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1.sin^2 5Â°+sin^2 10Â°+sin^2 15Â°+...+sin^2 90Â° is?

Answers (1)

R Ravindra Pindel

=sin²5+sin²10+.......sin²45+..........sin²85+sin²90

sin(90-x) =cosx, sin²(90-x)= cos²x

=sin²5+sin²10+.......sin²45+..........+cos²10+cos²5+sin²90

And sin²x+cos²x= 1

So in given series we get 8 cases where sin²x+cos²x= 1

=8+sin²45+sin²90

= 8+1/2+1

=9+1/2

= 19/2

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