# Need sin cos help

## Main Question or Discussion Point

Can any one give any hint to solve this ?

If sinx + sin2x + sin3x = 1,
then , cos6x - 4cos4x + 8cos2x =

(a) 1
(b) 4
(c) 2
(d) 3

uart
It's probably not in the spirit of the question but you can just solve that cubic numerically and get sin(x) = 0.5437 (to 4 dp). Then using cos^2(x) = 1- sin^2(x) you can easily evaluate it.

BTW. The answer is (b) 4.

Rewrite as: sinx + (sinx)^3 = (cos)^2. Then square both sides and use the identity (sin)^2=1-(cos)^2 and the answer will follow

Thanks for the hint . I got the answer .Now can you help me to solve this ?

If $$\frac{sin(2a + b)}{sin b}$$ = $$\frac{n}{m}$$ , then tan (a + b)cot a = ?

(A)$$\frac{n-m}{n+m}$$ (B) $$\frac{m-n}{m+n}$$

(C)$$\frac{n+m}{n-m}$$ (D)$$\frac{m+n}{m-n}$$

Last edited:
CRGreathouse