Question

Which sum or difference identity would you use to verify that cos (180° - q) = -cos q?

Answer 1

Answer:

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

Step-by-step explanation:

 Given : $\cos (180^{\circ}-q)=-\cos q$

We have to write which identity we will use to prove the given statement.

Consider $\cos (180^{\circ}-q)=-\cos q$

Take left hand side of given expression $\cos (180^{\circ}-q)$

We know

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

Comparing , we get, a= 180° and b = q

Substitute , we get,

$\cos (180^{\circ}-q)=\cos 180^{\circ} \cos (q)+\sin q \sin 180^{\circ}$

Also, we know $\sin 180^{\circ}=0$ and $\cos 180^{\circ}=-1$

Substitute, we get,

$\cos (180^{\circ}-q)=-1\cdot \cos (q)+\sin q \cdot 0$

Simplify , we get,

$\cos (180^{\circ}-q)=-\cos (q)$

Hence, use difference identity to  prove the given result.

Answer 2

cos (180° - q) = -cos q
First you would use the sum and difference formula of 
cos(a – b) = cos(a)cos(b) + sin(a)sin(b) because you have a difference inside the parentheses for cosine.

Hope this helps.