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SinA+ 2sin3A + sin5A/sin3A + 2sin5A + sin7A=sin3A/sin5A

SinA+ 2sin3A + sin5A/sin3A + 2sin5A + sin7A=sin3A/sin5A

Grade:11

1 Answers

Vikas TU
14149 Points
4 years ago
Dear Student,
By applying sin(x+y) and cos(x-y),
sin(x + y) cos(x - y) = [sinx cosy + siny cosx][cosx cosy + sinx siny]
=(sinx cosx cos²y + sinx cosx sin²y) + (siny cosy cos²x + sin²x siny cosy)
=(sinx cosx)(cos²y + sin²y) + (siny cosy)(cos²x + sin²x)
=sinx cosx + siny cosy
=(1/2)sin2x + (1/2)sin2y (By the formula of 2theta for sin)
=(1/2)(sin2x + sin2y) = sin(x + y) cos(x - y)  ----- equation(1)
the question given was-
(sina + 2sin3a + sin5a) / (sin3a + 2sin5a + sin7a)
=[(sina + sin3a) + (sin3a + sin5a)] / [(sin3a + sin5a) + (sin5a + sin7a)]
 
 
From equation (1): [2sin{(3a + a)/2} cos{(3a - a)/2} + 2sin{(5a + 3a)/2} cos{(5a - 3a)/2}] / [2sin{(5a + 3a)/2} cos{(5a - 3a)/2} + 2sin{(7a + 5a)/2} cos{(7a - 5a)/2}] =
[2sin2a cosa + 2sin4a cosa] / [2sin4a cosa + 2sin6a cosa] =
2cosa [sin2a + sin4a] / [{2cosa}{sin4a + sin6a}] =(sin2a + sin4a) / (sin4a + sin6a)
From equation (1):  2sin{(4a + 2a)/2} cos{(4a - 2a)/2} / [2sin{(6a + 4a)/2} cos{(6a - 4a)/2}]
=2sin3a cosa / [2sin5a cosa] =sin3a / sin5a (hence proved)
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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