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In Triangle ABC, (SIN 2A + SIN 2B + SIN 2C) / (COS A + COS B + COS C) -1 =

In Triangle ABC, (SIN 2A + SIN 2B + SIN 2C) / (COS A + COS B + COS C) -1 = 

Grade:11

1 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
9 years ago
Hello student,
Left Hand Side (LHS) = 2 sin A. cos A + 2 sin (B+C). cos (B - C)

=> 2 sin A. cos A + 2 sin (180 - A).cos ( B - C )

=> 2 sin A. cos A + 2 sin A. cos ( B - C )

=> 2 sin A [ cos A + cos ( B- C ) ]

( but cos A = cos { 180 - ( B + C ) } = - cos ( B + C )

=> 2 sin A [ 2 sin B. sin C ]

=> 4. sin A. sin B. sin C

Also
.. LHS

= ( cos A + cos B ) + cos C

= { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cos C

= { 2 · cos [ (π/2) - (C/2) ] · cos [ (A-B) / 2 ] } + cos C

= { 2 · sin( C/2 ) · cos [ (A-B) / 2 ] } + { 1 - 2 · sin² ( C/2 ) }

= 1 + 2 sin ( C/2 )· { cos [ (A -B) / 2 ] - sin ( C/2 ) }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - sin [ (π/2) - ( (A+B)/2 ) ] }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - cos [ (A+B)/ 2 ] }

= 1 + 2 sin ( C/2 )· 2 sin ( A/2 )· sin( B/2 ) ... ... ... (2)

= 1 + 4 sin(A/2) sin(B/2) sin(C/2)

Now you can simplify further

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