Recent content by makarov1901

  1. M

    How to solve (1+cosA)^2 - (1-cosA)^2 - sin^2A = ctgA*sinA*cosA

    My task is to transform both parts so that they're equal. I don't know why you say it's not an identity.
  2. M

    How to solve (1+cosA)^2 - (1-cosA)^2 - sin^2A = ctgA*sinA*cosA

    You assumed right. The problem is that both parts of the identity should be equal. As far as I know, 4cosA is not equal to 1.
  3. M

    How to solve (1+cosA)^2 - (1-cosA)^2 - sin^2A = ctgA*sinA*cosA

    Homework Statement (1+cosA)^2-(1-cosA)^2-sin^2A=ctgA*sinA*cosAHomework Equations The Attempt at a Solution I moved sin^2 to the right side, then expanded the left side and got to: 1+2cosA+cos^2A-1+2cosA-cos^2A=1 When I cancel the left side I get: 4cosA=1 I'd be very grateful if anyone help me.