Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Show that (cosx)^3 = (1/4)(3cosx + cos3x)?

Thanks heaps.

8 Answers

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  • 1 decade ago
    Favorite Answer

    cos³x = (1/4)[3cosx + cos(3x)]

    rewrite (3x) as (2x+x):

    cos³x = (1/4)[3cosx + cos(2x+ x)] →

    recall cosine addiction rule:

    cos(a + b) = cos(a) cos(b) - sin(a) sin(b)

    thus:

    cos³x = (1/4){3cosx + [cos(2x) cosx - sin(2x) sinx]} →

    find cos(2x) and sin(2x) using the double-angle identities:

    cos(2x) = cos²x - sin²x

    sin(2x) = 2sinx cosx

    cos³x = (1/4){3cosx + [(cos²x - sin²x)cosx - (2sinx cosx) sinx]} →

    cos³x = (1/4)[3cosx + (cos³x - sin²x cosx - 2sin²x cosx)] →

    cos³x = (1/4)(3cosx + cos³x - 3sin²x cosx) →

    replace sin²x with 1 - cos²x:

    cos³x = (1/4)[3cosx + cos³x - 3(1 - cos²x) cosx] →

    cos³x = (1/4)[3cosx + cos³x - 3(cosx - cos³x)] →

    cos³x = (1/4)(3cosx + cos³x - 3cosx + 3cos³x) →

    cos³x = (1/4)(4cos³x) →

    cos³x = cos³x

    thus your identity is verified

    I hope it helps...

    Bye!

  • 1 decade ago

    we know

    cos(A + B) = cos A cos B - sin A sin B ...1

    put B= A to get

    cos 2A = cos^2 A - sin ^2 A = 2 cos ^2 A - 1

    similarly sin 2A = 2 sin A cos A

    now in 1 put B = 2A

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

    = 2 cos ^2 A cos A - cos A - 2 cos A sin ^2A

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

    = 4 cos^2 A - 3 cos A

    cos 3A = 4 cos^3 A - 3 cos A

    so cos 3A + 3 cos A = 4 cos^3 A

    so cos^3 A= 1/4( cos 3A + 3 cos A)

    proved

  • Anonymous
    4 years ago

    Cos 3x Formula

  • 5 years ago

    When can I see it?

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  • 1 decade ago

    cos^3x = (1/4)(3cosx + cos3x)

    cos3x = 4cos^3x - 3cosx, so

    cos^3x = (1/4)(3cosx + 4cos^3x - 3cosx)

    cos^3x = 1/4(4cos^3x)

    cos^3x = cos^3x

  • 1 decade ago

    RHS = (1/4)(3cosx + cos(3x))

    = (1/4)(3cos(x) + cos(x + 2x))

    = (1/4)(3cos(x) + cos(x)cos(2x) - sin(x)sin(2x))

    = (1/4)(3cos(x) + cos(x)(cos²(x) - sin²(x)) - 2sin²(x)cos(x))

    = (1/4)(3cos(x) + cos³(x) - cos(x)sin²(x) - 2sin²(x)cos(x))

    = (1/4)(3cos(x) + cos³(x) - 3sin²(x)cos(x))

    = (1/4)(3cos(x)(1 - sin²(x))+ cos³(x))

    = (1/4)(3cos(x)cos²(x)+ cos³(x))

    = (1/4)(4cos³(x))

    = cos³(x)

    = LHS qed

  • Anonymous
    1 decade ago

    u need to use the formula of cos3x and expand it

  • 1 decade ago

    it doesn't...IDENTIES...(cosx)^2= 1/2(1+cos2x) (ANGLE NEVER CHANGES............so........(cosx)^3=1/3(1+cos2x')

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