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san4es73 [151]
3 years ago
5

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

Mathematics
2 answers:
Mazyrski [523]3 years ago
6 0

Answer:

The verification is in the explanation.

Step-by-step 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)

Mekhanik [1.2K]3 years ago
4 0

Answer:

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).

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