Question

If $\sin 27^\circ = p$, then the value of $\sqrt {1 + \sin 36^\circ }$

Answer 1

Here, we will use the trigonometric identities of sum of sine and cosine function and solve it further to get the form of the expression that needs to be found out. We will then simplify the equation further using trigonometric identities to get the required value. Trigonometric equation is defined as an equation involving the trigonometric ratios. Trigonometric identity is an equation which is always true.

Formula Used:
We will use the following formulas:

1. Trigonometric Identity: ${\sin ^2}\theta + {\cos ^2}\theta = 1$
2. Trigonometric Identity: $\sin 2\theta = 2\sin \theta \cos \theta$
3. Trigonometric Identity: $\sin \theta = \sin \left( {90^\circ - x} \right)$
4. Trigonometric Identity: $\sin \left( {90^\circ - x} \right) = \cos x$
5. Trigonometric Identity: $\sin \left( {90^\circ + x} \right) = \cos x$
6. Trigonometric Identity: $\cos x = \sqrt {1 + \sin x}$

Complete step by step solution:
We are given that $\sin 27^\circ = p$.
We know that ${\left( {\sin \theta + \cos \theta } \right)^2} = {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta$.
By substituting the trigonometric identity $\sin 2\theta = 2\sin \theta \cos \theta$ and ${\sin ^2}\theta + {\cos ^2}\theta = 1$, we get
$\Rightarrow {\left( {\sin \theta + \cos \theta } \right)^2} = 1 + \sin 2\theta$
By substituting the values of $\theta = 27^\circ$ , we get
$\Rightarrow {\left( {\sin 27^\circ + \cos 27^\circ } \right)^2} = 1 + \sin \left( {2 \cdot 27} \right)^\circ$
By multiplying the terms, we get
$\Rightarrow {\left( {\sin 27^\circ + \cos 27^\circ } \right)^2} = 1 + \sin 54^\circ$
By using the trigonometric identity $\sin \left( {90^\circ - x} \right) = \cos x$, we get
$\Rightarrow {\left( {\sin 27^\circ + \cos 27^\circ } \right)^2} = 1 + \sin \left( {90^\circ - 36^\circ } \right)$
$\Rightarrow {\left( {\sin 27^\circ + \cos 27^\circ } \right)^2} = 1 + \cos 36^\circ$
We know that $\cos 36^\circ = \dfrac{{1 + \sqrt 5 }}{4}$
$\Rightarrow {\left( {\sin 27^\circ + \cos 27^\circ } \right)^2} = 1 + \dfrac{{1 + \sqrt 5 }}{4}$
By taking the L.C.M, we get
$\Rightarrow {\left( {\sin 27^\circ + \cos 27^\circ } \right)^2} = 1 \times \dfrac{4}{4} + \dfrac{{1 + \sqrt 5 }}{4}$
$\Rightarrow {\left( {\sin 27^\circ + \cos 27^\circ } \right)^2} = \dfrac{{5 + \sqrt 5 }}{4}$
Taking square root on both the sides, we get
$\Rightarrow \left( {\sin 27^\circ + \cos 27^\circ } \right) = \sqrt {\dfrac{{5 + \sqrt 5 }}{4}} = 1 + \cos 36^\circ$ ……………………………………………………….$\left( 1 \right)$
By using the trigonometric identity, we get
$\Rightarrow {\left( {\sin \theta - \cos \theta } \right)^2} = {\sin ^2}\theta + {\cos ^2}\theta - 2\sin \theta \cos \theta$
By substituting the trigonometric identity, we get
$\Rightarrow {\left( {\sin \theta - \cos \theta } \right)^2} = 1 - \sin 2\theta$
By substituting the values of $\theta = 27^\circ$ , we get
$\Rightarrow {\left( {\sin 27^\circ - \cos 27^\circ } \right)^2} = 1 - \sin 2 \cdot 27^\circ$
By multiplying the terms, we get
$\Rightarrow {\left( {\sin 27^\circ - \cos 27^\circ } \right)^2} = 1 - \sin 54^\circ$
By using the trigonometric identity, we get
$\Rightarrow {\left( {\sin 27^\circ - \cos 27^\circ } \right)^2} = 1 - \sin \left( {90^\circ - 36^\circ } \right)$
$\Rightarrow {\left( {\sin 27^\circ - \cos 27^\circ } \right)^2} = 1 - \cos 36^\circ$
We know that $\cos 36^\circ = \dfrac{{1 + \sqrt 5 }}{4}$
$\Rightarrow {\left( {\sin 27^\circ - \cos 27^\circ } \right)^2} = 1 - \dfrac{{1 + \sqrt 5 }}{4}$
$\Rightarrow {\left( {\sin 27^\circ - \cos 27^\circ } \right)^2} = 1 \times \dfrac{4}{4} - \dfrac{{1 + \sqrt 5 }}{4}$
$\Rightarrow {\left( {\sin 27^\circ - \cos 27^\circ } \right)^2} = \dfrac{{3 - \sqrt 5 }}{4}$
Taking square root on both the sides, we get
$\Rightarrow \left( {\sin 27^\circ - \cos 27^\circ } \right) = \sqrt {\dfrac{{3 - \sqrt 5 }}{4}} = 1 - \cos 36^\circ$………………………………………………..$\left( 2 \right)$
By subtracting these equations, we get
$\Rightarrow 1 + \cos 36^\circ - 1 + \cos 36^\circ = \sqrt {\dfrac{{5 + \sqrt 5 }}{4}} - \sqrt {\dfrac{{3 - \sqrt 5 }}{4}}$
$\Rightarrow 2\cos 36^\circ = \dfrac{1}{2}\left[ {\sqrt {5 + \sqrt 5 } - \sqrt {3 - \sqrt 5 } } \right]$
$\Rightarrow \cos 36^\circ = \dfrac{1}{4}\left[ {\sqrt {5 + \sqrt 5 } - \sqrt {3 - \sqrt 5 } } \right]$
By using the trigonometric identity, we get
$\Rightarrow \sqrt {1 + \sin 36^\circ } = \dfrac{1}{4}\left[ {\sqrt {5 + \sqrt 5 } - \sqrt {3 - \sqrt 5 } } \right]$

Therefore, the value of $\sqrt {1 + \sin 36^\circ }$ is $\dfrac{1}{4}\left[ {\sqrt {5 + \sqrt 5 } - \sqrt {3 - \sqrt 5 } } \right]$.

Note:
We should know that we have many trigonometric identities which are related to all the other trigonometric equations. We should note that sine and tangent are odd functions since both the functions are symmetric about the origin. Cosine is an even function because the function is symmetric about the $y$ axis. So, we take the arguments in the negative sign for odd functions and positive signs for even functions.