Question

prove that,
cos a + cos 3a + cos 5a + cos 7a + cos 9a = (sin 10a)/ 2 sina

Answer 1

SOLUTION

TO PROVE

$\displaystyle \sf{ \cos A + \cos 3A + \cos 5A +\cos 7A +\cos 9A \: = \frac{\sin 10A \: }{2\sin A} }$

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric formula that

$\sf{2 \cos A \sin B = \sin(A + B) - \sin(A - B) }$

PROOF

$\displaystyle \sf{ \cos A + \cos 3A + \cos 5A +\cos 7A +\cos 9A \: }$

$= \displaystyle \sf{\frac{1 \: }{2\sin A} \bigg(2\sin A \cos A +2 \cos 3A \sin A+2 \cos 5A \sin A +2\cos 7A\sin A +2\cos 9A \sin A\: \bigg) }$

$= \displaystyle \sf{\frac{1 \: }{2\sin A} \bigg(\sin 2A +\sin4 A - \sin 2A +\sin 6A - \sin4A +\sin 8A - \sin 6A +\sin 10A - \sin 8A \: \bigg) }$

$\displaystyle \sf{ = \frac{\sin 10A \: }{2\sin A} }$

Hence proved

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