Question

If $f(x) = \cosh x - \sinh x$, then $f({x_1} + {x_2} + ... + {x_n})$ is equal to
$A)f({x_1}).f({x_2})...f({x_n})$
$B)f({x_1}) + f({x_2}) + ... + f({x_n})$
$C)0$
$D)1$

Answer 1

Hint: First, we need to analyze the given which is in the trigonometric form. We can equate the given expression into some form of exponent and then we will be using the power rule. The trigonometric functions are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Since from the given that we have two trigonometric relation $f(x) = \cosh x - \sinh x$
Formula used:
\[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}\]
\[\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}\]

Complete step-by-step solution:
Since from the given that we have $f(x) = \cosh x - \sinh x$ and we have to find the function $f({x_1} + {x_2} + ... + {x_n})$.
By the trigonometric functions, we have the formula for the \[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}\] and \[\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}\]
Substituting this in the above function we have, \[f(x) = \cosh x - \sinh x \Rightarrow \dfrac{{{e^x} + {e^{ - x}}}}{2} - (\dfrac{{{e^x} - {e^{ - x}}}}{2})\]
Further solving we get \[f(x) = \dfrac{{{e^x} + {e^{ - x}}}}{2} - (\dfrac{{{e^x} - {e^{ - x}}}}{2}) \Rightarrow \dfrac{{{e^x} + {e^{ - x}} - {e^x} + {e^{ - x}}}}{2}\]
Cancel out the common terms, we get \[f(x) = \dfrac{{{e^x} + {e^{ - x}} - {e^x} + {e^{ - x}}}}{2} \Rightarrow \dfrac{{2{e^{ - x}}}}{2} \Rightarrow {e^{ - x}}\]
Thus, we have \[f(x) = {e^{ - x}}\]
Since \[f(x) = {e^{ - x}}\]now put $x = {x_1}$ then we get \[f({x_1}) = {e^{ - {x_1}}}\]
Similarly, put $x = {x_2}$ then we get \[f({x_2}) = {e^{ - {x_2}}}\] and further proceeding we may put $x = {x_n}$ then we get \[f({x_n}) = {e^{ - {x_n}}}\]
Now adding every function, we get the resultant as $f({x_1} + {x_2} + ... + {x_n}) = {e^{ - ({x_1} + {x_2} + ... + {x_n})}}$
Which can be expressed in the form of $f({x_1} + {x_2} + ... + {x_n}) = {e^{ - ({x_1} + {x_2} + ... + {x_n})}} \Rightarrow {e^{ - {x_1}}}.{e^{ - {x_2}}}....{e^{ - {x_n}}}$
Hence by the use of above-found values, we get $f({x_1} + {x_2} + ... + {x_n}) = {e^{ - ({x_1} + {x_2} + ... + {x_n})}} \Rightarrow {e^{ - {x_1}}}.{e^{ - {x_2}}}....{e^{ - {x_n}}} \Rightarrow f({x_1}).f({x_2})...f({x_n})$ where \[f({x_1}) = {e^{ - {x_1}}}\] \[f({x_2}) = {e^{ - {x_2}}}\] and similarly \[f({x_n}) = {e^{ - {x_n}}}\]
Therefore, we get the option $A)f({x_1}).f({x_2})...f({x_n})$ is correct.

Note: We will use the basic concept of the power rule, which is ${e^{a + b}} = {e^a}{e^b}$ (we used this concept in the above solution)
Also, in the power rule, the inverse function can be expressed as ${e^{ - 1}} = \dfrac{1}{e}$
We also use simple algebra to simplify the expression that we will get after the power rule expansion.
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan $and $\tan = \dfrac{1}{{\cot }}$