Question

The value of \[\sin 18^\circ \] is
A). $\dfrac{{\sqrt 5 - 1}}{4}$
B). $\dfrac{{\sqrt 5 + 1}}{4}$
C). $\dfrac{{ - \sqrt 5 + 1}}{4}$
D). \[\dfrac{{1 - \sqrt 5 }}{4}\]

Answer 1

Hint: In this question, remember to take $A = {18^ \circ }$ and multiply both side by 5 and use the trigonometric identities such as $\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta $, using this information we can approach the solution of the problem.

Complete step-by-step solution:
According to the given information, we have to find out the value of $\sin 18^\circ $
Let the$A = {18^ \circ }$, multiply both the sides by 5, we get
Therefore, we can consider, 5A = ${90^ \circ }$
Now we can split the $5A = 3A + 2A$
SO, we have 3A + 2A = ${90^ \circ }$
2A = ${90^ \circ }$ – 3A
Taking sine on both sides, we get
$\sin 2A = \sin \left( {{{90}^ \circ } - 3A} \right) = \cos 3A$
Since we know that $\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta $ and $\cos 3a = 4{\cos ^3}a - 3\cos a$
Therefore $2\sin A\cos A = 4{\cos ^3}A - 3\cos A$
$ \Rightarrow 2\sin A\cos A = 4{\cos ^3}A - 3\cos A $
$ \Rightarrow \cos A\left( {2\sin A - 4{{\cos }^2}A + 3} \right) = 0$
Dividing both the sides by $\cos A = \cos {18^ \circ } \ne 0$, we get
\[ \Rightarrow 2\sin A - 4{\cos ^2}A + 3 = 0\]
Also, we know that \[{\cos ^2}\theta = 1 - {\sin ^2}\theta \]
Therefore $2\sin A - 4\left( {1 - {{\sin }^2}A} \right) + 3 = 0$
$4{\sin ^2}A + 2\sin A - 1 = 0$ which is a quadratic in $\operatorname{Sin} A$
By the quadratic formula i.e. $\dfrac{{ - b \pm \sqrt {{b^2} + 4ac} }}{{2a}}$
$\begin{gathered}
   \Rightarrow \sin A = \dfrac{{ - 2 \pm \sqrt {4 - \left( 4 \right)4.\left( { - 1} \right)} }}{{2 \times 4}} \\
                                     =\dfrac{{ - 2 \pm \sqrt {4 + 16} }}{8} \\
                                      = \dfrac{{ - 2 \pm 2\sqrt 5 }}{8} \\
\end{gathered} $
$\Rightarrow \sin A = \dfrac{{ - 1 \pm \sqrt 5 }}{4} \\ $
Now since ${18^ \circ }$ lies in the first quadrant therefore taking $\sin {18^ \circ }$ positive value
Therefore,$\sin A = \dfrac{{ - 1 + \sqrt 5 }}{4}$
Hence, “A” is the correct option.

Note: The sine, cosine, and tangent of an angle which we used in the above solution are all defined in terms of trigonometry, but they can also be expressed as functions. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature.