∫ 0 1 sin 5 x sin x ( cos 2 x − cos 2 5 π ) ( cos 2 x − cos 2 5 2 π ) . d x
The value of above integral is b a . Find a + b .
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What is the quick way to generalize the trigonometric identity that is stated?
IE Prove that for n = 2 k + 1 , sin n θ = sin θ × ∏ i = 1 k ( cos 2 k + 1 i π − cos θ ) × ∏ i = 1 k ( cos 2 k + 1 i π + cos θ ) .
Nice Question ! @Tanishq Varshney
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This question revolves around a couple of identities.
_Identity 1: _ r = 0 ∏ n − 1 sin ( x + n r π ) = 2 n − 1 sin ( n x )
Proof:
Consider the equation, z n = e 2 n i x . We can factorize the above equation using the concept of roots of unity.
z n − e 2 n i x ∣ 1 − e 2 n i x ∣ 2 sin ( n x ) ∣ ∣ sin ( n x ) − i cos ( n x ) ∣ ∣ 2 sin ( n x ) = ∏ r = 0 n − 1 ( z − e i ( 2 x + n 2 π r ) ) = ∏ r = 0 n − 1 ∣ ∣ ∣ ∣ ( 1 − e i ( 2 x + n 2 π r ) ) ∣ ∣ ∣ ∣ = ∏ r = 0 n − 1 2 sin ( x + n π ) ∣ ∣ ∣ ∣ sin ( x + n π ) − i cos ( x + n π ) ∣ ∣ ∣ ∣ = 2 n ∏ r = 0 n − 1 sin ( x + n r π )
And the identity follows.
Identity 2: cos 2 A − cos 2 B = sin ( B + A ) sin ( B − A )
Proof:
sin ( B + A ) sin ( B − A ) = ( sin A cos B + cos A sin B ) ( cos A sin B − sin A cos B ) = cos 2 A sin 2 B − sin 2 A cos 2 B = cos 2 A − cos 2 A cos 2 B − sin 2 A cos 2 B = cos 2 A − cos 2 B
Using the second identity, we can write, sin x ( cos 2 x − cos 2 y ) sin x ( cos 2 x − cos 2 2 y )
= sin x sin ( x + y ) sin ( y − x ) sin ( x + 2 y ) sin ( 2 y − x ) = sin ( x − 5 2 π ) sin ( x − 5 π ) sin ( x ) sin ( x + 5 π ) sin ( x + 5 2 π ) = sin ( x ) sin ( x + 5 π ) sin ( x + 5 2 π ) sin ( x + 5 3 π ) sin ( x + 5 4 π ) = 2 4 sin ( 5 x )
The last step follows from the first identity. Therefore,
∫ 0 1 sin ( 5 x ) sin x ( cos 2 x − cos 2 y ) sin x ( cos 2 x − cos 2 2 y ) d x
= ∫ 0 1 1 6 sin ( 5 x ) . sin ( 5 x ) 1 d x = ∫ 0 1 1 6 1 d x = 1 6 1
Compare the answer to find the values of a and b and add them up to get the final answer as 1 7 .