QUESTION

# Find the value of $\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}$.

Hint: Here, $\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}$ is of the form $\cos A\cos B+\sin A\sin B$ where $A={{60}^{\circ }}$ and $B={{30}^{\circ }}$. This is the expansion of $\cos (A-B)$ where $A-B={{60}^{\circ }}-{{30}^{\circ }}$. We also have to apply the trigonometric formulas:
$\cos ({{90}^{\circ }}-A)=\sin A$
$\sin ({{90}^{\circ }}-A)=\cos A$
$\sin 2A=2\sin A\cos A$

Here, we have to find the value of:
$\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}$
Hence, the above equation is of the form $\cos A\cos B+\sin A\sin B$, which is the expansion of $\cos (A-B)$. i.e. we have the formula:
$\cos (A-B)=\cos A\cos B+\sin A\sin B$
Since, we have$A={{60}^{\circ }}$ and $B={{30}^{\circ }}$. We can apply the above formula where:
$\cos (A-B)=\cos ({{60}^{\circ }}-{{30}^{\circ }})$
That is, we obtain the equation:
\begin{align} & \cos ({{60}^{\circ }}-{{30}^{\circ }})=\cos {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{60}^{\circ }}\sin {{30}^{\circ }} \\ & \cos {{30}^{\circ }}=\cos {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{60}^{\circ }}\sin {{30}^{\circ }} \\ \end{align}
We know that the value of $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$.
Therefore, we will get:
$\cos {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{60}^{\circ }}\sin {{30}^{\circ }}=\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$
Hence we can say that the value will be:
$\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$
OR
Here, there is another method to find the solution, i.e. by directly substituting the values for \begin{align} & \cos {{60}^{\circ }}=\dfrac{1}{2} \\ & \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} \\ & \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\ & \sin {{30}^{\circ }}=\dfrac{1}{2} \\ \end{align}
Hence by substituting all these values in $\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}$we get:
$\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}=\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}$
Next by simplification we get:
$\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}=\dfrac{\sqrt{3}}{4}+\dfrac{\sqrt{3}}{4}$
Now, by taking the LCM we get:
\begin{align} & \cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}=\dfrac{\sqrt{3}+\sqrt{3}}{4} \\ & \cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}=\dfrac{2\sqrt{3}}{4} \\ \end{align}
By cancellation, we get:
$\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$
OR
We can also solve this problem by using the formulas:
\begin{align} & \sin ({{90}^{\circ }}-A)=\cos A \\ & \cos ({{90}^{\circ }}-A)=\sin A \\ \end{align}
i.e. we can write:
\begin{align} & \cos {{30}^{\circ }}=\sin ({{90}^{\circ }}-{{30}^{\circ }}) \\ & \cos {{30}^{\circ }}=\sin {{60}^{\circ }}\text{ }.....\text{ (1)} \\ \end{align}
Similarly, we will get:
\begin{align} & \sin {{30}^{\circ }}=\cos ({{90}^{\circ }}-{{30}^{\circ }}) \\ & \sin {{30}^{\circ }}=\cos {{60}^{\circ }}\text{ }.....\text{ (2)} \\ \end{align}
By applying equation (1) and equation (2) in $\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}$we get:
\begin{align} & \cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}=\cos {{60}^{\circ }}\sin {{60}^{\circ }}+\sin {{60}^{\circ }}\cos {{60}^{\circ }} \\ & \cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}=2\cos {{60}^{\circ }}\sin {{60}^{\circ }}\text{ }.....\text{ (3)} \\ \end{align}
We know the formula that:
$\sin 2A=2\sin A\cos A$
That is, we will get:
\begin{align} & 2\cos {{60}^{\circ }}\sin {{60}^{\circ }}=\sin 2\times {{60}^{\circ }} \\ & 2\cos {{60}^{\circ }}\sin {{60}^{\circ }}=\sin {{120}^{\circ }} \\ & 2\cos {{60}^{\circ }}\sin {{60}^{\circ }}=\sin \left( {{180}^{\circ }}-{{60}^{\circ }} \right) \\ \end{align}
We, also know that $\sin \left( {{180}^{\circ }}-A \right)=\sin A$. i.e.
$\sin \left( {{180}^{\circ }}-{{60}^{\circ }} \right)=\sin {{60}^{\circ }}$
Therefore, we will get:
$2\cos {{60}^{\circ }}\sin {{60}^{\circ }}=\sin {{60}^{\circ }}\text{ }.....\text{ (4)}$
By substituting equation (4) in equation (3) we obtain:
By substituting this formula above we obtain:
$\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}=\sin {{60}^{\circ }}$
We know that $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$, therefore we get:
$\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$

Note: We can find the value of $\cos {{60}^{\circ }}\times \cos {{30}^{\circ }}+\sin {{60}^{\circ }}\times \sin {{30}^{\circ }}$ by using any one of the above methods. If we know the basic sine and cosine values, then directly by substituting the values we will get the answer easily.