how would you simplify 2sin3xcosx=1 i've been stuck on this question for some time ;[
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consider $\displaystyle sin(a)cos(b) = \frac{1}{2}[sin(a+b)+sin(a-b)] $ $\displaystyle 2sin(3x)cos(x)=1$ $\displaystyle 2(\frac{1}{2}[sin(3x+x)+sin(3x-x)])=1$ $\displaystyle sin(3x+x)+sin(3x-x)=1$ $\displaystyle sin(4x)+sin(2x)=1 $
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