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sinx+sin2x=1what is the value of sin10 x+sin8 x+3(sin6 x+sin4 x)+sin2 x-1

Rinkoo Gupta
6 years ago
sin8x (sin2x +1) +3sin4x(sin2x+1) +sin2x-1
sin8x(1-sinx+1)+3sin4x(1-sinx+1)+1-sinx-1
sin8x(2-sinx)+3sin4x(2-sinx)-sinx
(2-sinx)(sin8x+3sin4x) -sinx
(2-sinx)sin4x(sin4x+3) -sinx
(2-sinx)sin4x{(1-sinx)2+3} -sinx
sin4x(2-sinx){1+sin2x-2sinx+3} -sinx
sin4x(2-sinx)(1+1-sinx-2sinx+3) -sinx
sin4x(2-sinx)((5-3sinx) -sinx
sin4x(10-6sinx-5sinx+3sin2x) -sinx
sin4x[10-11sinx+3(1-sinx)]-sinx
sin4x[13-14sinx]-sinx
(1-sinx)2(13-14sinx) -sinx
(1+sin2x-2sinx)(13-14sinx)-sinx
(1+1-sinx-2sinx)(13-14sinx)-sinx
(2-3sinx)(13-14sinx)-sinx
26-28sinx-39sinx+42sin2x -sinx
26-67six+42(1-sinx) -sinx
26-68sinx+42-42sinx
68-110sinx
sin2x+sinx-1=0
solving by shridharacharya method we get sinx=(sqrt5-1)/2
so the value ofsin10x+sin8x+3(sin6x+sin4x)+sin2x-1
=68-110sinx
=68-110(sqrt5-1)/2
=68-55(sqrt5-1)
=68-55sqrt5+55
=123-55sqrt5 Ans.
Thanks & Regards
Rinkoo Gupta