How to proof this identity trigonometry?

cos 4x + cos 6x - cos 2x - 1 = -4 sin 2x sin 3x cos x

2 Answers

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  • 2 years ago
    Favorite Answer

    LHS=

    cos(4x)+cos(6x)-cos(2x)-1=

    cos(4x)+cos(6x)-2cos^2(x)=

    2cos(5x)cos(x)-2cos^2(x)=

    2cos(x)[cos(5x)-cos(x)]=

    2cos(x)[-2sin(3x)sin(2x)]=

    -4cos(x)sin(3x)sin(2x)=

    RHS

  • cidyah
    Lv 7
    2 years ago

    cos(a)+cos(b) = 2 cos ((a+b)/2) cos((a-b)/2)

    cos(-a) = cos(a)

    cos 4x + cos 6x = 2 cos (10x/2) cos( -2x/2) = 2 cos(5x) cos(x)

    2 cos (5x) cos(x) - cos(2x) - 1

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

    = 2 cos (5x) cos(x) - (2 cos^2(x) -1) -1

    = 2 cos(5x) cos(x) - 2 cos^2(x) + 1 - 1

    = 2 cos(5x) cos(x) - 2 cos^2(x)

    = 2 cos(x) ( cos(5x) - cos(x) )

    cos(a)-cos(b) = - 2 sin ( (a-b)/2 ) sin ((a+b)/2)

    cos(5x) - cos(x) = -2 sin (2x) sin(3x)

    2 cos(x) ( cos(5x) - cos(x) ) = 2 cos(x) [ -2 sin (2x) sin(3x)]

    = -4 cos(x) sin(2x) sin(3x)

    = -4 sin(2x) sin(3x) cos(x)

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