Question

Prove that
2 sin(a-c)cosb-sin(a-2 c)/2 sin(b-c)cosc-sin(b-2c)=sina/sinb

Answer 1

Given: The correct term is:

         2sin(a - c)cos c - sin(a - 2c) / 2sin(b - c) cos c - sin(b - 2c) = sina/sinb

To find: Prove LHS = RHS.

Solution:

• Now the given term is:

         2sin(a - c)cos c - sin(a - 2c) / 2sin(b - c) cos c - sin(b - 2c) = sina/sinb

• Lets consider LHS, we have:

         2sin(a - c)cos c - sin(a - 2c) / 2sin(b - c) cos c - sin(b - 2c)

• Consider numerator, we have:

         2sin(a - c)cos c - sin(a - 2c)

• Now using formulas:

         sin(x−y) = sinx cosy − cosx siny

         cos2y = 2cos2y − 1

         sin2x = 2sinx cosx

• We get:

         2(sina cosc - cosa sinc) x cosc - ( sina cos2c - cosa sin2c)

         2sinacos2c - 2cosa cosc x sinc - ( sina (2cos2c - 1) - cosa x 2 x sinc x cosc)

• Now after cancelation, it remains:

         sin a

• Consider denominator, we have:

         2sin(b - c) cos c - sin(b - 2c)

• Using same formulas, we get:

         2( sinbcosc - cosbsinc ) x cosc ( sinbcos2c - cosbsin2c )

         2sinbcos2c - 2cosbsinc x cosc - ( sinb (2cos2c - 1) - cosb x 2sinc x cosc)

• Now after cancellation, it remains:

         sin b

• So combining numerator and denominator, we get:

         sin a / sin b   = RHS

• Hence proved.

Answer:

      So in solution, we proved that :

     2sin(a - c)cos c - sin(a - 2c) / 2sin(b - c) cos c - sin(b - 2c) = sina/sinb

Answer 2

Answer:

iiokikplokukoliiikkkjjhujj