Sin 8theta -sin 10theta= cot 9theta (cos 10theta - cos 8 theta)
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
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.
Thank you
To solve the equation sin(8θ) - sin(10θ) = cot(9θ)(cos(10θ) - cos(8θ)), we will use trigonometric identities and simplification techniques.
Step 1: Simplify the cotangent term.
Using the identity cot(θ) = cos(θ) / sin(θ), we can rewrite cot(9θ) as cos(9θ) / sin(9θ).
Step 2: Expand the cos difference term.
Using the identity cos(a) - cos(b) = -2sin((a+b)/2)sin((a-b)/2), we can rewrite cos(10θ) - cos(8θ) as -2sin(9θ)sin(θ).
Substituting these simplifications into the original equation, we have:
sin(8θ) - sin(10θ) = (cos(9θ) / sin(9θ)) * (-2sin(9θ)sin(θ))
Step 3: Multiply both sides by sin(9θ) to eliminate the fractions.
Multiplying both sides by sin(9θ), we have:
sin(8θ)sin(9θ) - sin(10θ)sin(9θ) = -2sin^2(9θ)sin(θ)
Step 4: Use the sum-to-product identities to simplify the left side.
Using the identity sin(a)sin(b) = (cos(a-b) - cos(a+b)) / 2, we can rewrite the left side of the equation as:
(cos(8θ - 9θ) - cos(8θ + 9θ)) / 2 - (cos(10θ - 9θ) - cos(10θ + 9θ)) / 2 = -2sin^2(9θ)sin(θ)
Simplifying the expression gives us:
(cos(-θ) - cos(17θ)) / 2 - (cos(-θ) - cos(19θ)) / 2 = -2sin^2(9θ)sin(θ)
Step 5: Cancel out the common terms.
Cancelling out the common terms on both sides of the equation, we have:
cos(17θ) - cos(19θ) = -4sin^2(9θ)sin(θ)
Step 6: Use the double angle identity to simplify the left side.
Using the identity cos(a) - cos(b) = -2sin((a+b)/2)sin((a-b)/2), we can rewrite the left side of the equation as:
-2sin((17θ + 19θ) / 2)sin((17θ - 19θ) / 2) = -4sin^2(9θ)sin(θ)
Simplifying the expression gives us:
-2sin(18θ/2)sin(-θ/2) = -4sin^2(9θ)sin(θ)
Step 7: Simplify and solve for θ.
By simplifying further, we have:
-2sin^2(9θ)sin(θ) = -4sin^2(9θ)sin(θ)
Since both sides of the equation are identical, it means that the given equation is an identity. In other words, the equation is true for all values of θ.
Thus, this equation represents a trigonometric identity rather than an equation to be solved for θ.