Question

How do you find the exact value of \[\sin ({18^ \circ })\]?

Answer 1

Hint:In the above question, is based on the concept of trigonometry. The sine, cosine, tangent functions can be solved by using the multiple angle formula which is used inside trigonometric functions. By applying the formula of multiple angles on sine function we can get the exact value.

Complete step by step solution:
Trigonometric function means the function of the angle between the two sides. It tells us the relation between the angles and sides of the right-angle triangle. The trigonometric function having multiple angles is the multiple angle formula.

Double and triple angles formulas are there under the multiple angle formulas. Generally, it is written in the form \[\sin (nx)\] where n is a positive integer.

The sign of all the six trigonometric functions in the first quadrant is positive since x and y coordinates are both positive. In the second quadrant sine and cosecant are positive and in third only tangent and cotangent are positive. The fourth quadrant has only cosine and secant are positive. In the question above we need to evaluate the sine function.

So, Let \[A = {18^ \circ }\]

Therefore,\[5A = {90^ \circ }\]

Further bringing into 2A term,
\[2A = {90^ \circ } - 3A\]

Taking sine on both sides
\[\sin 2A = \sin (90 - 3A) = \cos 3A\] ( \[\because \sin ({90^ \circ } - A) = \cos A\])

By applying cos3A formula,
\[
\Rightarrow 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 = 0 \\

\Rightarrow \cos A(2\sin A - 4{\cos ^2}A + 3) = 0 \\
\]
But \[\cos A = \cos {18^ \circ } \ne 0\],so solving the second part,
\[
2\sin A - 4(1 - {\sin ^2}A) + 3 = 0 \\
4{\sin ^2}A + 2\sin A - 1 = 0 \\
\]
Since, we get quadratic equation by applying formula we get,
\[
x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\sin A = \dfrac{{ - 2 \pm \sqrt {4 + 16} }}{{2(4)}} \\
\sin A = \dfrac{{ - 1 \pm \sqrt 5 }}{4} \\
\]

Therefore, we get the above exact value.

Note: An important thing to note is that \[\sin (90 - 5A)\] which is \[\sin {18^ \circ }\], it lies in the first quadrant. Since in the first quadrant all the trigonometric functions are positive including sine function hence, we get the above value as positive.