Prove that $cos h (x - y) = cos h x cos h y - sin h x sin h y$

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Solution

We know that;
$cos h x cos h y = \frac{1}{2} (e^{x} + e^{- x}) \times \frac{1}{2} (e^{y} + e^{- y})$
$= \frac{1}{4} (e^{x + y} - e^{x - y} - e^{- (x + y)} + e^{- (x - y)})$
$sin h x sin h y = \frac{1}{2} (e^{x} - e^{- x}) \times \frac{1}{2} (e^{y} - e^{- y})$
$= \frac{1}{4} (e^{x + y} - e^{x - y} - e^{- (x - y)} + e^{- (x + y)})$
Subtracting gives
$cos h x cos h y - sin h x sin h y = 2 \times \frac{1}{4} (e^{x + y} - e^{- (x - y)}) = \frac{1}{2} (e^{x - y} - e^{- (x - y)}) = cos h (x - y)$

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