Trigonometric Identities

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

$\cot\theta$

$\sin\theta$

$\cos^2\theta$

$\sin^2\theta$

$\tan\theta$

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 $\tan\theta,$ $\cot\theta,$ $\csc\theta,$ 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.

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

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

$\cot\theta$

$\sin\theta$

$\cos^2\theta$

$\sin^2\theta$

$\tan\theta$

Explanation

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

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

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.

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

Simplify $\sin x\cot x$ .

$\csc x$

$\tan x$

$\sin x \cos x$

$\cos^2x$

Explanation

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 $\frac{\text{adjacent}}{\text{hypotenuse}}=\cos x$ .

Simplify

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

$\cos(x)$

$\tan(x)$

$\arcsin(\pi )$

$\csc(x)$

$\arcsin(x)$

Explanation

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

Simplify

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

$\cos(x)$

$\tan(x)$

$\arcsin(\pi )$

$\csc(x)$

$\arcsin(x)$

Explanation

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

Simplify $\sin x\cot x$ .

$\csc x$

$\tan x$

$\sin x \cos x$

$\cos^2x$

Explanation

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 $\frac{\text{adjacent}}{\text{hypotenuse}}=\cos x$ .

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

$-\sin x$

$\sin x\cos x$

$\tan x$

$\sin x$

$\sin^2 x$

Explanation

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

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

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

Simplify

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

$\sec(x)$

$\cos(x)$

$\cos(x)\times\tan(x)$

$\cot(x)/sec(x)$

$\csc(x)$

Explanation

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

and

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

Simplify

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

$\sec(x)$

$\cos(x)$

$\cos(x)\times\tan(x)$

$\cot(x)/sec(x)$

$\csc(x)$

Explanation

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

and

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

Simplify:

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

$\sec^2x$

$\sin x\cos x$

$\csc^2 x$

This is the most simplified version.

$2\cos x+\cot x$

Explanation

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

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