# Math Help - Learning how to solve trig Identites

1. ## Learning how to solve trig Identites

I'm trying to learn how to do trig identies and these questions are confusing me. Could someone show me step for step how to do these, please. Besides the identies is there any rule on how to do it? Plz and thk-u

1. 1+ 1\tan(squared)x = 1\sin(squared)x

2. 2sin(squared)x - 1= sin(squared)x - cos(squared)x

3. 1\ 1- sin(squared)x = 1 + tan(squared)x

4. cos(squared)x - sin(squared)x = 2cos(squared)x - 1

You don't have to do all of them although I'd really appericate if you did. Thanks

2. Originally Posted by SaRah<3
2. 2sin(squared)x - 1= sin(squared)x - cos(squared)x
$2\sin^2 x - (\sin ^2 x + \cos^2 x) = \sin^2 x - \cos^2 x$
4. cos(squared)x - sin(squared)x = 2cos(squared)x - 1
$2\cos^2 x - (\sin^2 x + \cos^2 x) = \cos^2x -\sin^2 x$

3. Originally Posted by SaRah<3
1. 1+ 1\tan(squared)x = 1\sin(squared)x
$1 + \frac{1}{tan^2(x)} = \frac{1}{sin^2(x)}$

My usual method is to convert everything to sin(x) and cos(x).

The LHS is:
$1 + \frac{cos^2(x)}{sin^2(x)}$ <-- Add the two terms

$= \frac{sin^2(x) + cos^2(x)}{sin^2(x)}$

$= \frac{1}{sin^2(x)}$
which is the RHS.

-Dan

4. Originally Posted by SaRah<3
3. 1\ 1- sin(squared)x = 1 + tan(squared)x
$\frac{1}{1 - sin^2(x)} = 1 + tan^2(x)$

The RHS:
$1 + tan^2(x) = 1 + \frac{sin^2(x)}{cos^2(x)}$

$= \frac{cos^2(x) + sin^2(x)}{cos^2(x)}$

$= \frac{1}{cos^2(x)}$

But $sin^2(x) + cos^2(x) = 1$ so...

$= \frac{1}{1 - sin^2(x)}$
which is the LHS.

-Dan

(You can shortcut a lot of the work here by noting that $sec^2(x) = 1 + tan^2(x)$.)

5. Hello, SaRah<3!

It certainly helps if you know all the basic identities and their variations:
. . $\sin\theta = \frac{1}{\csc\theta}\qquad\csc\theta = \frac{1}{\sin\theta}$
. . $\cos\theta = \frac{1}{\sec\theta}\qquad\sec\theta = \frac{1}{\cos\theta}$
. . $\tan\theta = \frac{1}{\cot\theta} \qquad \cot\theta = \frac{1}{\tan\theta}$

. . $\tan\theta = \frac{\sin\theta}{\cos\theta} \quad \cot\theta = \frac{\cos\theta}{\sin\theta}$

. . $\sin^2\theta + \cos^2\theta \:=\:1\qquad \tan^2\theta + 1 \:=\:\sec^2\theta$

$1)\;\;1+ \frac{1}{\tan^2x} \:=\: \frac{1}{\sin^2x}$
The left side is: . $1 + \cot^2x \;=\;\csc^2x\;=\;\frac{1}{\sin^2x}$

$2)\;\;2\sin^2\!x - 1 \:= \:\sin^2\!x - \cos^2\!x$
The right side is: . $\sin^2\!x - (1 - \sin^2\!x) \;=\;\sin^2\!x - 1 + \sin^2\!x \;=\;2\sin^2\!x - 1$

$3)\;\;\frac{1}{1- \sin^2x} \:= \:1 + \tan^2x$
The left side is: . $\frac{1}{\cos^2\!x} \;=\;\sec^2\!x\;=\;1 + \tan^2\!x$

$4)\;\;\cos^2\!x - \sin^2\!x \:= \:2\cos^2\!x - 1$
The left side is: . $\cos^2\!x - (1 - \cos^2\!x) \;=\;\cos^2\!x - 1 + \cos^2\!x \;=\;2\cos^2\!x - 1$