Question

The value of $sin{15^ \circ }$ is
A) $\dfrac{{\sqrt 3 + 1}}{{\sqrt 2 }}$
B) $\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
C) $\dfrac{{\sqrt 3 + 1}}{2}$
D) $\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$

Answer 1

Hint:
We can expand $\sin 15^\circ $ as $\sin \left( {45^\circ - 30^\circ } \right)$ . Then we can simplify it using the identity, $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . Then we can substitute this in the given expression. We can then give the values for trigonometric functions \[\cos 45^\circ ,\sin 45^\circ ,\cos 30^\circ \] and \[\sin 30^\circ \] . After further simplification, we will get the required solution. Then we can check the options and find the correct option.

Complete step by step solution:
We need to find the value of $\sin 15^\circ $
We can write 15 as 15 = 45 – 30. So, $\sin 15^\circ $ will become
$ \Rightarrow \sin 15^\circ = \sin \left( {45^\circ - 30^\circ } \right)$
 We know that, $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . On applying this identity, we get
$ \Rightarrow \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ $ .
We know that \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\] , \[\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}\] ,\[\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}\] and \[\sin 30^\circ = \dfrac{1}{2}\] . On substituting these in the above equation, we get
$ \Rightarrow \sin 15^\circ = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}$
On simplification, we get
$ \Rightarrow \sin 15^\circ = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} - \dfrac{1}{{2\sqrt 2 }}$
As the denominators are equal, we can add the numerators.
$ \Rightarrow \sin 15^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
Therefore, the value of $\sin 15^\circ $ is $\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$ .

So, the correct answer is option D.

Note:
Alternate method to solve this problem is given by,
 We need to find the value of $\sin 15^\circ $
We can write 15 as 15 = 60 – 45. So, $\sin 15^\circ $will become
$ \Rightarrow \sin 15^\circ = \sin \left( {60^\circ - 45^\circ } \right)$
 We know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . On applying this identity, we get
$ \Rightarrow \sin 15^\circ = \sin 60^\circ \cos 45^\circ - \cos 60^\circ \sin 45^\circ $.
We know that \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\] , \[\cos 60^\circ = \dfrac{1}{2}\] , \[\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}\] and \[\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}\] . On substituting these in the above equation, we get
$ \Rightarrow \sin 15^\circ = \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }}$
On simplification, we get
$ \Rightarrow \sin 15^\circ = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} - \dfrac{1}{{2\sqrt 2 }}$
As the denominators are equal, we can add the numerators.
$ \Rightarrow \sin 15^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
Therefore, the value of $\sin 15^\circ $ is $\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$ .
So, the correct answer is option D.
We must take care of the order while expanding the trigonometric function of the difference of the angle. We must know the values of trigonometric function at basic angles such as 0, 30, 45, 60 and 90 degrees.