Question

How do you find the exact values of $\sin 15$ degrees using the half-angle formula?

Answer 1

Here they have asked to find the exact value of a sine function by using half angle formula. We can make use of the half angle formula given by: $\sin \left( {\dfrac{x}{2}} \right) = \sqrt {\dfrac{{1 - \cos x}}{2}}$ . As the $15$ degrees is half of $30$ degrees, by putting $x = {30^ \circ }$ in the given half angle formula we can simplify the expression for the required answer.

Complete step by step answer:
In the given question they have asked for the exact value of sine function using half angle identities. So for that we can make use of the half angle formula for sine function given by: $\sin \left( {\dfrac{x}{2}} \right) = \sqrt {\dfrac{{1 - \cos x}}{2}}$.
As the $15$ degrees is half of $30$ degrees, by putting $x = {30^ \circ }$ in the given half-angle formula we can simplify the expression.
Therefore, we get
$\sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \sqrt {\dfrac{{1 - \cos {{30}^ \circ }}}{2}}$
From trigonometric ratio for standard functions, we know that $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$ , now by substituting this value in the above expression, we get
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \sqrt {\dfrac{{1 - \dfrac{{\sqrt 3 }}{2}}}{2}}$
Taking L.C.M. for the right hand side numerator, we get
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \sqrt {\dfrac{{\dfrac{{2 - \sqrt 3 }}{2}}}{2}}$
On simplification,
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \sqrt {\dfrac{{2 - \sqrt 3 }}{4}}$
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \dfrac{{\sqrt {2 - \sqrt 3 } }}{2}$
In the above step we have nested root which is root inside root, we can keep this as the answer, or else if you want to simplify further we can do as follows.
To remove nested root, multiply both the numerator and the denominator inside the square root by $2$, we get
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \sqrt {\dfrac{{2 - \sqrt 3 }}{4}.\dfrac{2}{2}}$
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \sqrt {\dfrac{{4 - 2\sqrt 3 }}{8}}$
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \sqrt {\dfrac{{4 - 2\sqrt 3 }}{8}}$
The above step can be written as
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \sqrt {\dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{{2^2}.\sqrt 2 }}}$
Square and the square root get cancel each other in the above step, we get
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \left( {\dfrac{{\left( {\sqrt 3 - 1} \right)}}{2}} \right).\dfrac{1}{{\sqrt 2 }}$
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \dfrac{{\left( {\sqrt 3 - 1} \right)}}{{2\sqrt 2 }}$
Now by rationalizing the denominator, that is multiply both numerators and denominator by $\sqrt 2$, we get
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \dfrac{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{4}$
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \dfrac{{\sqrt 6 - \sqrt 2 }}{4}$ , this is the required answer without the nested root.
If we want to keep this answer in a decimal form we can easily calculate using the calculator and find the value in decimal form.
$\Rightarrow \sin \left( {\dfrac{{{{30}^ \circ }}}{2}} \right) = \dfrac{{\sqrt 6 - \sqrt 2 }}{4} = 0.258$.

Note: This problem can also be solved in a simple way by using the identity $\sin (A - B) = \sin A\cos B - \sin B\cos A$ . Now, $\sin 15$ degrees can be written as $\sin (45 - 30)$ degrees.
Therefore now substituting this in the given formula we can have $\sin (45 - 30) = \sin 45\cos 30 - \sin 30\cos 45$ , now by making use of standard value of the functions, we get $\Rightarrow \sin (45 - 30) = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }}\dfrac{1}{2}$.
On simplification,
$\Rightarrow \sin (45 - 30) = \dfrac{{\sqrt 6 - \sqrt 2 }}{4} = 0.258$.