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

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

Answers (1)

=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

Posted by

Ravindra Pindel

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