Q 136 - Precalculus Enhanced with Graphing Utilities

Question

Explain how to find the value of $\sin {390}^o$ and $\cos \left(-{45}^o\right)$

using the unit circle.

Step-by-Step Solution

Answer

The value of :-

$\sin {390}^o=\frac{1}{2}$ and

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

Step 1. Given Information

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.

Step 2. Explanation of unit circle

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 :-

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

where $\theta$ is the angle between the  points $\left(1,0\right)$ and $(x,y)$.

Step 3. To find values

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

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

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

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 that  $\sin \theta =y$

So that :-

$\sin \left({390}^o\right)=\frac{1}{2}$

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 $\cos \theta =x$

So that :-

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