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Question

sinA+sin3A+sin5A+sin7A equals

A
4sinAcos2Acos4A
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B
4sinAcos2 Acos2 A
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C
4cosAsin2Asin4A
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D
4cosAcos2Asin4A
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Solution

The correct option is D 4cosAcos2Asin4A
sinA+sin3A+sin5A+sin7A
=2sin2AcosA+2sin6AcosA
=2[sin2A+sin6A]cosA
=2[2sin4Acos2A]cosA
=4sin4AcosAcos2A
=4cosAcos2Asin4A

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