$cos 40^{o} + cos 80^{o} + cos 160^{o} + cos 240^{o} =$

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Solution

The correct option is C $- 1 / 2$
cos 40 + cos 80 + cos 160 + cos 240
$= 2 c o s (\frac{40 + 80}{2}) c o s (\frac{40 - 80}{2}) + c o s (180 - 20) + c o s (180 + 60)$
$= 2 c o s 60. c o s (- 20) + (- c o s 20) + (- c o s 60)$
$∵ c o s C + c o s D = 2 c o s (\frac{C + D}{2}) . c o s (\frac{C - D}{2})$
$c o s (180 + θ) = - c o s θ$
$c o s (180 - θ) = - c o s θ$
$= 2 \times \frac{1}{2} c o s (20) - c o s (20) - \frac{1}{2}$ $(c o s (- θ) = c o s θ)$
$= c o s 20 - c o s 20 - \frac{1}{2}$
$= - \frac{1}{2}$

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