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SINA+SIN3A+SIN5A+SIN7A
Hey Manideep,the answer should be 16 sinA.cos22A.cos2Ayou will have to use these 3 formulas in order to get the answer:1)sinA+sinB=2sin(A+B)/2.cos(A-B)/22)cosA+cosB=2cos(A+B)/2.cos(A-B)/2 and lastly,3)sin2A=2sinA.cosA
And if you want the answer here it is: sinA + sin5A + sin3A + sin7A= (sinA + sin7A) + (sin5A + sin3A) After rearranging the terms use identity 1 from the above post,= 2 sin 4A cos 3A + 2 sin 4A cos AHere 2sin4A is common=2 sin 4A [ cos 3A + cos A ]Use Identity 2 from the above post,=2 sin 4A * 2 cos 2A cos Asin4A can be split into 2sin2A.cos2A from identity 3,= 4 sin 2A cos 2A * 2 cos 2A cos AAgain use sin2A formula= 8 sin A * cos A *cos 2A * 2 cos 2A cos Aand then just arrange the terms= 16 sin A cos² A * cos² (2A) Please approve the answer if you found it helpful
http://www.askiitians.com/forums/Trigonometry/sina-sin3a-sin5a-sin7a_125330.htm
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