A+B+C =180. The question is to prove sin2A-sin2B+sin2C=-4sinAcosBcosC. I get sin2A-sin2B+sin2C=-4sinBcosAcosC, am I right?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It is given that A + B + C = 180. We have to prove that sin 2A - sin 2B + sin 2C = -4*sin A*cos B*cos C

sin 2A - sin 2B + sin 2C

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

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

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

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

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

=> -4*cos C*sin[(A + A - 180)/2]*cos[(180 - B - B)/2]

=> -4*cos C*sin(A - 90)*cos(90 - B)

=> -4*cos C*cos A*sin B

The result that you have got is right.

eyshmuel's profile pic

eyshmuel | Student, Grade 9 | eNotes Newbie

Posted on

sin(A-90) = -sin(90-A) = -cosA

so:

-4 * cos C * sin(A - 90) * cos(90 - B)

=> -4 * cos C (-cos A) * sin B

=> 4 * cos C * cos C * sin B

 

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