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 \[\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{1}\] 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 \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] to simplify \[{{\cosh }^{2}}x\]. Check if the obtained expressions of \[\cosh 2x\] and \[\left( 2{{\cosh }^{2}}x-1 \right)\] 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 \[\left( \cos t,\sin t \right)\] form a circle with a unit radius, the points \[\left( \cosh t,\sinh t \right)\] form the right half of the equilateral hyperbola. These hyperbolic functions are written in exponential form as: -
(i) \[\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}\]
(ii) \[\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}\]
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,
\[\Rightarrow \cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}\]
Replacing ‘x’ with ‘2x’ in the above relation, we have,
\[\Rightarrow \cosh 2x=\dfrac{{{e}^{2x}}+{{e}^{-2x}}}{2}\] - (1)
Now, squaring both sides of relation (ii), we get,
\[\begin{align}
  & \Rightarrow {{\cosh }^{2}}2x=\dfrac{{{\left( {{e}^{2x}}+{{e}^{-2x}} \right)}^{2}}}{{{2}^{2}}} \\
 & \Rightarrow {{\cosh }^{2}}2x=\dfrac{{{\left( {{e}^{2x}}+{{e}^{-2x}} \right)}^{2}}}{4} \\
\end{align}\]
Applying the algebraic identity, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\], we get,
\[\begin{align}
  & \Rightarrow {{\cosh }^{2}}x=\dfrac{{{e}^{2x}}+{{e}^{-2x}}+2\times {{e}^{x}}\times {{e}^{-x}}}{4} \\
 & \Rightarrow {{\cosh }^{2}}x=\dfrac{{{e}^{2x}}+{{e}^{-2x}}+2}{4} \\
\end{align}\]
Multiplying both sides with 2, we get,
\[\begin{align}
  & \Rightarrow 2{{\cosh }^{2}}x=2\times \dfrac{{{e}^{2x}}+{{e}^{-2x}}+2}{4} \\
 & \Rightarrow 2{{\cosh }^{2}}x=\dfrac{{{e}^{2x}}+{{e}^{-2x}}+2}{2} \\
\end{align}\]
Subtracting 1 from both sides, we get,
\[\Rightarrow 2{{\cosh }^{2}}x-1=\dfrac{{{e}^{2x}}+{{e}^{-2x}}+2}{2}-1\]
Taking L.C.M we get,
\[\Rightarrow 2{{\cosh }^{2}}x-1=\dfrac{{{e}^{2x}}+{{e}^{-2x}}+2-2}{2}\]
\[\Rightarrow 2{{\cosh }^{2}}x-1=\dfrac{{{e}^{2x}}+{{e}^{-2x}}}{2}\] - (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.