Question

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

Answer 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

Answer 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 )

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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 ) ]
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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.

Answer 3

Thank you