Question

If $x$ is any non-zero real number, show that $\cos \theta $ and $\sin \theta $ can never be equal to $x + \dfrac{1}{x}$.

Answer 1

Hint: $x$ is any non-zero real number, it means that the value of $x$ can range from $( - \infty ,\infty ) - \{ 0\} $, that is any value except $0$. As in the question, we are also to deal with $\cos \theta $ and $\sin \theta $ but these values have the domain as real numbers but the range of $\cos \theta $ and $\sin \theta $ is $[ - 1,1]$. So, to show that $\left( {x + \dfrac{1}{x}} \right)$ is not equal to $\cos \theta $ or $\sin \theta $, we have to show that $\left( {x + \dfrac{1}{x}} \right)$ doesn’t lie in the range of $\cos \theta $ or $\sin \theta $.

Complete step by step answer:
Given, $x$ is any non-zero real number. We know, the range of $\cos \theta $ and $\sin \theta $ is $[ - 1,1]$. Let, $f(x) = x + \dfrac{1}{x}$.
Now, when $x > 0,$
$f(x) = x + \dfrac{1}{x}$
Now, x can be written as ${(\sqrt x )^2}$.
Applying this in$f(x)$, we get,
$ \Rightarrow f(x) = {(\sqrt x )^2} + \dfrac{1}{{{{(\sqrt x )}^2}}}$
Subtracting $2$ and adding $2$ in $f(x)$, we get,
$ \Rightarrow f(x) = {(\sqrt x )^2} + \dfrac{1}{{{{(\sqrt x )}^2}}} - 2 + 2$
$ \Rightarrow f(x) = {(\sqrt x )^2} + \dfrac{1}{{{{(\sqrt x )}^2}}} - 2 \times \sqrt x \times \dfrac{1}{{\sqrt x }} + 2$

Now, $f(x)$ can be condensed by using the algebraic identity $\left( {{a^2} + {b^2} - 2ab} \right) = {\left( {a - b} \right)^2}$. So, we get,
$ \Rightarrow f(x) = {\left( {\sqrt x - \dfrac{1}{{\sqrt x }}} \right)^2} + 2$
We know, a perfect square is always greater than or equal to zero.
That means, ${\left( {\sqrt x - \dfrac{1}{{\sqrt x }}} \right)^2} \geqslant 0$
That is, $f{(x)_{\min }} = 0 + 2 = 2$.
Therefore, $f(x)$ is always greater than or equal to $2$, which is not in the range of $\cos \theta $ or $\sin \theta $. So, when $x > 0$, $\cos \theta $ or $\sin \theta $ cannot be equal to $\left( {x + \dfrac{1}{x}} \right)$.

Now, when$x < 0$ we can write it as $y = - x$, where $y > 0$,
So, $f(x)$ can be written as,
$f(x) = x + \dfrac{1}{x}$
$ \Rightarrow f(y) = ( - y) + \dfrac{1}{{( - y)}}$
Taking $ - 1$ common from right hand side,
$f(y) = - \left( {y + \dfrac{1}{y}} \right)$
We can write $y$ as ${\left( {\sqrt y } \right)^2}$,
$f(y) = - \left[ {{{(\sqrt y )}^2} + \dfrac{1}{{{{\left( {\sqrt y } \right)}^2}}}} \right]$
Subtracting $2$ and adding $2$ in $f(y)$, we get,
$f(y) = - \left[ {{{(\sqrt y )}^2} + \dfrac{1}{{{{\left( {\sqrt y } \right)}^2}}} - 2 + 2} \right]$
$ \Rightarrow f(y) = - \left[ {{{(\sqrt y )}^2} + \dfrac{1}{{{{\left( {\sqrt y } \right)}^2}}} - 2 \times \sqrt y \times \dfrac{1}{{\sqrt y }} + 2} \right]$

Now, we condense the expression using the algebraic identity $\left( {{a^2} + {b^2} - 2ab} \right) = {\left( {a - b} \right)^2}$, we get,
$ \Rightarrow f(y) = - \left[ {{{\left( {\sqrt y - \dfrac{1}{{\sqrt y }}} \right)}^2} + 2} \right]$
We know, a perfect square is always greater than or equal to zero.
That means, ${\left( {\sqrt y - \dfrac{1}{{\sqrt y }}} \right)^2} \geqslant 0$
Therefore, $f{(y)_{\max }} = - [0 + 2] = - 2$.
Now, we know that the maximum value of $f(x) = x + \dfrac{1}{x}$ for $x < 0$ is $ - 2$.
Therefore, $f(x)$ is always $ - 2$ or less than $ - 2$, which is not in the range of $\cos \theta $ or $\sin \theta $. So, when $x < 0$, $\cos \theta $ or $\sin \theta $ cannot be equal to $\left( {x + \dfrac{1}{x}} \right)$.

Hence, it can be concluded that, $\cos \theta $ or $\sin \theta $ can never be equal to $\left( {x + \dfrac{1}{x}} \right)$, where $x$ is a non-zero real number.

Note:Every trigonometric function has their own domain and range within which they lie. For any function to lie in the range of a trigonometric function, the range of the function must coincide with the range of the trigonometric function. We must know the range of sine and cosine function to tackle the problem. We also must remember the algebraic whole square formulae of binomials in order to find the range of the function $\left( {x + \dfrac{1}{x}} \right)$.