sin3y/siny=1+2cos2y prove LHS=RHS?

trignometry sum

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  • Anonymous
    1 decade ago
    Favorite Answer

    sin3y/siny

    = sin(y+2y) / siny

    = ( sinycos2y + cosysin2y) / siny

    Using sin2y = 2sinycosy

    = [ sinycos2y + 2siny(cosy)^2 ] / siny

    = siny[ cos2y + 2(cosy)^2 ] / siny

    = cos2y + 2(cosy)^2

    = cos2y + 2(cosy)^2 - 1 + 1

    Using cos2y = 2(cosy)^2 - 1

    = cos2y + cos2y + 1

    = 2cos2y + 1

  • 1 decade ago

    sin3y/siny = 1 + 2cos2y

    LHS

    = sin3y/siny

    = (3 siny - 4 (siny)^3)/siny

    = 3 - 4(siny)^2

    = 3 - 2(1 - cos2y)

    = 1 + 2cos2y

    = RHS

  • 1 decade ago

    sin3y

    = 3siny - 4sin^3 y

    = siny (3 - 4sin^2 y)

    = siny (4 - 4sin^2 y - 1)

    = siny (4cos^2 y - 1)

    = siny [2(1 + cos2y) - 1]

    = siny (1 + 2cos2y)

    => sin3y/siny = 1 + 2cos2y

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