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.
