Prove that Cosh(2x)=2cos2x-1?

Question

15583 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T10:35:57

See the proof below

I am going to prove the following

$cosh 2 x = 2 {cosh}^{2} x - 1$ $cosh2x=2cosh^2x-1$

By the definition of hyperbolic functions

$cosh (2 x) = \frac{e^{2 x} + e^{- 2 x}}{2}$ $cosh(2x)=(e^(2x)+e^(-2x))/2$

$cosh x = \frac{e^{x} + e^{- x}}{2}$ $coshx=(e^x+e^-x)/(2)$

${cosh}^{2} x = {(\frac{e^{x} + e^{- x}}{2})}^{2}$ $cosh^2x=((e^x+e^-x)/(2))^2$

$= \frac{1}{4} {(e^{x} + e^{- x})}^{2}$ $=1/4(e^x+e^-x)^2$

$= \frac{1}{4} (e^{2 x} + e^{- 2 x} + 2)$ $=1/4(e^(2x)+e^(-2x)+2)$

Therefore,

$2 {cosh}^{2} x - 1 = 2 \cdot \frac{1}{4} (e^{2 x} + e^{- 2 x} + 2) - 1$ $2cosh^2x-1=2*1/4(e^(2x)+e^(-2x)+2)-1$

$= \frac{e^{2 x} + e^{- 2 x}}{2} + 1 - 1$ $=(e^(2x)+e^(-2x))/2+1-1$

$= cosh (2 x)$ $=cosh(2x)$

$Q E D$ $QED$