Sine-cosine offset identity
As well as the identity that \sin^2\theta+\cos^2\theta\equiv1, there is
another identity linking the \sin function with the \cos function:
\cos x\equiv\sin(x+90)
And, similarly, we can find that:
\sin x\equiv\cos(x-90)
Other derivatives
- \sin x\equiv\cos(90-x)\equiv\cos(x-90)
- \cos x\equiv\sin(90-x)\equiv\sin(x+90)
- \cos x=\sin(x+90)
- \sin x=\cos(x-90)
| Question | Answer |
|---|
| What is the sine-cosine offset identity for \cos x? | \cos x \equiv \sin(x + 90\degree) |
| What is the sine-cosine offset identity for \sin x? | \sin x \equiv \cos(x - 90\degree) |
| Express \cos x in terms of sine using a negative angle offset. | \cos x \equiv \sin(90\degree - x) |
| Express \sin x in terms of cosine using a negative angle offset. | \sin x \equiv \cos(90\degree - x) |
| Show the relationship \cos x = \sin(x + 90\degree) in another form. | \cos x = \sin(x + 90\degree) (this is the standard offset identity) |
| Show the relationship \sin x = \cos(x - 90\degree) in another form. | \sin x = \cos(x - 90\degree) (this is the standard offset identity) |