prove a trig equation is an identity

[sroulz13]

Hi I have to solve three problems on my math homework and cannot seem to get them.

1. Prove cos[sup:1rh9da9m]2[/sup:1rh9da9m](x/2)=(1+secx)/(2secx) is an identity.

2. Prove sin4x = 4sin(x)cos(x) - 8sin[sup:1rh9da9m]3[/sup:1rh9da9m](x)cos(x) is an identity.

3. Prove cos(pi + (x-y)) = -sin(x)sin(y) - cos(x)cos(y) is an identity.

For number three I tried using the sum and difference identities, but the sines just cancel out. And for number one I think I'm supposed to use the half angle identities, but I can't seem to get the right answer. Someone help please!

 

[BigGlenntheHeavy]

\(\displaystyle 1) \ Sometimes \ it \ behooves \ one \ to \ work \ both \ sides \ of \ the \ equation \ to \ prove \ it\)

\(\displaystyle is \ an \ identity, \ so \ here \ goes. \ cos^2(x/2) \ = \ \frac{1+sec(x)}{2sec(x)}\)

\(\displaystyle Now, \ cos(x/2) \ = \ \pm\sqrt{\frac{1+cos(x)}{2}}, \ hence \ cos^2(x/2) \ = \ \frac{1+cos(x)}{2}\)

\(\displaystyle Ergo, \ \frac{1+cos(x)}{2} \ = \ \frac{1+sec(x)}{2sec(x)}\)

\(\displaystyle =\frac{1+1/cos(x)}{2/cos(x)} \ = \ \frac{(1+cos(x))/cos(x)}{2/cos(x)} \ = \ \frac{1+cos(x)}{2}\)

\(\displaystyle Therefore, \ \frac{1+cos(x)}{2} \ = \ \frac{1+cos(x)}{2}, \ an \ identity.\)

\(\displaystyle Hint \ for \ 2): \ sin(4x) \ = \ sin(2x+2x)\)

 

[sroulz13]

Thank you very much!