If f(x) = sin x, show that f (2x)= 2f(x) f(1/2 π-x).
If "f(x)=sinx," then
"f(2x)=sin2x=2sinxcosx" (double angle formula).
Let's consider the right side of the equality:
"2f(x)f(\\frac{\\pi}{2}-x)=2sinxsin(\\frac{\\pi}{2}-x)=2sinxcos x,"
because "sin(\\frac{\\pi}{2}-x)=cosx" (co-function identity).
So we have:
"f(2x)=2sinxcosx=2f(x)f(\\frac{\\pi}{2}-x),"
and the statement is proved.
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