Trigonometric Identities - Math

Card 0 of 48

Question

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

Answer

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

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

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

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

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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Question

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

Answer

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

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

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

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

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.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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

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.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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

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

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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

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Question

Simplify

$\cot(x)\times{\sin(x) }$

Answer

$\cot(x) = \cos(x)/\sin(x)$ . Thus: $[\cos(x)/\sin(x)]\times\sin(x)=\cos(x)$

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Question

Simplify

$\frac{cot (x)\cdot tan(x)}{cos(x)}$

Answer

$\cot(x)\times\tan(x) = 1$

and

$1/(\cos(x)) = \sec(x)$ .

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Question

Simplify $\frac{\cos^2x-1}{\sin x}$ .

Answer

Remember that $\sin^2x+\cos^2 x=1$ . We can rearrange this to simplify our given equation:

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

$=\frac{-\sin^2x}{\sin x}=-\sin x$

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Question

Simplify:

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

Answer

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

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

Substitute and solve.

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

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

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

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Question

Simplify $\sin x\cot x$ .

Answer

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

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

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

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Question

Simplify $\csc x \tan x$ .

Answer

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

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

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

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Question

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

Answer

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

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

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

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

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.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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

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.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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

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.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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

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

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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

Compare your answer with the correct one above

Question

Simplify

$\cot(x)\times{\sin(x) }$

Answer

$\cot(x) = \cos(x)/\sin(x)$ . Thus: $[\cos(x)/\sin(x)]\times\sin(x)=\cos(x)$

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Question

Simplify

$\frac{cot (x)\cdot tan(x)}{cos(x)}$

Answer

$\cot(x)\times\tan(x) = 1$

and

$1/(\cos(x)) = \sec(x)$ .

Compare your answer with the correct one above

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