sin(270 degrees - theta) = -cos(theta)
*verify that it is an identity
That's what I thought. Thank you for posting a complete question.
To prove that sin(270 degrees - θ) = -cos(θ), we can use the fact that sin(θ) = cos(90 degrees - θ).
1. First, let's rewrite 270 degrees - θ as 90 degrees + (180 degrees - θ).
This is because 270 degrees is the same as 90 degrees + 180 degrees.
2. Now, using the property sin(θ) = cos(90 degrees - θ), we can rewrite the equation as
sin(90 degrees + (180 degrees - θ)) = -cos(θ).
3. By applying the property sin(θ) = cos(90 degrees - θ), the equation becomes
cos((90 degrees - θ) - 180 degrees) = -cos(θ).
4. Simplifying further, we have
cos(-θ) = -cos(θ).
5. Finally, using the property that cos(-θ) = cos(θ), our equation becomes
cos(θ) = -cos(θ).
This proves that sin(270 degrees - θ) = -cos(θ).
Prove the following identity:
sin(270 degrees - theta) = -cos(theta)
sin(270 degrees - theta)
= sin(270°)cos(θ) - cos(270°)sin(θ)
=(-1)cos(θ) - (0)sin(θ)
= -cos(θ)
Please post the complete question. It was not clear whether it was an equation to be solved or the proof of an identity required.