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Question Number 85620 by Roland Mbunwe last updated on 23/Mar/20

$$provethat$$$$$$$$\cosh (x-y)=\cosh x\cosh y-\sinh x\sinh y$$

Answered by Rio Michael last updated on 23/Mar/20

$$RHS=\cosh x\cosh y-\sinh x\sinh y$$$$=\frac{1}{2}(e^x+e^{-x})\frac{1}{2}(e^y+e^{-y})-\frac{1}{2}(e^x-e^{-x})\frac{1}{2}(e^y-e^{-y})$$$$=\frac{1}{4}[(e^x+e^{-x})(e^y+e^{-y})-(e^x-e^{-x})(e^y-e^{-y})]$$$$=\frac{1}{4}(2e^{(x-y)}+2e^{-(x-y)})=\frac{1}{2}(e^{(x-y)}+e^{-(x-y)})=\cosh (x-y)$$

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