Math helppp

Sin 8theta -sin 10theta= cot 9theta (cos 10theta - cos 8 theta)

  1. 0
  2. 1
asked by Jes
  1. Just use the sum-to-product formulas...

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

    So, let
    a=8θ
    b=10θ
    and you have

    sin(8θ)-sin(10θ) = 2cos(9θ)sin(-θ)
    cos(10θ)-cos(8θ) = -2sin(9θ)sin(θ)

    now it is clear to see that
    2cos(9θ)sin(-θ) = cot(9θ)(-2sin(9θ)sin(θ))
    The -2sin(θ) factors cancel, and you are left with

    cos(9θ) = cot(9θ)sin(9θ)
    which is true, since cot = cos/sin

    1. 0
    posted by Steve
  2. sin ( 8 θ ) - sin ( 10 θ ) = cot ( 9 θ ) [ cos ( 10 θ ) - cos ( 8 θ ) ]

    sin ( 8 θ ) - sin ( 10 θ ) = [ cos ( 9 θ ) / sin ( 9 θ ) ] ∙ [ cos ( 10 θ ) - cos ( 8 θ ) ]

    Multiply both sides by sin ( 9 θ )

    sin ( 9 θ ) ∙ [ sin ( 8 θ ) - sin ( 10 θ ) ] = cos ( 9 θ ) ∙ [ cos ( 10 θ ) - cos ( 8 θ ) ]

    sin ( 9 θ ) ∙ sin ( 8 θ ) - sin ( 9 θ ) ∙ sin ( 10 θ ) = cos ( 9 θ ) ∙ [ cos ( 10 θ ) - cos ( 8 θ ) ]

    sin ( 9 θ ) ∙ sin ( 8 θ ) - sin ( 9 θ ) ∙ sin ( 10 θ ) = cos ( 9 θ ) ∙ cos ( 10 θ ) - cos ( 9 θ ) ∙ cos ( 8 θ )

    _______________________________________
    sin (A ) ∙ sin (B ) = ( 1 / 2 ) [ cos ( A - B ) - cos ( A + B ) ]

    cos ( A ) ∙ cos ( B ) = ( 1 / 2 ) [ cos ( A - B ) + cos ( A + B ) ]


    sin ( 9 θ ) ∙ sin ( 8 θ ) = ( 1 / 2 ) [ cos ( 9 θ - 8 θ ) - cos ( 9 θ + 8 θ ) ]

    sin ( 9 θ ) ∙ sin ( 8 θ ) = ( 1 / 2 ) [ cos ( θ ) - cos ( 17 θ ) ]


    sin ( 9 θ ) ∙ sin ( 10 θ ) = ( 1 / 2 ) [ cos ( 9 θ - 10 θ ) - cos ( 9 θ + 10 θ ) ]

    sin ( 9 θ ) ∙ sin ( 10 θ ) = ( 1 / 2 ) [ cos ( - θ ) - cos ( 19 θ ) ]

    Since:

    cos ( - θ ) = cos ( θ )

    sin ( 9 θ ) ∙ sin ( 10 θ ) = ( 1 / 2 ) [ cos ( θ ) - cos ( 19 θ ) ]


    cos ( 9 θ ) ∙ cos ( 10 θ ) = ( 1 / 2 ) [ cos ( 9 θ - 10 θ ) + cos ( 9 θ + 10 θ ) ]

    cos ( 9 θ ) ∙ cos ( 10 θ ) = ( 1 / 2 ) [ cos ( - θ ) + cos ( 19 θ ) ]

    Since:

    cos ( - θ ) = cos ( θ

    cos ( 9 θ ) ∙ cos ( 10 θ ) = ( 1 / 2 ) [ cos ( θ ) + cos ( 19 θ ) ]


    cos ( 9 θ ) ∙ cos ( 8 θ ) = ( 1 / 2 ) [ cos ( 9 θ - 8 θ ) + cos ( 9 θ + 8 θ ) ]

    cos ( 9 θ ) ∙ cos ( 8 θ ) = ( 1 / 2 ) [ cos ( θ ) + cos ( 17 θ ) ]
    _______________________________________

    Replace this values in equation:

    sin ( 9 θ ) ∙ sin ( 8 θ ) - sin ( 9 θ ) ∙ sin ( 10 θ ) = cos ( 9 θ ) ∙ cos ( 10 θ ) - cos ( 9 θ ) ∙ cos ( 8 θ )

    ( 1 / 2 ) [ cos ( θ ) - cos ( 17 θ ) ] - ( 1 / 2 ) [ cos ( θ ) - cos ( 19 θ ) ] = ( 1 / 2 ) [ cos ( θ ) + cos ( 19 θ ) ] - ( 1 / 2 ) [ cos ( θ ) + cos ( 17 θ ) ]

    Multiply both sides by 2

    cos ( θ ) - cos ( 17 θ ) - [ cos ( θ ) - cos ( 19 θ ) ] = cos ( θ ) + cos ( 19 θ ) - [ cos ( θ ) + cos ( 17 θ ) ]

    cos ( θ ) - cos ( 17 θ ) - cos ( θ ) + cos ( 19 θ ) = cos ( θ ) + cos ( 19 θ ) - cos ( θ ) - cos ( 17 θ )

    - cos ( 17 θ ) + cos ( 19 θ ) = cos ( 19 θ ) - cos ( 17 θ )

    cos ( 19 θ ) - cos ( 17 θ ) = cos ( 19 θ ) - cos ( 17 θ )

    This mean identity is true.

    1. 0
    posted by Bosnian
  3. Thank you

    1. 0
    posted by Jes

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