Trigonometry

Problem : 15) Prove that :
sinA + sin5A + sin3A + sin7A = 16 sin A cos²A cos² (2A) 


Solution :


Problem : sinA + sin5A + sin3A + sin7A = 16 sin A cos²A cos² (2A)

L.H.S. = sinA + sin5A + sin3A + sin7A
=> (sinA + sin7A) + (sin5A + sin3A) - { Rearranging the terms }
=> 2 sin 4A cos 3A + 2 sin 4A cos A
=> 2 sin 4A [ cos 3A + cos A ]

=> 2 sin 4A * 2 cos 2A cos A
=> 4 sin 2A cos 2A * 2 cos 2A cos A
=> 8 sin A * cos A *cos 2A * 2 cos 2A cos A
=> 16 sin A cos² A * cos² (2A) ==> R.H.S.

Note : 1) sin C + sin D = 2 * sin [(C + D)/2 ] * cos [(C - D)/2]
. . . . . 2)cos C + cos D = 2 * cos [(C + D)/2 ] * cos [(C - D)/2]
. . . . . 3) sin 2A = 2sin A * cos A


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