Question

Evaluate the values of $\sin {15^ \circ }$ and $\cos {15^ \circ }$.

Answer 1

Hint:For the value like $\sin {15^ \circ }$, $\cos {15^ \circ }$, $\sin {75^ \circ }$ and $\cos {75^ \circ }$ we need to split the angle into the binomial terms of angles for which the values of trigonometric terms are known. In this problem we need to split \[{15^ \circ }\] into ${45^ \circ }$ and ${30^ \circ }$.

Complete step-by-step answer:
We are asked to find out the values of $\sin {15^ \circ }$ and $\cos {15^ \circ }$. So, firstly let us find the value of and $\sin {15^ \circ }$later on $\cos {15^ \circ }$.
Firstly, split \[{15^ \circ }\] into ${45^ \circ }$ and ${30^ \circ }$.
\[ \Rightarrow \sin \left( {{{15}^ \circ }} \right) = \sin \left( {{{45}^ \circ } - {{30}^ \circ }} \right)\]
Here, we need to use the most popular trigonometric formula to expand for further simplification
$\sin \left( {A - B} \right) = \sin A\cos B - \sin B\cos A$
Substituting the $A$ value as ${45^ \circ }$ and $B$ value as ${30^ \circ }$. We get,
$ \Rightarrow \sin {45^ \circ }\cos {30^ \circ } - \sin {30^ \circ }\cos {45^ \circ }$
And we know that $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, $\sin {30^ \circ } = \dfrac{1}{2}$, $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ and $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Substituting the above values, we will get as follows:
$
   \Rightarrow \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) - \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
   = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} - \dfrac{1}{{2\sqrt 2 }} \\
   = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} \\
 $
Hence, we get $\sin {15^ \circ } = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
Now, let’s find out $\cos {15^ \circ }$ in a similar way.
Firstly, split \[{15^ \circ }\] into ${45^ \circ }$ and ${30^ \circ }$.
\[ \Rightarrow \cos \left( {{{15}^ \circ }} \right) = \cos \left( {{{45}^ \circ } - {{30}^ \circ }} \right)\]
Here, we need to use the most popular trigonometric formula to expand for further simplification
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
Substituting the $A$value as ${45^ \circ }$ and $B$ value as ${30^ \circ }$. We get,
$ \Rightarrow \cos {45^ \circ }\cos {30^ \circ } - \sin {45^ \circ }\sin {30^ \circ }$
And we know that $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, $\sin {30^ \circ } = \dfrac{1}{2}$, $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ and $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Substituting the above values, we will get as follows:
$
   \Rightarrow \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{1}{2}} \right) - \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) \\
   = \dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} \\
   = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }} \\
 $
Hence, we get $\cos {15^ \circ } = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$

Note:This problem can be solved in other ways also. We know the formula $\cos 2\theta = 1 - 2{\sin ^2}\theta $ and $\cos 2\theta = 2{\cos ^2}\theta - 1$. By substituting the $\theta = {15^ \circ }$ in the above equations, we can get the same values.