Given:
[tex]\sin (\frac{3\pi}{2})=-1[/tex]
Using cofunction identities
[tex]\sin \theta=\cos (90-\theta)[/tex]
This implies
[tex]\sin (\frac{3\pi}{2})=\cos (90-\frac{3\pi}{2})[/tex]
Hence
[tex]\cos (90-\frac{3\pi}{2})=-1[/tex]
By radian properties
[tex]90=\frac{\pi}{2}[/tex]
Hence the equation becomes
[tex]\begin{gathered} \cos (\frac{\pi}{2}-\frac{3\pi}{2})=-1 \\ \cos (\frac{-2\pi}{2})=-1 \\ \cos (-\pi)=-1 \end{gathered}[/tex]
Therefore, one of the values for which cos(Ф)=-1 is
[tex]-\pi[/tex]
The second value is
[tex]-\pi+2\pi=\pi[/tex]
In degree form, two of the possible values are - 180 and 180
By the cofunction relation
[tex]\sin (\theta)=\cos (90-\theta)[/tex]
And
[tex]\cos (\theta)=\sin (90-\theta)[/tex]
The cofunction relation shows that the graph of one trigonometric function is 90 degrees horizontal shift of the other
