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If A + B + C = 180°, prove that cos A + cos B − cos C = ABC-1+4cos A2cos B2sin C2

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प्रश्न

If A + B + C = 180°, prove that cos A + cos B − cos C = `- 1 + 4cos  "A"/2 cos  "B"/2 sin  "C"/2`

योग
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उत्तर

(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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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.7 [पृष्ठ १२४]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.7 | Q 1. (ii) | पृष्ठ १२४

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