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High School Math : Trigonometry

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Example Questions

Example Question #1 : Trigonometric Identities

Simplify:

$\sin^2x+\cos^2x+\tan^2x$ $\displaystyle \sin^2x+\cos^2x+\tan^2x$

Possible Answers:

$\csc^2 x$ $\displaystyle \csc^2 x$

$\sin x\cos x$ $\displaystyle \sin x\cos x$

$\sec^2x$ $\displaystyle \sec^2x$

$2\cos x+\cot x$ $\displaystyle 2\cos x+\cot x$

This is the most simplified version.

Correct answer:

$\sec^2x$ $\displaystyle \sec^2x$

Explanation:

Whenever you see a trigonometric function squared, start looking for a Pythagorean identity.

The two identities used in this problem are $\sin^2 x + \cos ^2 x=1$ $\displaystyle \sin^2 x + \cos ^2 x=1$ and $\tan^2 x +1 =\sec^2 x$ $\displaystyle \tan^2 x +1 =\sec^2 x$.

Substitute and solve.

$\sin^2x+\cos^2x+\tan^2x=?$ $\displaystyle \sin^2x+\cos^2x+\tan^2x=?$

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

$(1)+\tan^2x=\sec^2 x$ $\displaystyle (1)+\tan^2x=\sec^2 x$

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Example Question #1 : Using Identities Of Squared Functions

Factor and simplify $\frac{\sin^2x-9}{\sin x-3}$ $\displaystyle \frac{\sin^2x-9}{\sin x-3}$.

Possible Answers:

$\sin x \cos x$ $\displaystyle \sin x \cos x$

$\sin x+3$ $\displaystyle \sin x+3$

$\cos^2 x+3$ $\displaystyle \cos^2 x+3$

$\csc x$ $\displaystyle \csc x$

This is already it's most reduced form.

Correct answer:

$\sin x+3$ $\displaystyle \sin x+3$

Explanation:

To reduce $\frac{\sin^2x-9}{\sin x-3}$ $\displaystyle \frac{\sin^2x-9}{\sin x-3}$, factor the numerator: $\frac{(\sin x-3)(\sin x+3)}{\sin x-3}$ $\displaystyle \frac{(\sin x-3)(\sin x+3)}{\sin x-3}$

Notice that we can cancel out a $\sin x -3$ $\displaystyle \sin x -3$.

This leaves us with $\sin x +3$ $\displaystyle \sin x +3$.

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Example Question #1 : Using Sum And Product Identities

Simplify $\sin x\cot x$ $\displaystyle \sin x\cot x$.

Possible Answers:

$\sin x \cos x$ $\displaystyle \sin x \cos x$

$cos \, x$ $\displaystyle cos \, x$

$\cos^2x$ $\displaystyle \cos^2x$

$\csc x$ $\displaystyle \csc x$

$\tan x$ $\displaystyle \tan x$

Correct answer:

$cos \, x$ $\displaystyle cos \, x$

Explanation:

To simplify $\sin x\cot x$ $\displaystyle \sin x\cot x$, break them into their SOHCAHTOA parts:

$\frac{\text{opposite}}{\text{hypotenuse}}*\frac{\text{adjacent}}{\text{opposite}}$ $\displaystyle \frac{\text{opposite}}{\text{hypotenuse}}*\frac{\text{adjacent}}{\text{opposite}}$.

Notice that the opposite's cancel out, leaving $\frac{\text{adjacent}}{\text{hypotenuse}}=\cos x$ $\displaystyle \frac{\text{adjacent}}{\text{hypotenuse}}=\cos x$.

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Example Question #1 : Using Sum And Product Identities

Simplify $\csc x \tan x$ $\displaystyle \csc x \tan x$.

Possible Answers:

$\sec x$ $\displaystyle \sec x$

$\cos x$ $\displaystyle \cos x$

This is the most simplified version.

$\displaystyle 1$

$\sin x \cos x$ $\displaystyle \sin x \cos x$

Correct answer:

$\sec x$ $\displaystyle \sec x$

Explanation:

Break down $\csc x \tan x$ $\displaystyle \csc x \tan x$ into SOHCAHTOA to solve:

$\csc=\frac{\text{hypotenuse}}{\text{opposite}}$ $\displaystyle \csc=\frac{\text{hypotenuse}}{\text{opposite}}$ and $\tan=\frac{\text{opposite}}{\text{adjacent}}$ $\displaystyle \tan=\frac{\text{opposite}}{\text{adjacent}}$.

Therefore, $\csc x \tan x=\frac{\text{hypotenuse}}{\text{opposite}}*\frac{\text{opposite}}{\text{adjacent}}$ $\displaystyle \csc x \tan x=\frac{\text{hypotenuse}}{\text{opposite}}*\frac{\text{opposite}}{\text{adjacent}}$. Note that the opposite's cancel out, leaving $\frac{\text{hypotenuse}}{\text{adjacent}}$ $\displaystyle \frac{\text{hypotenuse}}{\text{adjacent}}$, which is the same as $\sec x$ $\displaystyle \sec x$.

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Example Question #1 : Using Identities Of Inverse Operations

$\textup{Which of the following expressions is equal to }\frac{\tan \theta }{\sec \theta }\:\textup{?}$ $\displaystyle \textup{Which of the following expressions is equal to }\frac{\tan \theta }{\sec \theta }\:\textup{?}$

Possible Answers:

$\sin\theta$ $\displaystyle \sin\theta$

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

$\csc\theta$ $\displaystyle \csc\theta$

$\cos\theta$ $\displaystyle \cos\theta$

$\frac{1}{\cos ^{2}\theta }$ $\displaystyle \frac{1}{\cos ^{2}\theta }$

Correct answer:

$\sin\theta$ $\displaystyle \sin\theta$

Explanation:

$\textup{Use identities to solve: }\tan\theta=\frac{\sin \theta }{\cos \theta }\:\:\:\:\:\sec\theta=\frac{1}{\cos \theta }$ $\displaystyle \textup{Use identities to solve: }\tan\theta=\frac{\sin \theta }{\cos \theta }\:\:\:\:\:\sec\theta=\frac{1}{\cos \theta }$

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

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High School Math : Trigonometry

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Trigonometric Identities

Example Question #1 : Using Identities Of Squared Functions

Example Question #1 : Using Sum And Product Identities

Example Question #1 : Using Sum And Product Identities

Example Question #1 : Using Identities Of Inverse Operations

All High School Math Resources

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