Need sin cos help

  • Thread starter johncena
  • Start date
  • #1
131
1

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
 

Answers and Replies

  • #2
uart
Science Advisor
2,776
9
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
55
0
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
131
1
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]
 
Last edited:
  • #5
CRGreathouse
Science Advisor
Homework Helper
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I suppose the easiest way would be to choose values for a and b and see which fail.
 
  • #6
55
0
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
131
1
Thank you very much for your help sir....I got the answer easily
 

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