Question

How do you evaluate \[\sin \left( {\dfrac{\pi }{5}} \right)\].

Answer 1

Hint: In this question, we have a trigonometric expression, which values find out. First, we simplify this expression in the form of a double angle and triple angle (trigonometric formula). Then we use the double angle and triple angle formula and solve it. After that, we convert them into quadratic equations. Then we find the roots of the quadratic equation. After that, we find the value of the given trigonometric expression.

Complete step by step solution:
In this question, we use the words double angle, triple angle. First, we know about the double angle formula.
Double angle formula: the double angle is called because they have a double angle of trigonometric functions like \[\sin 2\theta,\;\cos 2\theta \], etc.
The double angle formula is written as below in the form of\[\sin \].
\[ \Rightarrow \sin 2\theta = 2\sin \theta \cos \theta \]
Now we know about the triple angle formula.
Triple angle formula: the triple angle is called because they have a triple angle of trigonometric functions like \[\sin 3\theta,\;\cos 3\theta \], etc. the triple angle formula is written as below in the form of\[\sin \].
\[ \Rightarrow \sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta \]
Now we come to the question. In the question, the trigonometric expression is given as below.
\[ \Rightarrow \sin \left( {\dfrac{\pi }{5}} \right)\]
Let we assume that,
\[ \Rightarrow \theta = \dfrac{\pi }{5}\]
Then,
\[5\theta = \pi \]
It can be written as below.
\[ \Rightarrow 3\theta = \pi - 2\theta \]
We write both sides of the trigonometric function\[\sin \].
Then,
\[ \Rightarrow \sin 3\theta = \sin \left( {\pi - 2\theta } \right)\]
We know that\[\sin \left( {\pi - \theta } \right) = \sin \theta \].
Then, the above expression is written as.
\[\sin 3\theta = \sin 2\theta \]
Now we use the double angle and triple angle formula.
Then,
\[ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta = 2\sin \theta \cos \theta \]
We divided both sides with\[\sin \theta \].
Then,
\[ \Rightarrow 3 - 4{\sin ^2}\theta = 2\cos \theta \]
We know that\[{\sin ^2}\theta = 1 - {\cos ^2}\theta \].
Then the above expression is written as.
\[ \Rightarrow 3 - 4\left( {1 - {{\cos }^2}\theta } \right) = 2\cos \theta \]
Now we solve the above expression.
\[3 - 4 + 4{\cos ^2}\theta = 2\cos \theta \]
\[ \Rightarrow - 1 + 4{\cos ^2}\theta - 2\cos \theta = 0\]
We arrange the above expression as.
\[ \Rightarrow 4{\cos ^2}\theta - 2\cos \theta - 1 = 0\]
It is a quadratic equation. Hence the root of the quadratic equation is.
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Then from the above quadratic equation, the root is
\[\cos \theta = \dfrac{{2 \pm \sqrt {{2^2} - 4 \times 4 \times \left( { - 1} \right)} }}{{2 \times 4}}\]
\[ \Rightarrow \cos \theta = \dfrac{{2 \pm \sqrt {4 + 16} }}{8}\]
\[ \Rightarrow \cos \theta = \dfrac{{1 \pm \sqrt 5 }}{4}\]
We know that \[\cos \theta \] cannot take negative value, then
\[ \Rightarrow \cos \theta = \dfrac{{1 + \sqrt 5 }}{4}\]
Then we find the value of\[\sin \theta \].
\[ \Rightarrow \sin \theta = \sqrt {1 - {{\cos }^2}\theta } \]
Put the \[\cos \theta \]from above.
Then,
\[ \Rightarrow \sin \theta = \sqrt {1 - {{\left( {\dfrac{{1 + \sqrt 5 }}{4}} \right)}^2}} \]
\[ \Rightarrow \sin \theta = \sqrt {1 - \left( {\dfrac{{1 + 5 + 2\sqrt 5 }}{{16}}} \right)} \]
\[\therefore \sin \theta = \dfrac{{\sqrt {10 - 2\sqrt 5 } }}{4}\]
Put the value of \[\theta \] in the above.
Then,
\[\sin \dfrac{\pi }{5} = \dfrac{{\sqrt {10 - 2\sqrt 5 } }}{4}\]

Therefore, the value of \[\sin \dfrac{\pi }{5}\] is \[\dfrac{{\sqrt {10 - 2\sqrt 5 } }}{4}\].

Note:
If you have a trigonometric function. And you want to find the value of that function. Then you can convert that function into double angle triple angle format. After applying the double angle and triple angle formula you can find the value of a trigonometric function.