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

a. sin (a -b) = sin a cos b – cos a sin b
b. cos (a -b) = cos a cos b – sin a sin b
c. cos (a -b) = cos a cosb + sin a sin b
d. sin (a + b) = sin a cos b + cos a sin b

1 Answer

6 votes

Answer:

c. cos (a -b) = cos a cosb + sin a sin b

Explanation:

The given identity is
\cos (180\degree-q)=-\cos q

Using the formula:
\cos (a-b)=\cos a\cos b+\sin a \sin b

We put
a=180\degree and
b=q to get:


\cos (180\degree-q)=\cos 180\degree\cos q+\sin 180\degree \sin q


\cos (180\degree-q)=-(1)\cos q+(0)\sin q


\cos (180\degree-q)=-\cos q+0


\cos (180\degree-q)=-\cos q

The correct answer is C

User Niladri
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