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Posted by on Thursday, February 12, 2009 at 10:04pm.

How would you verify the identity?
sin^1/2x-cosx-sin^5/2xcosx=cos^3xradsinx

  • Pre-Cal (verifying trig identities) - , Thursday, February 12, 2009 at 10:12pm

    what is rad?

  • Pre-Cal (verifying trig identities) - , Thursday, February 12, 2009 at 10:14pm

    radical

  • thinking needed - , Thursday, February 12, 2009 at 10:31pm

    multiply both sides by the sqrt sinx

    sinx-cosxsinx-sin^3x cosx=cos^3x sinx
    sinx-cosxsinx-(1-cos^2x)sinxcosx=
    sinx(1-cosx)-sinxcosx+sinxcox^3x=
    sinx(1-cosx-cosx)+sinx cox^3x=
    Hmmm. This is telling me the identity does not exist, because sinx(1-2cosx) is not zero.

    check my work. There has to be an error here, but I am certain this is (was ) leading somewhere.

  • Pre-Cal (verifying trig identities) - , Thursday, February 12, 2009 at 10:38pm

    I read your identity as
    (sinx)^(1/2) - cosx - (sinx)^(5/2)*cosx = (cosx)^3 *(sinx)^(1/2)

    I picked 30 degrees, and the
    left side was NOT equal to the right side.

    So the way you typed it, it is not an identity.

  • Pre-Cal (verifying trig identities) - , Thursday, February 12, 2009 at 10:48pm

    Thanks, Reiny.

  • Pre-Cal (verifying trig identities) - , Monday, February 8, 2010 at 9:59am

    I think I have the same book as you Crystal... I have no clue how to solve this thing... :p

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