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Question
Evaluate-
cos 20° + cos 100° + cos 140°
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Solution
LHS = (cos 20° + cos 100°) + cos 140°
= 2 cos `((20^circ + 100^circ)/2) cos ((20^circ - 100^circ)/2)` + cos 140°
`[∵ cos "C" + cos "D" = 2 cos (("C + D")/2) cos (("C + D")/2)]`
= 2 cos 60° cos(-40°) + cos 140°
`= 2 xx 1/2 × cos(-40°) + cos(180° – 140°)`
`[∵ cos(-θ) = cos θ, cos 60° = 1/2]`
= cos 40° – cos 40°
= 0
Hence Proved.
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