Question

Use the definitions of $$\sinh x$$ and $$\cosh x$$ in terms of exponential functions to prove that $$\cosh 2x=2{\cosh }^{2}x-1$$.

Answer 1

Hint: Use the formula of hyperbolic given as $$\cosh x={\displaystyle \frac{e^x+e^{-x}}{1}}$$ and replace ‘x’ with ‘2x’ to find the expression for $$\cosh 2x$$. Now, square both sides of the expression of $$\cosh x$$, multiply with 2 and subtract 1 to find the value of $$2{\cosh }^{2}x-1$$. Use the algebraic identity $${(a+b)}^2=a^2+b^2+2ab$$ to simplify $${\cosh }^{2}x$$. Check if the obtained expressions of $$\cosh 2x$$ and $$(2{\cosh }^{2}x-1)$$ are equal or not.

Complete step by step answer:
Here, we have been provided with the hyperbolic functions $$\sinh x$$ and $$\cosh x$$ and we are asked to prove the relation $$\cosh 2x=2{\cosh }^{2}x-1$$.
Now, we know that in mathematics, hyperbolic functions are analogs of the ordinary trigonometric functions defined for the hyperbola rather than on the circle. Just like the points $$(\cos t,\sin t)$$ form a circle with a unit radius, the points $$(\cosh t,\sinh t)$$ form the right half of the equilateral hyperbola. These hyperbolic functions are written in exponential form as: -
(i) $$\sinh x={\displaystyle \frac{e^x-e^{-x}}{2}}$$
(ii) $$\cosh x={\displaystyle \frac{e^x+e^{-x}}{2}}$$
Now, let us come to the question, we have to prove $$\cosh 2x=2{\cosh }^{2}x-1$$. So, considering relation (ii) from the above listed relations, we have,
$$\Rightarrow \cosh x={\displaystyle \frac{e^x+e^{-x}}{2}}$$
Replacing ‘x’ with ‘2x’ in the above relation, we have,
$$\Rightarrow \cosh 2x={\displaystyle \frac{e^{2x}+e^{-2x}}{2}}$$ - (1)
Now, squaring both sides of relation (ii), we get,
$$\begin{array}{rl}& \Rightarrow {\cosh }^{2}2x={\displaystyle \frac{{(e^{2x}+e^{-2x})}^2}{2^2}}\\ & \Rightarrow {\cosh }^{2}2x={\displaystyle \frac{{(e^{2x}+e^{-2x})}^2}{4}}\end{array}$$
Applying the algebraic identity, $${(a+b)}^2=a^2+b^2+2ab$$, we get,
$$\begin{array}{rl}& \Rightarrow {\cosh }^{2}x={\displaystyle \frac{e^{2x}+e^{-2x}+2\times e^x\times e^{-x}}{4}}\\ & \Rightarrow {\cosh }^{2}x={\displaystyle \frac{e^{2x}+e^{-2x}+2}{4}}\end{array}$$
Multiplying both sides with 2, we get,
$$\begin{array}{rl}& \Rightarrow 2{\cosh }^{2}x=2\times {\displaystyle \frac{e^{2x}+e^{-2x}+2}{4}}\\ & \Rightarrow 2{\cosh }^{2}x={\displaystyle \frac{e^{2x}+e^{-2x}+2}{2}}\end{array}$$
Subtracting 1 from both sides, we get,
$$\Rightarrow 2{\cosh }^{2}x-1={\displaystyle \frac{e^{2x}+e^{-2x}+2}{2}}-1$$
Taking L.C.M we get,
$$\Rightarrow 2{\cosh }^{2}x-1={\displaystyle \frac{e^{2x}+e^{-2x}+2-2}{2}}$$
$$\Rightarrow 2{\cosh }^{2}x-1={\displaystyle \frac{e^{2x}+e^{-2x}}{2}}$$ - (2)
Clearly, we can see that the R.H.S of equation (1) and (2) are same, so comparing and equating their L.H.S, we get,
$$\Rightarrow \cosh 2x=2{\cosh }^{2}x-1$$
Hence, proved

Note:
 You must not think that $$\sinh x$$ and $$\cosh x$$ are trigonometric functions with angle (h, x) as it will be a wrong assumption. Here, ‘h’ always denotes hyperbolic function. Now, you may see that the proven statement is analogous to the identity in trigonometry given as: - $$\cos 2x=2{\cos }^{2}x-1$$. So, you may remember the identities of hyperbolic functions as they are used as a formula in topics like limits, integration, differentiation, etc.