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
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