Please do the last part of the qn "By using the relation sin2x=cos(90-2x), find one of these solutions by calculation."
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Originally Posted by maybeline9216 Please do the last part of the qn "By using the relation sin2x=cos(90-2x), find one of these solutions by calculation." we have $\displaystyle \sin 2x = \cos 3x$ then we are told $\displaystyle \sin 2x = \cos(90 - 2x)$ this means that $\displaystyle \cos (90 - 2x) = \cos 3x$ $\displaystyle \Rightarrow 90 - 2x = 3x$ now continue
O.o so easy arh?? thanks
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