(sin2a)(cosa)/(1+cos2a)(1+cosa)= tana/2. prove this
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sin2a = 2sina . cosa
And cos2a = 2cos^2a -1 (cos2a= cos^2a-sin^2a)
So
the equation transforms to
sina/(1+cosa), cancelling 2cos^2a on both numerator and denominator..
Now sina=2sinx. cosx, considering a =2(x) where x=a/2
and cosa=2cos^2x-1
So the equation transforms to
(sinx)/(cosx)
or tanx
or
tana/2...
And cos2a = 2cos^2a -1 (cos2a= cos^2a-sin^2a)
So
the equation transforms to
sina/(1+cosa), cancelling 2cos^2a on both numerator and denominator..
Now sina=2sinx. cosx, considering a =2(x) where x=a/2
and cosa=2cos^2x-1
So the equation transforms to
(sinx)/(cosx)
or tanx
or
tana/2...
sksssksk:
Can u explain it
Yeah
The whole answer or some part of it?
Bro the whole answer
See the identity sin2x = sin(x+x)= sinxcosx+cosxsinx =2sinxcosx.. See the identity sin(x+y) is sinxcosy+cosxsiny
Secondly, cos(x+y)=cosxcosy-sinsiny and here cos(x+x)=cosxcosx-sinxsinx=cos^2x - sin^2x..
or 1-sin^2x-sin^2x or 1-2sin^2x or we can say that is equivalent to 2cos^2x - 1
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