The coordinates of the point on the unit circle at an angle -135° are [tex](-\frac{\sqrt{2} }{2} ,-\frac{\sqrt{2} }{2} )[/tex]
What are coordinates of point on the circle?
"The coordinates of any point on the circle [tex]x^{2}+ y^{2}= r^{2}[/tex] are [tex](r~cos\theta,r~sin\theta)[/tex]"
Formula of sin(A ± B)
sin(A ± B) = sin(A)cos(B) ± cos(A)sin(B)
Formula of cos(A - B)
cos(A - B) = cos(A)cos(B) + sin(A)sin(B)
For given question,
We need to find the coordinates of the point on the unit circle at an angle -135°
We know, for unit circle radius = 1
So, the coordinates of any point on the unit circle would be [tex](cos\theta, sin\theta)[/tex]
Here, [tex]\theta=-135^{\circ}[/tex]
Now, we find the the value of [tex]sin(-135^{\circ}),~cos(-135^{\circ})[/tex]
[tex]sin(-135^{\circ})\\\\=-sin(135^{\circ})\\\\=-sin(180^{\circ}-45^{\circ})\\\\=-[sin(180^{\circ})cos(45^{\circ})-cos(180^{\circ})sin(45^{\circ})]\\\\=-[0\times \frac{1}{\sqrt{2} }-(-1)\times \frac{1}{\sqrt{2} } ]\\\\=-(0+\frac{1}{\sqrt{2} } )\\\\=-\frac{1}{\sqrt{2} }\\\\=-\frac{1\times \sqrt{2} }{\sqrt{2} \times \sqrt{2} }\\\\=-\frac{\sqrt{2} }{2}[/tex]
and cos(-135°) would be,
[tex]cos(-135^{\circ})\\\\=cos(135^{\circ})\\\\=cos(180^{\circ}-45^{\circ})\\\\=cos(180^{\circ})sin(45^{\circ})+sin(180^{\circ})sin(45^{\circ})\\\\=((-1)\times \frac{1}{\sqrt{2}} )+(0\times \frac{1}{\sqrt{2} }) \\\\=-\frac{1}{\sqrt{2} }+0\\\\ =-\frac{1}{\sqrt{2}}\\\\=-\frac{1\times \sqrt{2} }{\sqrt{2}\times \sqrt{2}}\\\\=-\frac{\sqrt{2} }{2}[/tex]
Therefore, the coordinates of the point on the unit circle at an angle -135° are [tex](-\frac{\sqrt{2} }{2} ,-\frac{\sqrt{2} }{2} )[/tex]
Learn more about the coordinates of the point on the unit circle here:
https://brainly.com/question/12100731
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