Evaluate $\cos {{15}^{\circ }}$ using the formula: $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$

Hint: Put $A={{60}^{\circ }}$ and $B={{45}^{\circ }}$in the given expression so that difference of A and B is ${{15}^{\circ }}$ and get value of $\cos {{15}^{\circ }}$. Use the values of $\cos {{60}^{\circ }},\cos {{45}^{\circ }},sin{{60}^{\circ }},sin{{45}^{\circ }}$ which are given as

$\begin{align}
  & \cos {{60}^{\circ }}=\dfrac{1}{2},\cos 45=\sin 45=\dfrac{1}{\sqrt{2}} \\
 & sin{{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\
\end{align}$

Complete step-by-step answer:
Here, we have given a formula $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ and hence, we need to determine the value of $\cos {{15}^{\circ }}$ using the above identity. So we have,

$\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ …………………… (i)

Now, we know that number 15 can be written in the form of a difference of two numbers i.e.

 (60 – 45). So, we can put values of A and B as ${{60}^{\circ }}$ and ${{50}^{\circ

}}$respectively.

So, putting $A={{60}^{\circ }}$ and $B={{45}^{\circ }}$ in equation (i); we get

$\cos \left( 60-45 \right)=\cos 60\cos 45+sin60sin45$ ……………. (ii)

Now, we know the values of $\cos {{60}^{\circ }},cos45,sin60,sin{{45}^{\circ }}$ can be given

as

$\begin{align}
  & \cos {{60}^{\circ }}=\dfrac{1}{2} \\
 & \cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} \\
 & \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\
 & \sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} \\
\end{align}$

Hence, putting the above values in the equation (ii), we get

$\begin{align}
  & \cos {{15}^{\circ }}=\dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}} \\
 & \cos {{15}^{\circ }}=\dfrac{1}{2\sqrt{2}}+\dfrac{\sqrt{3}}{2\sqrt{2}} \\
\end{align}$

Now, taking L.C.M of both fractions, We get


$\cos {{15}^{\circ }}=\dfrac{1+\sqrt{3}}{2\sqrt{2}}$

Now, we can remove irrational form in denominator to rational form by rationalizing the given fraction.

So, we can multiply and divide the fraction by $2\sqrt{2}$ to get the rational form of the denominator.

Hence, we get

$\begin{align}
  & \cos {{15}^{\circ }}=\left( \dfrac{1+\sqrt{3}}{2\sqrt{2}} \right)\times \dfrac{2\sqrt{2}}{2\sqrt{2}} \\
 & =\dfrac{\left( 1+\sqrt{3} \right)2\sqrt{2}}{{{\left( 2\sqrt{2} \right)}^{2}}}=\dfrac{2\sqrt{2}+2\sqrt{6}}{8} \\
 & \cos {{15}^{\circ }}=\dfrac{\sqrt{2}+\sqrt{6}}{4}=\dfrac{\sqrt{2}\left( 1+\sqrt{3} \right)}{4} \\
\end{align}$

Hence, value of $\cos {{15}^{\circ }}$ is $\dfrac{1+\sqrt{3}}{2\sqrt{2}}$ or $\dfrac{\sqrt{2}\left( 1+\sqrt{3} \right)}{4}$.

Note: One can put infinite values of A and B with difference as ${{15}^{\circ }}$. But we need to use those values of A and B of which, we already knew the cosines and sines of them. So, we can choose those values of which cosines and sines are known. Another possible values of A and B are (45, 30), (135, 120), (150, 135) etc.
Don’t use any other identity to get the value of $\cos {{15}^{\circ }}$. We need to use the given identity only otherwise the method will not be correct. We need not rationalize $\dfrac{\sqrt{3}+1}{2\sqrt{2}}$. We did rationalization only to put the value of $\cos {{15}^{\circ }}$ in more simplified form.