Question

If the function $E:R\to$ defines by $f\left(x\right)\frac{3^x+3^{-x}}{2}$ than show that $f\left(x+y\right)+f\left(x-y\right)=2f\left(x\right)f\left(y\right)$.

Answer 1

Given

$f\left(x\right)=\frac{3^x+3^{-x}}{2}$

$2f\left(x\right)=3^x+3^{-x}$

SImilarly

$f\left(y\right)=\frac{3^y+3^{-y}}{2}$

$2f\left(y\right)=3^y+3^{-y}$

If we put $x+y$ in the place of $x$ we get

$f(x+y)=\frac{3^{x+y}+3^{-(x+y)}}{2}$

If we put $x-y$ in the place of $x$ we get

$f(x-y)=\frac{3^{x-y}+3^{-(x-y)}}{2}$

Adding $f(x+y)$ and $f(x-y)$ we get

$\frac{3^{x+y}+3^{-(x+y)}}{2}+$ $\frac{3^{x-y}+3^{-(x-y)}}{2}$= $\frac{3^{x+y}+3^{x-y}+3^{-x-y}+3^{-x+y}}{2}$

Taking common from first two and last two terms we get

$=\frac{3^x(3^y+3^{-y})+3^{-x}(3^{-y}+3^y)}{2}$

$=\frac{(3^x+3^{-x})(3^{-y}+3^y)}{2}$

$=\frac{2f\left(x\right)\times 2f\left(y\right)}{2}$

$=2f\left(x\right)f\left(y\right)$