Question

The value of $\cos {{9}^{0}}-\sin {{9}^{0}}$ is:
(a) $-\dfrac{\sqrt{5-\sqrt{5}}}{2}$
(b) $\dfrac{5+\sqrt{5}}{4}$
(c) $\dfrac{1}{2}\sqrt{5-\sqrt{5}}$
(d) None of these

Answer 1

Hint:First of all multiply and divide the given expression by $\sqrt{2}$ then the expression will look like $\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\cos {{9}^{0}}-\dfrac{1}{\sqrt{2}}\sin {{9}^{0}} \right)$.Now, in this expression we can write $\dfrac{1}{\sqrt{2}}$ as $\cos {{45}^{0}}$ or $\sin {{45}^{0}}$ so we can write the expression as $\sqrt{2}\left( \sin {{45}^{0}}\cos {{9}^{0}}-\cos {{45}^{0}}\sin {{9}^{0}} \right)$ which can be further written as $\sqrt{2}\left( \sin \left( {{45}^{0}}-{{9}^{0}} \right) \right)$ which is equal to $\sqrt{2}\sin {{36}^{0}}$ then put the value of $\sin {{36}^{0}}$.

Complete step-by-step answer:
The expression that we have to evaluate is:
$\cos {{9}^{0}}-\sin {{9}^{0}}$
Multiplying and dividing the above expression by $\sqrt{2}$ will give us the following expression.
$\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\cos {{9}^{0}}-\dfrac{1}{\sqrt{2}}\sin {{9}^{0}} \right)$
We know from the value of trigonometric ratios that the value of $\sin {{45}^{0}}\And \cos {{45}^{0}}$ is equal to $\dfrac{1}{\sqrt{2}}$ so substituting the value of $\dfrac{1}{\sqrt{2}}$ in the above expression we get,
$\sqrt{2}\left( \sin {{45}^{0}}\cos {{9}^{0}}-\cos {{45}^{0}}\sin {{9}^{0}} \right)$ …………Eq. (1)
If you can see carefully the above expression, you recall that the above expression apart from $\sqrt{2}$ is the expansion of $\sin \left( {{45}^{0}}-{{9}^{0}} \right)$ which is equal to $\sin {{45}^{0}}\cos {{9}^{0}}-\cos {{45}^{0}}\sin {{9}^{0}}$ so writing this expression in as $\sin \left( {{45}^{0}}-{{9}^{0}} \right)$ in the above we get,
$\sqrt{2}\sin \left( {{45}^{0}}-{{9}^{0}} \right)$
Simplifying the above expression we get,
$\sqrt{2}\sin {{36}^{0}}$
We know from the trigonometric ratios that the value of $\sin {{36}^{0}}$ is equal to $\dfrac{\sqrt{10-2\sqrt{5}}}{4}$ so substituting this value of $\sin {{36}^{0}}$ in the above expression we get,
$\sqrt{2}\left( \dfrac{\sqrt{10-2\sqrt{5}}}{4} \right)$
Now, taking $\sqrt{2}$ as common from the expression in the under root we get,
$\begin{align}
  & \sqrt{2}\left( \dfrac{\sqrt{2\left( 5 \right)-2\sqrt{5}}}{4} \right) \\
 & =\sqrt{2}\left( \sqrt{2} \right)\left( \dfrac{\sqrt{5-\sqrt{5}}}{4} \right) \\
 & =2\left( \dfrac{\sqrt{5-\sqrt{5}}}{4} \right) \\
 & =\dfrac{\sqrt{5-\sqrt{5}}}{2} \\
\end{align}$
From the above solution, the evaluation of the given expression is $\dfrac{\sqrt{5-\sqrt{5}}}{2}$.
Hence, the correct option is (c).

Note: The alternate way of solving the above problem is to rewrite the eq. (1) in the above solution as:
$\sqrt{2}\left( \sin {{45}^{0}}\cos {{9}^{0}}-\cos {{45}^{0}}\sin {{9}^{0}} \right)$
In place of this expression we can write the above expression as:
$\sqrt{2}\left( \cos {{45}^{0}}\cos {{9}^{0}}-\sin {{45}^{0}}\sin {{9}^{0}} \right)$
Now, the above trigonometric expression i.e. $\cos {{45}^{0}}\cos {{9}^{0}}-\sin {{45}^{0}}\sin {{9}^{0}}$ is the expansion of $\cos \left( {{45}^{0}}+{{9}^{0}} \right)$ so we can substitute this value of expansion in the above expression.
$\sqrt{2}\cos \left( {{45}^{0}}+{{9}^{0}} \right)$
Simplifying the above expression we get,
$\sqrt{2}\cos {{54}^{0}}$
We know from the trigonometric ratios that the value of $\cos {{54}^{0}}$ is equal to $\dfrac{\sqrt{10-2\sqrt{5}}}{4}$ we get,
$\begin{align}
  & \sqrt{2}\left( \dfrac{\sqrt{10-2\sqrt{5}}}{4} \right) \\
 & =\dfrac{\sqrt{5-\sqrt{5}}}{2} \\
\end{align}$
Hence, we have got the same value of the expression in the given question as we have solved above.We should remember the formula of $\sin(A-B)=\sin A \cos B-\cos A \sin B$ and $\cos(A+B)=\cos A \cos B-\sin A \sin B$.Also we have to memorize the trigonometric angle value of $\sin {{36}^{0}}$ is $\dfrac{\sqrt{10-2\sqrt{5}}}{4}$ and $\cos {{54}^{0}}$ is $\dfrac{\sqrt{10-2\sqrt{5}}}{4}$