(sin theta)(2^ncos theta)=?
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=(2^n)sin2tita
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thank u
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Answered by
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hello users
solution
we know that
sin 2θ = 2 sinθ cosθ
Here
sinθ * 2^n cosθ
= 2^n sinθ cosθ
= 2^n * 2* sinθ cosθ / 2 .....divide and multiply by 2 ....
= 2^n / 2 * 2sinθ cosθ
= 2^(n -1) sin2θ Answer
# hope it helps :)
solution
we know that
sin 2θ = 2 sinθ cosθ
Here
sinθ * 2^n cosθ
= 2^n sinθ cosθ
= 2^n * 2* sinθ cosθ / 2 .....divide and multiply by 2 ....
= 2^n / 2 * 2sinθ cosθ
= 2^(n -1) sin2θ Answer
# hope it helps :)
Ankit1408:
hopeit helps
^^'
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