Question

Verify sin(360º - θ)= -sin θ
Show your work

Answer 1

The verification is in the explanation.

Explanation:

To solve this I'm going to use the difference identity for sine:

`\sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)`.

`\sin(360^\circ-\theta)=\sin(360^\circ)\cos(\theta)-\sin(\theta)\cos(360^\circ)`

We are going to apply that
`\sin(360^\circ)=0 \text{ while } \cos(360^\circ)=1`

`\sin(360^\circ-\theta)=0 \cdot \cos(\theta)-\sin(\theta)\cdot 1`

`\sin(360^\circ-\theta)=-\sin(\theta)`

Answer 2

A nice way to show it is through the unit circle.

In the unit circle, the point at angle theta from the origin has a y value of sin theta.

If you rotate a point 360-theta degrees from the origin, that is like rotating it theta degrees "backwards", or downwards, which is going to yield the same exact point, reflected through the x-axis. In other words, the y value, or sin(360-theta), is exactly -sin(theta).