Question

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

Answer 1

Hint: Here, we have $\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}$ which is of the form $\sin A\cos B+\cos A\sin B$ where $A={{60}^{\circ }}$ and $B={{30}^{\circ }}$, which is the expansion of $\sin (A+B)$ where $A+B={{60}^{\circ }}+{{30}^{\circ }}$, hence we will get $\sin ({{60}^{\circ }}+{{30}^{\circ }})$ and also apply the formulas:
$\begin{aligned} & \cos ({{90}^{\circ }}-A)=\sin A \\\ & \sin ({{90}^{\circ }}-A)=\cos A \\\ \end{aligned}$

Complete step-by-step answer:
Here, we have to find the value of $\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}$.
Now, we can rewrite the equation as:
$\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }}$
Hence, the above equation is of the form $\sin A\cos B+\cos A\sin B$, which is the expansion of $\sin (A+B)$. i.e. we can write:
$\sin (A+B)=\sin A\cos B+\cos A\sin B$
Since, we have$A={{60}^{\circ }}$ and $B={{30}^{\circ }}$. We can apply the above formula where:
$\sin (A+B)=\sin ({{60}^{\circ }}+{{30}^{\circ }})$
i.e. we obtain the equation:
$\begin{aligned} & \sin ({{60}^{\circ }}+{{30}^{\circ }})=\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{30}^{\circ }}\sin {{60}^{\circ }} \\\ & \sin {{90}^{\circ }}=\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{30}^{\circ }}\sin {{60}^{\circ }} \\\ \end{aligned}$
We know that the value of $\sin {{90}^{\circ }}=1$.
Therefore, we will get:
$\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{30}^{\circ }}\sin {{60}^{\circ }}=\sin {{90}^{\circ }}=1$
Hence we can say that the value of,
$\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}=1$
or
Here, there is another method to find the solution, i.e, by directly substituting the values for $\begin{aligned} & \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\\ & \sin {{30}^{\circ }}=\dfrac{1}{2} \\\ & \cos {{60}^{\circ }}=\dfrac{1}{2} \\\ & \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$
Hence by substituting all these values in $\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }}$we get:
$\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}$
We know that $\sqrt{3}\times \sqrt{3}=3$. Hence, we get:
$\sin {{60}^{{}^\circ }}\cos {{30}^{{}^\circ }}+\cos {{60}^{{}^\circ }}\sin {{30}^{{}^\circ }}=\dfrac{3}{4}+\dfrac{1}{4}$
Now, by taking the LCM we get:
$\begin{aligned} & \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }}=\dfrac{3+1}{4} \\\ & \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }}=\dfrac{4}{4} \\\ & \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }}=1 \\\ \end{aligned}$
or
Here, we can also solve this by converting everything into sine. i.e.
We have the formulas:
$\begin{aligned} & \cos ({{90}^{\circ }}-A)=\sin A \\\ & \sin ({{90}^{\circ }}-A)=\cos A \\\ \end{aligned}$
That is, we can write:
$\begin{aligned} & \cos {{30}^{\circ }}=\sin ({{90}^{\circ }}-{{30}^{\circ }}) \\\ & \cos {{30}^{\circ }}=\sin {{60}^{\circ }}\text{ }.....\text{ (1)} \\\ \end{aligned}$
Similarly, we will get:
$\begin{aligned} & \sin {{30}^{\circ }}=\cos ({{90}^{\circ }}-{{30}^{\circ }}) \\\ & \sin {{30}^{\circ }}=\cos {{60}^{\circ }}\text{ }.....\text{ (2)} \\\ \end{aligned}$
By applying equation (1) and equation (2) in $\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}$we get:
$\begin{aligned} & \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}=\sin {{60}^{\circ }}\sin {{60}^{\circ }}+\cos {{60}^{\circ }}\cos {{60}^{\circ }} \\\ & \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}={{\sin }^{2}}{{60}^{\circ }}+{{\cos }^{2}}{{60}^{\circ }} \\\ \end{aligned}$
We also know that ${{\cos }^{2}}A+{{\sin }^{2}}A=1$.
Therefore, we can say that ${{\sin }^{2}}{{60}^{\circ }}+{{\cos }^{2}}{{60}^{\circ }}=1$
Hence, we will get:
$\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}=1$
Hence we got the value of $\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}=1$ in three different ways.
We can apply any one of these methods to obtain the solution.

Note: Here, three different methods are given to find the value of $\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}$. If we know the trigonometric values of sine and cosine angles, then it is the easiest method to solve the problem. But, if we have doubt regarding any values, then go for the other two alternate methods.