Question

Find the value of
(i)$\sin {{120}^{\circ }}$
(ii)$\cos (-{{300}^{\circ }})$

Answer 1

Hint: Here, $\sin {{120}^{\circ }}$ can be written as $\sin {{120}^{{}^\circ }}=\sin ({{90}^{\circ }}+{{30}^{\circ }})$ or $\sin {{120}^{{}^\circ }}=\sin ({{180}^{\circ }}-{{60}^{\circ }})$ and $\cos (-{{300}^{{}^\circ }})=\cos ({{300}^{{}^\circ }})$ can be written as $\cos ({{300}^{{}^\circ }})=\cos ({{180}^{\circ }}+{{120}^{\circ }})$, then apply the formulas:
$\begin{aligned} & \sin (A+B)=\sin A\cos B+\cos A\sin B \\\ & \sin (A-B)=\sin A\cos B-\cos A\sin B \\\ & \cos (A+B)=\cos A\cos B-\sin A\sin B \\\ & \cos (A-B)=\cos A\cos B+\sin A\sin B \\\ \end{aligned}$

Complete step-by-step answer:
(i) $\sin {{120}^{\circ }}$
Here, we can write:
$\sin {{120}^{{}^\circ }}=\sin ({{90}^{\circ }}+{{30}^{\circ }})\text{ }.....\text{ (1)}$
The above equation is of the form $\sin (A+B)$ where $A={{90}^{\circ }}$ and $B={{30}^{\circ }}$ We have the formula:
$\sin (A+B)=\sin A\cos B+\cos A\sin B$
By applying the formula we get:
$\Rightarrow \sin ({{90}^{\circ }}+{{30}^{\circ }})=\sin {{90}^{\circ }}\cos {{30}^{\circ }}+\cos {{90}^{\circ }}\sin {{30}^{\circ }}$
Since, we know that:
$\begin{aligned} & \sin {{90}^{\circ }}=1 \\\ & \cos {{90}^{\circ }}=0 \\\ & \sin {{30}^{\circ }}=\dfrac{1}{2} \\\ & \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$
Our equation becomes:
$\begin{aligned} & \Rightarrow \sin ({{90}^{\circ }}+{{30}^{\circ }})=1\times \dfrac{\sqrt{3}}{2}+0\times \dfrac{1}{2} \\\ & \Rightarrow \sin ({{90}^{\circ }}+{{30}^{\circ }})=\dfrac{\sqrt{3}}{2}+0 \\\ & \Rightarrow \sin ({{90}^{\circ }}+{{30}^{\circ }})=\dfrac{\sqrt{3}}{2}\text{ }.....\text{ (2)} \\\ \end{aligned}$
Hence, by substituting equation (2) in equation (1) we obtain:
$\sin {{120}^{{}^\circ }}=\dfrac{\sqrt{3}}{2}$
OR

Alternatively, you can also prove this by applying the formula:
$\sin ({{90}^{\circ }}+A)=\cos A$.
Therefore, we will get:
$\Rightarrow \sin ({{90}^{\circ }}+{{30}^{\circ }})=\cos {{30}^{\circ }}$
We know that $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$. Hence, we will get:
$\begin{aligned} & \Rightarrow \sin ({{90}^{\circ }}+{{30}^{\circ }})=\dfrac{\sqrt{3}}{2} \\\ & \Rightarrow \sin {{120}^{\circ }}=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$
OR

We can also write $\sin {{120}^{\circ }}$as,
$\sin {{120}^{{}^\circ }}=\sin ({{180}^{\circ }}-{{60}^{\circ }})$
The above equation is of the form $\sin (A+B)$ where $A={{90}^{\circ }}$ and $B={{30}^{\circ }}$, we have the formula:
$\sin (A-B)=\sin A\cos B-\cos A\sin B$
By applying the above formula we get:
$\Rightarrow \sin ({{180}^{\circ }}-{{60}^{\circ }})=\sin {{180}^{\circ }}\cos {{60}^{\circ }}-\cos {{180}^{\circ }}\sin {{60}^{\circ }}\text{ }.....\text{ (4)}$
Next, we have to find $\sin {{180}^{\circ }}$ and $\cos {{180}^{\circ }}$
Now we can write:
$\sin {{180}^{\circ }}=\sin ({{90}^{\circ }}+{{90}^{\circ }})$
We know that $\sin ({{90}^{\circ }}+A)=\cos A$
Therefore, by applying this we will get:
$\begin{aligned} & \sin {{180}^{\circ }}=\sin ({{90}^{\circ }}+{{90}^{\circ }})=\cos {{90}^{\circ }} \\\ & \Rightarrow \sin {{180}^{\circ }}=\cos {{90}^{\circ }} \\\ \end{aligned}$
Since, $\cos {{90}^{\circ }}=0$, we can say that
$\sin {{180}^{\circ }}=0$.
Similarly, we can write:
$\cos {{180}^{\circ }}=\cos ({{90}^{\circ }}+{{90}^{\circ }})$
We know that:
$\cos ({{90}^{\circ }}+A)=-\sin A$.
Therefore, by applying this we will get:
$\begin{aligned} & \Rightarrow \cos {{180}^{\circ }}=\cos ({{90}^{\circ }}+{{90}^{\circ }}) \\\ & \Rightarrow \cos {{180}^{\circ }}=-\sin {{90}^{\circ }} \\\ \end{aligned}$
We have,
$\sin {{90}^{\circ }}=1$
Therefore, we will get:
$\Rightarrow \cos {{180}^{\circ }}=-1$
Therefore, we have:
$\begin{aligned} & \cos {{60}^{\circ }}=\dfrac{1}{2} \\\ & \text{sin}{{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$
$\sin {{180}^{\circ }}=0$
$\cos {{180}^{\circ }}=-1$
By substituting all the above values in equation (4) we obtain:
$\begin{aligned} & \Rightarrow \sin ({{180}^{\circ }}-{{60}^{\circ }})=0\times \dfrac{1}{2}-(-1)\times \dfrac{\sqrt{3}}{2} \\\ & \Rightarrow \sin ({{180}^{\circ }}-{{60}^{\circ }})=0+\dfrac{\sqrt{3}}{2} \\\ & \Rightarrow \sin ({{180}^{\circ }}-{{60}^{\circ }})=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$
Hence, we can say that $\sin {{120}^{{}^\circ }}=\dfrac{\sqrt{3}}{2}$.

$\cos \left( -{{300}^{\circ }} \right)$.
Next, we have to find the value of $\cos \left( -{{300}^{\circ }} \right)$
We know that,
$\cos (-A)=\cos A$
Therefore, we can write:
$\cos (-{{300}^{\circ }})=\cos ({{300}^{\circ }})$
Now, we can write:
$\cos ({{300}^{\circ }})=\cos ({{180}^{\circ }}+{{120}^{\circ }})$
The above equation is of the form $\cos (A+B)$ where $A={{180}^{\circ }}$ and $B={{120}^{\circ }}$. By expansion we have:
$\cos (A+B)=\cos A\cos B-\sin A\sin B$
By applying the formula we get:
$\cos ({{180}^{\circ }}+{{120}^{\circ }})=\cos {{180}^{\circ }}\cos {{120}^{\circ }}-\sin {{180}^{\circ }}\sin {{120}^{\circ }}\text{ }....\text{ (5)}$
Now, we have to find, $\cos {{120}^{\circ }}$.
$\cos {{120}^{\circ }}=\cos ({{90}^{\circ }}+{{30}^{\circ }})$
We know that $\cos ({{90}^{\circ }}+A)=-\sin {{90}^{\circ }}$.
Therefore, we can say that:
$\begin{aligned} & \cos ({{90}^{\circ }}+{{30}^{\circ }})=-\sin {{30}^{\circ }} \\\ & \Rightarrow \cos {{120}^{\circ }}=-\sin {{30}^{\circ }} \\\ \end{aligned}$
We have $\sin {{30}^{\circ }}=\dfrac{1}{2}$.
Therefore, we will get:
$\cos {{120}^{\circ }}=-\dfrac{1}{2}$
We know that:
$\begin{aligned} & \cos {{180}^{\circ }}=-1 \\\ & \text{sin18}{{0}^{\circ }}=0 \\\ & \text{sin}{{120}^{\circ }}=\dfrac{\sqrt{3}}{2} \\\ & \cos {{120}^{\circ }}=-\dfrac{1}{2} \\\ \end{aligned}$
By applying all these values in equation (5) we obtain:
$\begin{aligned} & \cos ({{180}^{\circ }}+{{120}^{\circ }})=-1\times -\dfrac{1}{2}-0\times \dfrac{\sqrt{3}}{2} \\\ & \cos ({{180}^{\circ }}+{{120}^{\circ }})=\dfrac{1}{2} \\\ & \cos ({{300}^{\circ }})=\dfrac{1}{2} \\\ \end{aligned}$
Therefore, we can say that $\cos (-{{300}^{\circ }})=\dfrac{1}{2}$.

Note: Here, for finding $\cos ({{300}^{\circ }})$ you can also apply the formula $\cos ({{360}^{\circ }}-{{60}^{\circ }})=\cos {{60}^{\circ }}=\dfrac{1}{2}$
Instead of expanding $\cos ({{180}^{\circ }}+{{120}^{\circ }})$ if you know the formula, you can directly write:
$\begin{aligned} & \cos ({{180}^{\circ }}+{{120}^{\circ }})=-\cos {{120}^{\circ }} \\\ & \Rightarrow \cos ({{180}^{\circ }}+{{120}^{\circ }})=-\left( -\dfrac{1}{2} \right) \\\ & \Rightarrow \cos ({{180}^{\circ }}+{{120}^{\circ }})=\dfrac{1}{2} \\\ \end{aligned}$.