Answer:
[tex]sin\frac{3\pi}{2}=-1[/tex] and [tex]cos\frac{3\pi}{2}=0[/tex]
Step-by-step explanation:
We are given that a unit circle
We have to find the value of [tex]sin\frac{3\pi}{2}[/tex] and [tex]cos\frac{3\pi}{2}[/tex] by using the unit circle
Radius of circle=r=1 unit
We know that
[tex]x=r cos\theta[/tex] and [tex]y=r sin\theta[/tex]
We [tex]\theta=\frac{3\pi}{2}[/tex]
Then x=[tex]1\cdot cos\frac{3\pi}{2}[/tex]
[tex]x=cos (2\pi-\frac{\pi}{2})[/tex]
[tex]x=cos \frac{\pi}{2}[/tex] ([tex]cos(2\pi-\theta)=cos\theta[/tex])
[tex]x=0 (cos\frac{\pi}{2}=0)[/tex]
[tex]y=1\cdot sin\frac{3\pi}{2}[/tex]
[tex]y=sin(2\pi-\frac{\pi}{2})[/tex]
[tex]y=-sin\frac{\pi}{2}[/tex] ([tex]sin(2\pi-\theta)=-sin\theta[/tex])
[tex]y=-1[/tex] ([tex]sin\frac{\pi}{2}=1[/tex])
Hence, [tex]sin\frac{3\pi}{2}=-1[/tex] and [tex]cos\frac{3\pi}{2}=0[/tex]
