`cos(x + y) cos(x - y) = cos^2(x) - sin^2(y)` Prove the identity.

$\displaystyle{\cos{{\left({x}+{y}\right)}}}{\cos{{\left({x}-{y}\right)}}}={{\cos}}^{{2}}{\left({x}\right)}-{{\sin}}^{{2}}{\left({y}\right)}$

We will use the following product formulas to prove the identity,

$\displaystyle{2}{\cos{{A}}}{\cos{{B}}}={\cos{{\left({A}+{B}\right)}}}+{\cos{{\left({A}-{B}\right)}}}$

LHS= $\displaystyle{\cos{{\left({x}+{y}\right)}}}{\cos{{\left({x}-{y}\right)}}}$

$\displaystyle=\frac{{{\cos{{\left({x}+{y}+{x}-{y}\right)}}}+{\cos{{\left({x}+{y}-{\left({x}-{y}\right)}\right)}}}}}{{2}}$

$\displaystyle=\frac{{{\cos{{\left({2}{x}\right)}}}+{\cos{{\left({2}{y}\right)}}}}}{{2}}$

Now we will use $\displaystyle{\cos{{\left({2}\theta\right)}}}={2}{{\cos}}^{{2}}{\left(\theta\right)}-{1},{\cos{{\left({2}\theta\right)}}}={1}-{2}{{\sin}}^{{2}}{\left(\theta\right)}$

$\displaystyle=\frac{{{2}{{\cos}}^{{2}}{\left({x}\right)}-{1}+{1}-{2}{{\sin}}^{{2}}{\left({y}\right)}}}{{2}}$

$\displaystyle=\frac{{{2}{\left({{\cos}}^{{2}}{\left({x}\right)}-{{\sin}}^{{2}}{\left({y}\right)}\right)}}}{{2}}$

$\displaystyle={{\cos}}^{{2}}{\left({x}\right)}-{{\sin}}^{{2}}{\left({y}\right)}$

LHS=RHS , Hence proved.

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Monday, September 21, 2015

$\displaystyle{\cos{{\left({x}+{y}\right)}}}{\cos{{\left({x}-{y}\right)}}}={{\cos}}^{{2}}{\left({x}\right)}-{{\sin}}^{{2}}{\left({y}\right)}$ Prove the identity.

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