Question

How do you prove $\sinh x+\cosh x={{e}^{x}}$ ?

Answer 1

Hint: In this question we have trigonometric functions in the left-hand side, which we have to prove to the exponential term in the right-hand side. We will consider the left hand-side first and then we will substitute the value of the trigonometric terms in terms of exponential terms. We will substitute the value of $\sinh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$ and the value of $\cosh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$ in the expression and simplify it to prove the expression.

Complete step by step solution:
We have the expression as $\sinh x+\cosh x={{e}^{x}}$.
Consider the left-hand side of the expression:
$\Rightarrow \sinh x+\cosh x$
On substituting the value of $\sinh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$ and the value of $\cosh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$ in the expression, we get:
$\Rightarrow \dfrac{{{e}^{x}}+{{e}^{-x}}}{2}+\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$
Since we have the same denominator on both the fractions, we can take the lowest common multiple and write it as:
$\Rightarrow \dfrac{{{e}^{x}}+{{e}^{-x}}+{{e}^{x}}-{{e}^{-x}}}{2}$
Now the same term with opposite signs, cancel out each other, we can write the expression as:
$\Rightarrow \dfrac{{{e}^{x}}+{{e}^{x}}}{2}$
On simplifying the terms in the numerator, we get:
$\Rightarrow \dfrac{2{{e}^{x}}}{2}$
On simplifying, we get:
$\Rightarrow {{e}^{x}}$, which is the right-hand side, hence proved.

Note: In this question we are using the hyperbolic trigonometric functions and not the general trigonometric functions. It is to be remembered that the graphs of hyperbolic functions are different from the traditional trigonometric functions. In this we have the term ${{e}^{-x}}$ which is in the form of a negative exponent. It can also be written in the reciprocal format as $\dfrac{1}{{{e}^{x}}}$. The symbol $e$ is the Euler’s number which is a recurring decimal number which has the value as $2.7182...$