# Thread: Proving Trig Identities help

1. ## Proving Trig Identities help

I'm stumped on these two problems for my homework, if someone could help me out I'd appreciate it.

1.) (tan2x+1)(cos2x-1)=-tan2x

2.) 2+cos2x-3cos4x=sin2x(2+3cos2x)

note the #'s in front of the x are supposed to be exponents

2. ## (:

It took me a little while to think of the first one, but since I'm studying for a test, I decided to just try it until it worked, and I finally got it.

1) (tan^2x+1)(cos^2x-1) = -tan^2x

Using the Left Hand Side

(tan^2x+1)(cos^2x-1)
= **(sec^2x)(cost^2x-1)
= (1/cos^2x)(cos^2x-(cos^2x+sin^2x)
= (1/cos^2x)(cos^2x-cos^2x-sin^2x)
= (1/cos^2x)(-sin^2x)
= *(-sin^2x/cos^2x)
= -tan^2x

* where: tan^2x =sin^2x/cos^2x
** sec^2= 1 + tan^2x
derived from:
1 = sin^2x + cos^2x (divided by cos^2x)

2) 2+cos^2x-3cos^4x = sin^2x(2+3cos^2x)

Using the Right hand side

sin^2x(2+3cos^2x)
= 2sin^2x + 3sin^2cos^2x
= 2(1-cos^2x)+3(1-cos^2x)cos^2x
= 2-2cos^2x+3(cos^2-cos^4)
= 2-2cos^2x+3cos^2-3cos^4)
= 2+cos^2x-3cos^4x

Using the Identities:
1 = sin^2x + cos^2x

3. Hello, WolfMV!

$\displaystyle 1)\;\;(\tan^2\!x+1)(\cos^2\!x-1)\:=\:-\tan^2\!x$

We have: .$\displaystyle \underbrace{(\tan^2\!x+1)}_{\sec^2\!x}\,\bigg[-\underbrace{(1-\cos^2\!x)}_{\sin^2\!x}\bigg] \;=\;\sec^2\!x(-\sin^2\!x)$

. . . $\displaystyle = \;\frac{1}{\cos^2\!x}(-\sin^2\!x) \;=\;-\frac{\sin^2\!x}{\cos^2\!x} \;=\;-\left(\frac{\sin x}{\cos x}\right)^2 \;=\;-\tan^2\!x$

$\displaystyle 2)\;\; 2+\cos^2\!x-3\cos^4\!x\;=\;\sin^2\!x(2+3\cos^2\!x)$

$\displaystyle \text{Factor: }\;2 + \cos^2\!x -3\cos^4\!x \;=\;\underbrace{(1-\cos^2\!x)}_{\sin^2\!x}(2 + 3\cos^2\!x) \;=\;\sin^2\!x(2 + 3\cos^2\!x)$

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# prove that 3cos^2x sin^2 x=2

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