91Ó°ÊÓ

We have to find the value of $\sin {390}^o$and $\cos \left(-{45}^0\right)$.

We have to use unit circle to find these values.

The unit circle is a circle whose radius is $1$unit.Also center of unit circle is at the origin. (As shown in following figure) :-

For any point $\left(x,y\right)$lies on a unit circle :-

$$\begin{gathered} x=\cos \mathrm{θ} \\ y=\sin \mathrm{θ} \end{gathered}$$

where $\mathrm{θ}$is the angle between the points$\left(1,0\right)$and $(x,y)$.

We know that maximum angle at unit circle is ${360}^o$.

By using periodic property of sine function we can write :-

$$\begin{gathered} \sin \left({390}^o\right)=\sin ({360}^o+{30}^o) \\ ⇒\sin \left({390}^o\right)=\sin \left({30}^o\right) \end{gathered}$$

The angle ${30}^o$lies in first quadrant and the point associated with this angle is $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$

We know thatrole="math" localid="1647263311620" $\sin \mathrm{θ}=y$

So that :-

role="math" localid="1647263247507" $\sin \left({390}^o\right)=\frac{1}{2}$

Also we know that cosine function is an even function. This gives us :-

$\cos (-{45}^o)=\cos {45}^o$

The angle ${45}^o$lies in first quadrant and the point associated with this angle is $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$

We know that $\cos \mathrm{θ}=x$

So that :-

$\cos (-{45}^0)=\frac{1}{\sqrt{2}}$.