Question

Using the formula \[\sin (A - B) = \sin A \cdot \cos B - \cos A \cdot \sin B\] find the value of $\sin {15^ \circ }$
A) \[\sqrt 3 \]
B) $\sqrt 3 + 1$
C) $\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
D) $\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$

Answer 1

Hint: Generally we know the values of $\sin $, $\cos $and $\tan $ of ${0^ \circ }$, ${30^ \circ }$, ${45^ \circ }$, ${60^ \circ }$ and ${90^ \circ }$. To obtain the values in between then use the formula of $\sin (A - B)$, $\sin (A + B)$, $\cos (A - B)$ and $\cos (A + B)$. Here we have to find the value of ${15^ \circ }$ and we know that ${15^ \circ } = {45^ \circ } - {30^ \circ } = {60^ \circ } - {45^ \circ }$, so use any one combination and use given formula \[\sin (A - B) = \sin A \cdot \cos B - \cos A \cdot \sin B\]. Put the standard values, simplify the equation to obtain the value of $\sin {15^ \circ }$.

Complete step-by-step answer:
In this case we have to find the value of $\sin {15^ \circ }$.
Generally we know the values of $\sin $, $\cos $and $\tan $ of ${0^ \circ }$, ${30^ \circ }$, ${45^ \circ }$, ${60^ \circ }$ and ${90^ \circ }$. To obtain the values in between then use the formula for $\sin (A - B)$, $\sin (A + B)$, $\cos (A - B)$ ad $\cos (A + B)$.
As we know that ${15^ \circ } = {45^ \circ } - {30^ \circ } = {60^ \circ } - {45^ \circ }$, so use any one combination to obtain value of $\sin {15^ \circ }$.
Let’s take ${15^ \circ } = {60^ \circ } - {45^ \circ }$. However you can take ${15^ \circ } = {45^ \circ } - {30^ \circ }$ and solve the equation similarly as solved below. You will get the same answer.
Here, we can say that the value of A and B as $A = {60^ \circ }$ and $B = {45^ \circ }$.
Using the formula of sin of difference of angle, \[\sin (A - B) = \sin A \cdot \cos B - \cos A \cdot \sin B\]
Put value of $A = {60^ \circ }$ and $B = {45^ \circ }$ in above equation,
So, \[\sin ({60^ \circ } - {45^ \circ }) = \sin {60^ \circ } \cdot \cos {45^ \circ } - \cos {60^ \circ } \cdot \sin {45^ \circ }\],
So, \[\sin ({15^ \circ }) = \sin {60^ \circ } \cdot \cos {45^ \circ } - \cos {60^ \circ } \cdot \sin {45^ \circ }\].
As we know that $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, $\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$, $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ and $\cos {60^ \circ } = \dfrac{1}{2}$,
So putting the values in \[\sin ({15^ \circ }) = \sin {60^ \circ } \cdot \cos {45^ \circ } - \cos {60^ \circ } \cdot \sin {45^ \circ }\] equation,
\[\sin ({15^ \circ }) = \dfrac{{\sqrt 3 }}{2} \cdot \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2} \cdot \dfrac{1}{{\sqrt 2 }}\]
Taking $\dfrac{1}{{\sqrt 2 }}$ term common in right of the equation,

So, \[\sin ({15^ \circ }) = \dfrac{1}{{\sqrt 2 }} \cdot (\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2})\]
Simplifying, \[\sin ({15^ \circ }) = \dfrac{1}{{\sqrt 2 }} \cdot (\dfrac{{\sqrt 3 - 1}}{2})\]
So, \[\sin ({15^ \circ }) = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\]
So the value of $\sin {15^ \circ }$ is $\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$.

So, Option (C) is the correct answer.

Note: In similar type questions we can find the value of $\sin {75^ \circ }$, $\cos {15^ \circ }$, $\cos {75^ \circ }$ and so on. As ${75^ \circ } = {30^ \circ } + {45^ \circ }$ we can use similar type formula to obtain value of $\sin {75^ \circ }$ or $\cos {75^ \circ }$. There are some equations for solving such type of questions, like \[\sin (A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B\], \[\cos (A + B) = \cos A \cdot \cos B - \sin A \cdot \sin B\] and \[\cos (A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B\].