means that tan(-x) $\displaystyle tan(-x)$ is our answer.

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Trigonometry : Identities of Inverse Operations

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Identities Of Inverse Operations

Example Question #1 : Identities Of Inverse Operations

Example Question #3 : Identities Of Inverse Operations

Example Question #4 : Identities Of Inverse Operations

Example Question #5 : Identities Of Inverse Operations

Example Question #6 : Identities Of Inverse Operations

Example Question #71 : Trigonometry

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