If A - B - C - prove that cos A -cos B -cos C -4 cos A -2cos B -2sinC-2-1

CosA+cosB cosC =2 cos ( A+B/2 )cos ( A B/2 ) ( 1 2 sin^2 C/2 ) =2 cos ( π/2 C/2 ) cos ( A B/2 ) 1+2 sin^2 C/2 =2 sinC/2 [ cos ( A B/2 )+ sinC/2 ] 1 =2 sinC/2 [ ...

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Byju's Answer

Standard XI

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If A + B + C ...

Question

If A+B+C = $π$ , prove that cos A + cos B - cos C = $(4 cos \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}) - 1$

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Solution

$cos A + cos B - cos C$

= $2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2}) - (1 - 2 {sin}^{2} \frac{C}{2})$

= $2 cos (\frac{π}{2} - \frac{C}{2}) cos (\frac{A - B}{2}) - 1 + 2 {sin}^{2} \frac{C}{2}$

= $2 sin \frac{C}{2} [cos (\frac{A - B}{2}) + sin \frac{C}{2}] - 1$

= $2 sin \frac{C}{2} [cos (\frac{A - B}{2}) + sin {\frac{π}{2} - \frac{(A + B)}{2}}] - 1$

= $2 sin \frac{C}{2} [cos \frac{A - B}{2} + cos \frac{A + B}{2}] - 1$

$= 2 sin \frac{C}{2} {2 cos \frac{A}{2} cos \frac{B}{2}} - 1$

= ${4 cos \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}} - 1 = RHS.$

$∴ cos A + cos B - cos C = (4 cos \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}) - 1$

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If A+B+C =π, prove that cos A + cos B - cos C = (4cosA2cosB2sinC2)−1

If A+B+C = $π$ , prove that cos A + cos B - cos C = $(4 cos \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}) - 1$