Question

I just learned about unit circle, and not really sure..

Answer 1

In the unit circle below:

In Quadrant I:

$\begin{gathered} At\text{ 60, we have: (x,y)=(}\frac{1}{2},\frac{\sqrt[]{3}}{2}) \\ \implies\text{Opposite}=\frac{\sqrt[]{3}}{2} \\ \text{Adjacent}=\frac{1}{2} \\ \text{Hypotenuse}=1 \\ \cos 60\degree=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{1}{2} \end{gathered}$

Similarly, in Quadrant II:

$\begin{gathered} At\text{150, we have: (x,y)=(}-\frac{\sqrt[]{3}}{2},\frac{1}{2}) \\ \implies\text{Opposite}=\frac{1}{2} \\ \text{Adjacent}=-\frac{\sqrt[]{3}}{2} \\ \text{Hypotenuse}=1 \\ \sin 150\degree=\frac{\text{Opposite}}{\text{Hypotenuse}}=\frac{1}{2} \end{gathered}$

So, we have:

$\begin{gathered} \sin 150\degree=0.5 \\ \cos 60\degree=0.5 \end{gathered}$

In general, to get a positive value, pick the cosine of the angle in Quadrant 1, add 90 to the angle and take the sine of the result to obtain another positive value:

$\cos \theta=\sin (90\degree+\theta)$