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High School Math : Using Basic and Definitional Identities

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Example Questions

Example Question #1 : Trigonometric Identities

Trig_id

What is the $\tan$ $\displaystyle \tan$ of $\theta$ $\displaystyle \theta$?

Possible Answers:

$\frac{a}{h}$ $\displaystyle \frac{a}{h}$

$\frac{a}{o}$ $\displaystyle \frac{a}{o}$

$\frac{o}{a}$ $\displaystyle \frac{o}{a}$

$\frac{o}{h}$ $\displaystyle \frac{o}{h}$

Correct answer:

$\frac{o}{a}$ $\displaystyle \frac{o}{a}$

Explanation:

When working with basic trigonometric identities, it's easiest to remember the mnemonic: $\displaystyle SOHCAHTOA$.

$\sin=\frac{opposite}{hypotenuse}$ $\displaystyle \sin=\frac{opposite}{hypotenuse}$

$\cos=\frac{adjacent}{hypotenuse}$ $\displaystyle \cos=\frac{adjacent}{hypotenuse}$

$\tan=\frac{opposite}{adjacent}$ $\displaystyle \tan=\frac{opposite}{adjacent}$

When one names the right triangle, the opposite side is opposite to the angle, the adjacent side is next to the angle, and the hypotenuse spans the two legs of the right angle.

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Example Question #1 : Trigonometric Identities

Simplify $(1-\cos^2\theta )\cdot\cot\theta\cdot\csc^2\theta$ $\displaystyle (1-\cos^2\theta )\cdot\cot\theta\cdot\csc^2\theta$.

Possible Answers:

$\tan\theta$ $\displaystyle \tan\theta$

$\cos^2\theta$ $\displaystyle \cos^2\theta$

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

$\sin^2\theta$ $\displaystyle \sin^2\theta$

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

Correct answer:

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

Explanation:

Simplifying trionometric expressions or identities often involves a little trial and error, so it's hard to come up with a strategy that works every time. A lot of times you have to try multiple strategies and see which one helps.

Often, if you have any form of $\tan\theta,$ $\displaystyle \tan\theta,$ $\cot\theta,$ $\displaystyle \cot\theta,$ $\csc\theta,$ $\displaystyle \csc\theta,$ or $\sec\theta,$ $\displaystyle \sec\theta,$ in an expression, it helps to rewrite it in terms of sine and cosine. In this problem, we can use the identities $\cot\theta=\frac{\cos\theta }{\sin\theta }$ $\displaystyle \cot\theta=\frac{\cos\theta }{\sin\theta }$ and $\csc\theta=\frac{1}{\sin\theta }$ $\displaystyle \csc\theta=\frac{1}{\sin\theta }$.

$(1-\cos^2\theta )\cdot\cot\theta\cdot\csc^2\theta=(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot(\frac{1}{\sin\theta})^2$ $\displaystyle (1-\cos^2\theta )\cdot\cot\theta\cdot\csc^2\theta=(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot(\frac{1}{\sin\theta})^2$

$=(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}$ $\displaystyle =(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}$.

This doesn't seem to help a whole lot. However, we should recognize that $(1-\cos^2\theta )=\sin^2\theta$ $\displaystyle (1-\cos^2\theta )=\sin^2\theta$ because of the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ $\displaystyle \sin^2\theta+\cos^2\theta=1$.

$(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}=(\sin^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}$ $\displaystyle (1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}=(\sin^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}$

We can cancel the $\sin^2\theta$ $\displaystyle \sin^2\theta$ terms in the numerator and denominator.

$(\sin^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}=\frac{\cos\theta }{\sin\theta }=\cot\theta$ $\displaystyle (\sin^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}=\frac{\cos\theta }{\sin\theta }=\cot\theta$.

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Example Question #1 : Using Basic And Definitional Identities

Trig_id

What is the $\sin$ $\displaystyle \sin$ of $\theta$ $\displaystyle \theta$?

Possible Answers:

$\frac{h}{a}$ $\displaystyle \frac{h}{a}$

$\frac{a}{h}$ $\displaystyle \frac{a}{h}$

$\frac{o}{a}$ $\displaystyle \frac{o}{a}$

$\frac{o}{h}$ $\displaystyle \frac{o}{h}$

$\frac{h}{o}$ $\displaystyle \frac{h}{o}$

Correct answer:

$\frac{o}{h}$ $\displaystyle \frac{o}{h}$

Explanation:

When working with basic trigonometric identities, it's easiest to remember the mnemonic: $\displaystyle SOHCAHTOA$.

$\sin=\frac{opposite}{hypotenuse}$ $\displaystyle \sin=\frac{opposite}{hypotenuse}$

$\cos=\frac{adjacent}{hypotenuse}$ $\displaystyle \cos=\frac{adjacent}{hypotenuse}$,

$\tan=\frac{opposite}{adjacent}$ $\displaystyle \tan=\frac{opposite}{adjacent}$

When one names the right triangle, the opposite side is opposite to the angle, the adjacent side is next to the angle, and the hypotenuse spans the two legs of the right angle.

Report an Error

Example Question #2 : Using Basic And Definitional Identities

Trig_id

What is the $\cos$ $\displaystyle \cos$ of $\theta$ $\displaystyle \theta$?

Possible Answers:

$\frac{a}{h}$ $\displaystyle \frac{a}{h}$

$\frac{o}{a}$ $\displaystyle \frac{o}{a}$

$\frac{o}{h}$ $\displaystyle \frac{o}{h}$

$\frac{h}{o}$ $\displaystyle \frac{h}{o}$

$\frac{h}{a}$ $\displaystyle \frac{h}{a}$

Correct answer:

$\frac{a}{h}$ $\displaystyle \frac{a}{h}$

Explanation:

When working with basic trigonometric identities, it's easiest to remember the mnemonic: $\displaystyle SOHCAHTOA$.

$\sin=\frac{opposite}{hypotenuse}$ $\displaystyle \sin=\frac{opposite}{hypotenuse}$

$\cos=\frac{adjacent}{hypotenuse}$ $\displaystyle \cos=\frac{adjacent}{hypotenuse}$,

$\tan=\frac{opposite}{adjacent}$ $\displaystyle \tan=\frac{opposite}{adjacent}$

When one names the right triangle, the opposite side is opposite to the angle, the adjacent side is next to the angle, and the hypotenuse spans the two legs of the right angle.

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High School Math : Using Basic and Definitional Identities

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Trigonometric Identities

Example Question #1 : Trigonometric Identities

Example Question #1 : Using Basic And Definitional Identities

Example Question #2 : Using Basic And Definitional Identities

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