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)=(}(1)/(2),\frac{\sqrt[]{3}}{2}) \\ \implies\text{Opposite}=\frac{\sqrt[]{3}}{2} \\ \text{Adjacent}=(1)/(2) \\ \text{Hypotenuse}=1 \\ \cos 60\degree=\frac{\text{Adjacent}}{\text{Hypotenuse}}=(1)/(2) \end{gathered}`

Similarly, in Quadrant II:

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