# Need help

1. Oct 4, 2009

### johncena

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

2. Oct 4, 2009

### 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.

3. Oct 4, 2009

### defunc

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

4. Oct 12, 2009

### johncena

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: Oct 12, 2009
5. Oct 12, 2009

### CRGreathouse

I suppose the easiest way would be to choose values for a and b and see which fail.

6. Oct 12, 2009

### defunc

Write sin(2a+b) as sin [a+(a+b)] and sin(b) as sin[a+(b-a)]. Then use the double angle identities and you should obtain the answer easily.

7. Oct 13, 2009

### johncena

Thank you very much for your help sir....I got the answer easily