SinA+Sin3A+Sin5A+Sin7A+.......+Sin29A=????????????????~
Prakash , 9 Years ago
Grade 12
Trigonometry> SinA+Sin3A+Sin5A+Sin7A+.......+Sin29A=???...
1 AnswersPUSHPAK
sin(A) + sin(A+2A) +sin(A+4A) …............+sin(A+28A)
A, 3A , 5A............... 29A are in AP with common difference 2A.
hence sum ={ sin[( 1st + last term)/2] X sin (n x d/2)} / sin d/2
on solving we get
(sin(15A) X sin 29A ) / sin A
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