Solve sin cos: Find cos6x - 4cos4x + 8cos2x =?

[johncena]

Can anyone 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.

 

[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

 

[johncena]

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

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

(A)[tex]\frac{n-m}{n+m}[/tex] (B) [tex]\frac{m-n}{m+n}[/tex]

(C)[tex]\frac{n+m}{n-m}[/tex] (D)[tex]\frac{m+n}{m-n}[/tex]

 

[CRGreathouse]

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

 

[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.

 

[johncena]

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