$c o s 40^{\circ} + c o s 80^{\circ} + c o s 160^{\circ} + c o s 240^{\circ} =$

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Solution

$= 2 c o s (\frac{40^{\circ} + 80^{\circ}}{2}) c o s (\frac{40^{\circ} - 80^{\circ}}{2}) + c o s 160^{\circ} - c o s (180^{\circ} + 60^{\circ}) [∵ c o s A + c o s B = 2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2})] = 2 c o s 60^{\circ} c o s (- 20^{\circ}) + c o s 160^{\circ} - \frac{1}{2} = 2 \times \frac{1}{2} c o s 20^{\circ} + c o s 160^{\circ} - \frac{1}{2} = - c o s (180 - 20)^{\circ} + c o s 160^{\circ} - \frac{1}{2} - \frac{1}{2}$

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cos 20^{\circ} \cdot cos 40^{\circ} \cdot cos 60^{\circ} \cdot cos 80^{\circ} = . . . . .

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cos 80^{\circ} + cos 40^{\circ} - cos 20^{\circ}

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