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Question Number 22102 by Parvez akhter last updated on 11/Oct/17

$$\cosh \left(15\right)$$

Answered by $@ty@m last updated on 11/Oct/17

$$\cos hx=\frac{e^x+e^{-x}}{2}$$$$\cos h\left(15\right)=\frac{e^{15}+e^{-15}}{2}$$$$=\frac{e^{15}+\frac{1}{e^{15}}}{2}$$

Commented by $@ty@m last updated on 11/Oct/17

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