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Need help

  1. Oct 4, 2009 #1
    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. jcsd
  3. Oct 4, 2009 #2

    uart

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    Science Advisor

    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.
     
  4. Oct 4, 2009 #3
    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
     
  5. Oct 12, 2009 #4
    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: Oct 12, 2009
  6. Oct 12, 2009 #5

    CRGreathouse

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    Science Advisor
    Homework Helper

    I suppose the easiest way would be to choose values for a and b and see which fail.
     
  7. Oct 12, 2009 #6
    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.
     
  8. Oct 13, 2009 #7
    Thank you very much for your help sir....I got the answer easily
     
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