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Cos(A+B)Cos(A-B)

Show that c...

Question

Show that $cos h x + sin h x = e^{x}$ and simplify $cos h x - sin h x = ? .$ By considering ${(cos h x + sin h x)}^{2} + {(cos h x - sin h x)}^{2}$ show that ${cos}^{2} h x - {sin}^{2} h x = cos h 2 x$ .

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Solution

We know that;
$c o s h x = \frac{e^{x} + e^{- x}}{2}$

$s i n h x = \frac{e^{x} - e^{- x}}{2}$
Thus;
$c o s h x + s i n h x = \frac{e^{x} + e^{- x}}{2} + \frac{e^{x} - e^{- x}}{2} = e^{x}$

$c o s h x - s i n h x = \frac{e^{x} + e^{- x}}{2} - \frac{e^{x} - e^{- x}}{2} = e^{- x}$

Thus;
$(c o s h x + s i n h x)^{2} = e^{2 x}$

$(c o s h x - s i n h x)^{2} = e^{- 2 x}$

$\to$ $(c o s h x + s i n h x)^{2} +$ $(c o s h x - s i n h x)^{2} = e^{2 x} + e^{- 2 x}$

$\to 2 (c o s h^{2} h x + s i n^{2} h x) = e^{2 x} + e^{- 2 x}$

$\to (c o s h^{2} h x + s i n^{2} h x) = \frac{e^{2 x} + e^{- 2 x}}{2} = c o s h 2 x$

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Cos(A+B)Cos(A-B)

Standard XII Mathematics

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