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# Find the numerical value of each expression.(a) $\sin h 0$ (b) $\cos h 0$

## (a) $\sinh 0=\frac{1}{2}\left(e^{0}-e^{-0}\right)=0$(b) $\cosh 0=\frac{1}{2}\left(e^{0}+e^{-0}\right)=\frac{1}{2}(1+1)=1$

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### Video Transcript

Okay. We know that sign of zero is gonna be eating zero minus eat of the negatives here, over to which is one minus one over to which is zero part B. Either the zero plus even the negative zero over to is one plus one over to which is to over two, which is gonna give us one.

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