To me this appears to be a silly question, if your 0 is indeed a zero.
Why even write it as (3π/2 + 0)
it would simply be cos 3π/2
and you should know that cos 3π/2 = 0 which is also the value of the right side.
End of problem.
how do you establish the identity of cos(3pi/2+0)=sin0
2 answers
is the "0" in the problem, the angle theta? if it is,
cos(3pi/2+[theta]) = sin [theta]
*expand cos(3pi/2+[theta]),
cos(3pi/2)cos(theta) - sin(3pi/2)sin(theta) = sin [theta]
*cos(3pi/2) = 0 and sin(3pi/2) = -1, thus:
-(-sin [theta]) = sin [theta]
sin [theta] = sin [theta]
so there,, :)
cos(3pi/2+[theta]) = sin [theta]
*expand cos(3pi/2+[theta]),
cos(3pi/2)cos(theta) - sin(3pi/2)sin(theta) = sin [theta]
*cos(3pi/2) = 0 and sin(3pi/2) = -1, thus:
-(-sin [theta]) = sin [theta]
sin [theta] = sin [theta]
so there,, :)