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Question
If A + B + C = 180°, prove that cos A + cos B − cos C = `- 1 + 4cos "A"/2 cos "B"/2 sin "C"/2`
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Solution
(cos A + cos B) − cos C = `2 cos ("A" + "B")/2 cos ("A" - "B")/2- [1 - 2 sin^2 "C"/2]`
Hint: `[cos ("A" + "B")/2 = sin "C"/2]`
= `2sin "C"/2 cos ("A" - "B")/2 - 1 + 2sin^2 "C"/2`
= `- 1 + 2sin "C"/2[cos ("A" - "B")/2 + sin "C"/2]`
= `- 1 + 2 sin "C"/2[cos ("A" - "B")/2 + cos ("A" + "B")/2]`
= `- 1 + 2sin "C"/2[2cos (2"A")/4 + cos (2"B")/4]`
= `- 1 + 2sin "C"/2[2cos "A"/2 cos "B"/2]`
= `- 1 + 4 cos "A"/2 cos "B"/2 sin "C"/2`
= R.H.S
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