How do you simplify #2 sin 35 cos 35#?

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Answer 1 · 2015-05-12T21:08:42

Use #e^(i theta) = cos theta + i sin theta#

#cos 2theta + i sin 2theta#

#= e^(2 i theta)#

#= e^(i theta)e^(i theta)#

#= (cos theta + i sin theta)(cos theta + i sin theta)#

#= (cos^2 theta - sin^2 theta) + i (2 cos theta sin theta)#

Looking at the coefficients of #i#, we get

#sin (2 theta) = 2 cos theta sin theta = 2 sin theta cos theta#

So #2 sin 35^o cos 35^o = sin ( 2*35^o ) = sin 70^o#