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December 18, 2014

December 18, 2014

Posted by **briane** on Monday, April 7, 2014 at 3:12pm.

*please show the calculation

- math -
**Steve**, Monday, April 7, 2014 at 3:29pmthe two legs are a and b, and the hypotenuse is √(a^2+b^2). So, assuming a,b > 0,

So, in QII,

sin2x = b/√(a^2+b^2)

cos2x = -a/√(a^2+b^2)

sin^2(x) = (1-cos2x)/2 = (1+a/√(a^2+b^2))/2

= (a+√(a^2+b^2)) / 2√(a^2+b^2)

and similarly for cosx

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