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Q. The value of $ cos\text{ }20{}^\circ +cos\text{ }100{}^\circ +cos\text{ }140{}^\circ $ is equal to

KEAMKEAM 2011

Solution:

$ cos\text{ }20{}^\circ +cos\text{ }100{}^\circ +cos\text{ }140{}^\circ $
$ \Rightarrow $ $ cos\text{ }20{}^\circ +cos(90{}^\circ +10{}^\circ )+cos(90{}^\circ +50{}^\circ ) $
$=cos\text{ }20{}^\circ -sin\text{ }10{}^\circ -sin\text{ }50{}^\circ $
$=cos\text{ }20{}^\circ -(sin\text{ }10{}^\circ +sin\text{ }50{}^\circ ) $
$=cos\text{ }20{}^\circ -2\text{ }sin\text{ }30{}^\circ \,cos\text{ }20{}^\circ $ $ (\because \sin C+\sin D=2\sin \frac{(C+D)}{2}.\cos \frac{(C-D)}{2}) $
$=\cos 20{}^\circ -2.\frac{1}{2}\cos 20{}^\circ $
$=\cos 20{}^\circ -\cos 20{}^\circ =0 $