Question

Prove these Sin3A-SinA/Cos3A+Cos5A=Cot2A

Answer 1

Step-by-step explanation:

Write the left side of the given expression as N/D, where

N = sinA - sin3A + sin5A - sin7A

D = cosA - cos3A - cos5A + cos7A

Therefore we want to show that N/D = cot2A.

We shall use these identities:

sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)

cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A

    = -2cos4A*sin3A + 2cos4A*sinA

    = 2cos4A(sinA - sin3A)

    = 2cos4A*2cos(2A)sin(-A)

    = -4cos4A*cos2A*sinA

D = cos7A + cosA - (cos5A + cos3A)

   = 2cos4A*cos3A - 2cos4A*cosA

   = 2cos4A(cos3A - cosA)

   = 2cos4A*(-2)sin2A*sinA

   = -4cos4A*sin2A*sinA

Therefore

N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]

       = cos2A/sin2A

      = cot2A

This verifies the identity.

HOPE IT HELPS.....!