# precal!

posted by
**linds**
.

Prove: (3cosx + 4sinx)^2 + (4cosx - 3sinx)^2=5 ...this is all under the square root sign.

I get 9cos^2x + 16sin^2x +16cos^2x + 9sin^2x

now i know that sin^2 equals 1-cos^2....so would that be 9cos^2x + 17-cos^2x+16cos^2+ 10cos^2x? if not please help..and if this is correct..what would come after this step?

continuing from your left side you had:

9cos^2x + 16sin^2x +16cos^2x + 9sin^2x

this is ok so far.

group this into 9(sin^2 x + cos^2 x) + 16(sin^2 x + cos^2 x)

which is 9(1) + 16(1), because sin^2 x + cos^2 x = 1

So the left side is 25

You said that there was a square root over everything?

so...., how about that!