Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Sin (5x)=16 sin^5 x-20 sin^3 x-5 sin x How Do I Prove this equation?

Any math genius, please help me out

6 Answers

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  • 1 decade ago
    Favorite Answer

    u are gonna need the foll :

    sin(a+b) = sinacosb +cosasinb

    cos(a+b) = cosacosb-sinasinb

    sin2a=2sinacosa

    cos2a=1-2sin^2a

    cos3a= cosa(1-4sin^2a)

    sin3a =3sina-4sin^3a

    sin(5x) = sin(3x +2x) = sin3xcos2x + sin2xcos3x

    = (3sinx-4sin^3x)(1-2sin^2x) + 2sinxcosx*cosx(1-4sin^2x)

    = (3sinx-4sin^3x)(1-2sin^2x) + 2sinx(1-sin^2x)(1-4sin^2x)

    = 16sin^5x - 20sin^3x -5sinx

  • 1 decade ago

    This cab be proved as

    sin 5A = SIN(A+4A)

    = sin A cos 4A + cos A sin 4A

    = sin A (1- 2 Sin^2A) + cos A 2. Sin 2A cos 2A

    = sin A(1- 8 Sin^2 A cos^2 A) + Cos A 4 Cos A sin A (1- 2 Sin^2 A)

    = sin A - 8 sin ^3A (1-sin^2 A) + 4 sin A Cos ^2 A(1- 2 sin ^2 A)

    = sin A - 8 sin ^3 A + 8 Sin ^5 A + 4 sin A(1-sin^2A)(1-2 SIn ^2 A)

    = sin A - 8 sin ^3 A + 8 sin ^5 A + 4 sin A ( 1- 3 sin ^2A + 2 Sin ^5 A)

    = 5 SIn A - 20 sin ^3 A + 16 sin ^5 A

    please follow the above step

    by putting A = pi/2 I find my ans correct 5 in place of -5 .

    please check the same.

  • 1 decade ago

    The form of this suggests that you should be familiar with the Euler identity:

    e^(i*x) = cos(x) + i * sin(x)

    Where i = sqrt(-1)

    If you arent familiar with this, the problem is very difficult. Using the Euler identity, notice that:

    cos(5*x) + i * sin(5*x) = e^(i*5*x) = (e^(i*x))^5 = (cos(x) + i * sin(x))^5

    If you carry out the expansion and equate the real and immaginary parts separately, the identity you are trying to prove will be obvious.

    By the way, the identity as you typed it is incorrect. The last term needs a + sign instead of a - sign. The Euler identity is so powerful it allowed me to spot the error immediately without having to calculate anything.

  • 1 decade ago

    Listen dear it is too long..

    u have 2 do it urself

    hints are here

    sin(5x)=sin(4x+1)

    sin(x+y)=sinxcosy+cosxsiny

    sin(2x)=2sinxcosx

    cos(2x)=cos^2(x)-sin^2(x)

    cos^2(x)=1-sin^2(x)

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  • 1 decade ago

    think of sin(5x) as sin[(2x) + ((2x) + x )]

    sin(2x)=2sinxcosx

    sin(a + b)=sinacosb +sinbcosa

    cos(2x)=2(cos^2)(x) - 1

    just replace and cancel until you get the final answer.

  • raj
    Lv 7
    1 decade ago

    sin5A=sin(3A+2A)=sin3Acos2A

    +cos3Asin2A)

    (3sinA-4sin^3A)(1-2sin^2A)

    +(4cos^3A-3cosA)(2sinAcosA)

    =3sinA-6sin^3A-4sin^3A

    +8sin^5A+8sinAcos^4A

    -6sinAcos^2A

    8sin^5A-10sin^3A+3sinA

    +8sinA(1-sin^2A)^2-6sinA

    (1-sin^2A)

    8sin^5A-10sin^3A+3sinA+8sinA

    +8sin^5A-16sin^3A-6sinA+6sin^3A

    16sin^5A-20sin^3A+5sinA

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