\( \cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240^\circ \)
Options:
(a) \(0\)
(b) \(1\)
(c) \(\frac12\)
(d) \(-\frac12\)
Solution:
We know that,
\[
\cos(180^\circ-\theta)=-\cos\theta
\]
and
\[
\cos(180^\circ+\theta)=-\cos\theta
\]
Therefore,
\[
\cos160^\circ=-\cos20^\circ
\]
and
\[
\cos240^\circ=-\cos60^\circ=-\frac12
\]
Hence,
\[
\cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240^\circ
\]
\[
=\cos40^\circ+\cos80^\circ-\cos20^\circ-\frac12
\]
Using identity,
\[
\cos20^\circ=\cos40^\circ+\cos80^\circ
\]
Substituting,
\[
=\cos40^\circ+\cos80^\circ-(\cos40^\circ+\cos80^\circ)-\frac12
\]
\[
=-\frac12
\]
Therefore,
\[
\boxed{-\frac12}
\]
Correct option: (d)