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Trigonometry : Trigonometric Identities

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Example Question #71 : Trigonometric Identities

Simplify using identities. Leave no fractions in your answer.

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

Possible Answers:

$\cot\theta$

$\sin\theta - \cos\theta$

$\cos\theta - \sec\theta$

$\sin\theta - \sec\theta$

$\cos\theta-\sin\theta$

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 #1 : Identities Of Inverse Operations

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

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

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 #3 : Identities Of Inverse Operations

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

-cos^4(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 #1 : 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:

cos^2(x)

sin^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 #71 : Trigonometry

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:

sin(x)

2+sin(x)

2-sin(x)

-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 #6 : Identities Of Inverse Operations

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

tanx = cotx

Possible Answers:

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

$0, 2\pi$

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

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

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 #71 : Trigonometric Identities

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

Possible Answers:

-tan(-x)

$cot(x)^{-1}$

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

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

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 #1 : Complete Basic Trigonometry Proofs

Using trigonometric identities determine whether the following is valid:

$\tan2A = \frac{2\sin \left(-A\right ) \cos \left(-A\right )}{\cos^{2}\left(-A\right ) + \sin^{2}\left(-A\right)}$

Possible Answers:

Only valid in the range of: $0 < A < \frac{\pi}{2}$

False

Only valid in the range of: $0 < A < 2\pi$

Uncertain

True

Correct answer:

False

Explanation:

We can choose either side to work with to attempt to obtain the equivalency. Here we will work with the right side as it is the more complex. First, we want to eliminate the negative angles using the appropriate relations. Sine is odd and therefore, the negative sign comes out front. Cosine is even which is interpreted by dropping the negative out of the equation:

$\tan 2A = \frac{-2\sin A\cos A}{\cos^{2}A + \sin^{2}A}$

The squaring of the sine in the denominator makes the sine term positive, i.e.

$\sin^{2}\left(-A \right ) = \left(\sin-A \right )\times\left(\sin -A \right ) = -\sin A \times -\sin A = \sin^{2}A$

The numerator is the double angle formula for sine:

$\tan 2A = \frac{-\sin 2A}{\cos^{2}A + \sin^{2}A}$

The denominator is recognized to be the pythagorean theorem as it applies to trigonometry:

$\tan 2A = \frac{-2\sin 2A}{1}$

The final reduced equation is:

$\tan 2A = -2\sin 2A$

Thus proving that the equivalence is false.

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Example Question #2 : Complete Basic Trigonometry Proofs

You can derive the formula $1+\tan^2\theta=\sec^2\theta$ by dividing the formula $\sin^2\theta+\cos^2\theta=1$ by which of the following functions?

Possible Answers:

$\sin\theta$

$\sin^2\theta$

$\cos\theta$

$\cos^2\theta$

$\tan^2\theta$

Correct answer:

$\cos^2\theta$

Explanation:

The correct answer is $\cos^2\theta$ . Rather than memorizing all three Pythagorean Relationships, you can memorize only $\sin^2\theta+\cos^2\theta=1$ , then simply divide all terms by $\cos^2\theta$ to get the formula that relates $\tan^2\theta$ and $\sec^2\theta$ . Alternatively, you can divide all terms of $\sin^2\theta+\cos^2\theta=1$ by $\sin^2\theta$ to get the formula that relates $\cot^2\theta$ and $\csc^2\theta$ . The former is demonstrated below.

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

$\frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}$

$\tan^2\theta+1=\sec^2\theta$

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Example Question #3 : Complete Basic Trigonometry Proofs

You can derive the formula $1+\cot^2\theta=\csc^2\theta$ by dividing the formula $\sin^2\theta+\cos^2\theta=1$ by which of the following functions?

Possible Answers:

$\sin^2\theta$

$\sin\theta$

$\tan^2\theta$

$\cos\theta$

$\cos^2\theta$

Correct answer:

$\sin^2\theta$

Explanation:

The correct answer is $\sin^2\theta$ . Rather than memorizing all three Pythagorean Relationships, you can memorize only $\sin^2\theta+\cos^2\theta=1$ , then simply divide all terms by $\sin^2\theta$ to get the formula that relates $\cot^2\theta$ and $\csc^2\theta$ . Alternatively, you can divide all terms of $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ to get the formula that relates $\tan^2\theta$ and $\sec^2\theta$ . The former is demonstrated below.

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

$\frac{\sin^2\theta}{\sin^2\theta}+\frac{\cos^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}$

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

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Trigonometry : Trigonometric Identities

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #71 : Trigonometric Identities

Example Question #1 : Identities Of Inverse Operations

Example Question #3 : Identities Of Inverse Operations

Example Question #1 : Identities Of Inverse Operations

Example Question #71 : Trigonometry

Example Question #6 : Identities Of Inverse Operations

Example Question #71 : Trigonometric Identities

Example Question #1 : Complete Basic Trigonometry Proofs

Example Question #2 : Complete Basic Trigonometry Proofs

Example Question #3 : Complete Basic Trigonometry Proofs

All Trigonometry Resources

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