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Answer:
[tex]\sin \frac{3\pi}{2} = -1[/tex]
[tex]\cos\frac{3\pi}{2} =0[/tex]
Step-by-step explanation:
Please refer to the image attached.
Here we have a circle with unit radius. At some angle Ф the radius = 1 , and it is the hypotenuse (shown by green line in the image attached) of the ΔPQR thus formed. As our angle Ф increases, the hypotenuse gets closer to the positive y axis and at 90°, it overlap the y axis. Hypotenuse (H) and Opposite site (O) becomes same and Adjacent (A) becomes 0.
As our angle move further and reaches 180, the Hypotenuse and adjacent becomes same and overlap negative x axis. As we move further at 270 i.e [tex]\frac{3\pi}{2}[/tex] , the hypotenuse and opposite side overlap on y axis and Adjacent side become 0. However the opposite side becomes negative here .
Our sine ratio says
[tex]\sin \frac{3\pi}{2} =\frac{opposite}{Hypotenuse}[/tex]
[tex]\sin \frac{3\pi}{2} =\frac{-1}{1}[/tex]
[tex]\sin \frac{3\pi}{2} =-1[/tex]
Hence we have our [tex]\sin \frac{3\pi}{2} = -1[/tex]
Now
[tex]\cos\frac{3\pi}{2} =\frac{Adjacent}{Hypotenuse}[/tex]
Adjacent as we discussed is 0 at [tex]\frac{3\pi}{2}[/tex]
[tex]\cos\frac{3\pi}{2} =\frac{0}{1}[/tex]
[tex]\cos\frac{3\pi}{2} =0[/tex]