Answer:
[tex]\cos (a-b)=\cos a \cos b+\sin a \sin b[/tex]
Step-by-step explanation:
Given : [tex]\cos (180^{\circ}-q)=-\cos q[/tex]
We have to write which identity we will use to prove the given statement.
Consider [tex]\cos (180^{\circ}-q)=-\cos q[/tex]
Take left hand side of given expression [tex]\cos (180^{\circ}-q)[/tex]
We know
[tex]\cos (a-b)=\cos a \cos b+\sin a \sin b[/tex]
Comparing , we get, a= 180° and b = q
Substitute , we get,
[tex]\cos (180^{\circ}-q)=\cos 180^{\circ} \cos (q)+\sin q \sin 180^{\circ}[/tex]
Also, we know [tex]\sin 180^{\circ}=0[/tex] and [tex]\cos 180^{\circ}=-1[/tex]
Substitute, we get,
[tex]\cos (180^{\circ}-q)=-1\cdot \cos (q)+\sin q \cdot 0[/tex]
Simplify , we get,
[tex]\cos (180^{\circ}-q)=-\cos (q)[/tex]
Hence, use difference identity to prove the given result.
