If Cos(a+b) =0 then sin(a-b) is reduced to Cos 2b . How?
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Hi,
cos(a+b)=0 is equivalent to a+b=(2k+1) pi/2 thus b=-a+(2k+1) pi/2, k is an integer.
Then sin(a-b)=sin(2a-(2k+1) pi/2)=sin(2a-k pi-pi/2)=-sin(pi/2-(2a-k pi))=-cos(2a-k pi)
= -cos(2b) if k is even ; cos(2b) if k is odd.
Hope this is it....
cos(a+b)=0 is equivalent to a+b=(2k+1) pi/2 thus b=-a+(2k+1) pi/2, k is an integer.
Then sin(a-b)=sin(2a-(2k+1) pi/2)=sin(2a-k pi-pi/2)=-sin(pi/2-(2a-k pi))=-cos(2a-k pi)
= -cos(2b) if k is even ; cos(2b) if k is odd.
Hope this is it....
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