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Question: Prove that \({{\cosh }^{2}}x-{{\sinh }^{2}}x=1\) ....

Prove that cosh⁡2x−sinh⁡2x=1{{\cosh }^{2}}x-{{\sinh }^{2}}x=1 .

Explanation

Solution

Hint: cosh⁡x andsinh⁡x\cosh x\text{ and}\sinh x in the equation given in the question represents hyperbolic functions. So, to prove the equation just put the values of coshx and sinhx and solve the left-hand side of the equation using the basic algebraic formulas.
Complete step-by-step answer:
In the equation given in the question the terms cosh⁡x andsinh⁡x\cosh x\text{ and}\sinh x represent the hyperbolic function and we know that the values of hyperbolic functions can be represented as given below:
cosh⁡x=ex+e−x2 sinh⁡x=ex−e−x2 \begin{aligned} & \cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \\\ & \sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \\\ \end{aligned}
So, if we start by solving the left-hand side of the equation given in the question by putting the values of hyperbolic functions, we get
cosh⁡2x−sinh⁡2x{{\cosh }^{2}}x-{{\sinh }^{2}}x
=(ex+e−x2)2−(ex−e−x2)2={{\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)}^{2}}-{{\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \right)}^{2}}
Now, we know that a2−b2{{a}^{2}}-{{b}^{2}} can be written as (a+b)(a−b)\left( a+b \right)\left( a-b \right) . So, if we use this in our expression, we get
((ex+e−x2)−(ex−e−x2))((ex+e−x2)+(ex−e−x2))\left( \left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)-\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \right) \right)\left( \left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)+\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \right) \right)
Now if we take LCM of each part to be 2 and open the brackets with proper signs, we get
(ex+e−x−ex+e−x2)(ex+e−x+ex−e−x2)\left( \dfrac{{{e}^{x}}+{{e}^{-x}}-{{e}^{x}}+{{e}^{-x}}}{2} \right)\left( \dfrac{{{e}^{x}}+{{e}^{-x}}+{{e}^{x}}-{{e}^{-x}}}{2} \right)
=(2e−x2)(2ex2)=\left( \dfrac{2{{e}^{-x}}}{2} \right)\left( \dfrac{2{{e}^{x}}}{2} \right)
=exe−x={{e}^{x}}{{e}^{-x}}
Now, we know that aka−k{{a}^{k}}{{a}^{-k}} is always equal to 1. So, we can conclude that exe−x{{e}^{x}}{{e}^{-x}} is equal to 1.
So, we have shown that the left-hand side of the equation given in the question is equal to the right-hand side of the equation. Hence, we have proved that cosh⁡2x−sinh⁡2x=1{{\cosh }^{2}}x-{{\sinh }^{2}}x=1 .

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, you need to remember the values of the hyperbolic functions sinhx and coshx to be able to solve such problems and make sure that you are not confused between their values, as they differ by only a sign.