what is the value of cosine at 135°? 1/2 √2 /2 0 -√2 /2

Explanation:

Step1: Express 135° as sum of known - angle

We know that (135^{\circ}=90^{\circ} + 45^{\circ}).

Step2: Use the cosine addition formula (\cos(A + B)=\cos A\cos B-\sin A\sin B)

Let (A = 90^{\circ}) and (B = 45^{\circ}). Then (\cos(90^{\circ}+45^{\circ})=\cos90^{\circ}\cos45^{\circ}-\sin90^{\circ}\sin45^{\circ}). Since (\cos90^{\circ}=0) and (\sin90^{\circ}=1), (\cos(90^{\circ}+45^{\circ})=0\times\cos45^{\circ}-1\times\sin45^{\circ}). And (\sin45^{\circ}=\frac{\sqrt{2}}{2}), so (\cos(90^{\circ}+45^{\circ})=-\frac{\sqrt{2}}{2}). Another way:

1. Recall the unit - circle definition. The coordinates of a point ((x,y)) on the unit - circle (x^{2}+y^{2}=1) corresponding to an angle (\theta) are ((\cos\theta,\sin\theta)). For (\theta = 135^{\circ}), the angle is in the second quadrant. The reference angle (\theta_{r}=180^{\circ}-135^{\circ}=45^{\circ}). In the second quadrant, the (x) - coordinate (which is (\cos\theta)) is negative and the value of (\cos45^{\circ}=\frac{\sqrt{2}}{2}), so (\cos135^{\circ}=-\frac{\sqrt{2}}{2}).

Answer:

D. (-\frac{\sqrt{2}}{2})