Prove that (cos 2x - cos x) / (cos x - 1) = 2 cos x + 1 ?

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  • 1 year ago
    Best Answer

    LHS = (cos 2x - cos x) / (cos x - 1)

    = [ 2cos^2(x) - 1 - cos(x)]/[cos(x) - 1] ................ using the identity, cos(2x) = 2cos^2(x) - 1

    = [ 2cos^2(x) - cos(x) - 1]/[cos(x) - 1] ........... rearranging terms of numerator

    = [ ( 2cos(x) + 1)(cos(x) - 1)]/[cos(x) - 1] ....... factorising numerator

    = 2cos(x) + 1 .................. cos(x) - 1 cancel out

    = RHS

  • Mike G
    Lv 7
    1 year ago

    [cos(2x)-cosx]/(cosx-1) =

    [2cos^2(x)-cosx- 1]/(cosx-1) =

    (2cosx+1)(cosx-1)/cosx-1) =

    2cosx + 1

  • Raj K
    Lv 7
    1 year ago

    Prove that (cos 2x - cos x) / (cos x - 1) = 2 cos x + 1

    LHS=(cos 2x - cos x) / (cos x - 1)

    ={(2cos²x-1)-cosx}/(cosx-1) ............ Use cos2x=2cos²x-1

    =(2cos²x-cosx-1)(cosx-1)

    =(2cos²x-2cosx+cosx-1)(cosx-1) .......... Factorise numerator

    ={2cosx(cosx-1)+(cosx-1)}(cosx-1)

    ={(2cosx+1)(cosx-1)}(cosx-1)......... Cancel out (cosx-1)........... condition........... cosx -1 not equal to 0

    =(2cosx+1)

    =RHS Hence Proved

  • cos(2x) - cos(x) =>

    cos(x)^2 - sin(x)^2 - cos(x) =>

    cos(x)^2 - (1 - cos(x)^2) - cos(x) =>

    cos(x)^2 + cos(x)^2 - cos(x) - 1 =>

    2 * cos(x)^2 - cos(x) - 1

    2 * cos(x)^2 - cos(x) - 1 = 0

    cos(x) = (1 +/- sqrt(1 + 8)) / 4

    cos(x) = (1 +/- 3) / 4

    cos(x) = -2/4 , 4/4

    cos(x) = -1/2 , 1

    cos(x) = -1/2

    2 * cos(x) = -1

    2 * cos(x) + 1 = 0

    cos(x) = 1

    cos(x) - 1 = 0

    cos(2x) - cos(x) =>

    2 * cos(x)^2 - cos(x) - 1 =>

    (2 * cos(x) + 1) * (cos(x) - 1)

    (cos(2x) - cos(x)) / (cos(x) - 1) =>

    (2 * cos(x) + 1) * (cos(x) - 1) / (cos(x) - 1) =>

    2 * cos(x) + 1

    This is true as long as cos(x) does not equal 1

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