How do you simplify trig identities with cosθ + sinθ = √2 cos(θ-∏/4)?

[weirdobomb]

cosθ + sinθ = √2 cos(θ-∏/4)

what are the steps in between?

 

[ehild]

Expand cos(θ-∏/4). What do you get?

ehild

 

[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.

 

[MrWarlock616]

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

 

[NascentOxygen]

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.

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.

 

[eumyang]

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.

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.

 

[Chestermiller]

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