# Homework Help: Trig Identity or something

1. Feb 21, 2013

### weirdobomb

cosθ + sinθ = √2 cos(θ-∏/4)

what are the steps in between?

2. Feb 21, 2013

### ehild

Expand cos(θ-∏/4). What do you get?

ehild

3. Feb 21, 2013

### weirdobomb

But how would I get from cosθ + sinθ to √2 cos(θ-∏/4)
I can expand and get the original expression but don't understand the other way around.

Last edited: Feb 21, 2013
4. Feb 21, 2013

### MrWarlock616

Multiply and divide by $\sqrt{2}$.

5. Feb 21, 2013

### Staff: Mentor

Expand cos(θ-/4) means to apply the well-known trig expansion cos (A - B) = (cos A)(cos B) + (sin A)(sin B)

This formula (along with a few others) should be well-known by the time you sit for your next closed-book exam.

6. Feb 21, 2013

### eumyang

Are you saying that you can get from
$\sqrt{2} \cos \left( \theta - \frac{\pi}{4} \right)$
to
$\cos \theta + \sin \theta$
but not the other way around? Just take the steps you get from the RHS to the LHS and go backwards.

7. Feb 21, 2013

### Staff: Mentor

Write $$Acos(\theta)+Bsin(\theta)=\sqrt{A^2+B^2}(\frac{A}{\sqrt{A^2+B^2}}cos(\theta)+\frac{B}{\sqrt{A^2+B^2}}sin(\theta))$$

Then let $\phi=\arctan{\frac{B}{A}}$
Then,$$Acos(\theta)+Bsin(\theta)=\sqrt{A^2+B^2}(cos(\phi)cos(\theta)+sin(\phi)sin(\theta))=\sqrt{A^2+B^2}cos(\theta-\phi)$$