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March 31, 2015

March 31, 2015

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

sin^1/2x-cosx-sin^5/2xcosx=cos^3xradsinx

- Pre-Cal (verifying trig identities) -
**bobpursley**, Thursday, February 12, 2009 at 10:12pmwhat is rad?

- Pre-Cal (verifying trig identities) -
**Crystal**, Thursday, February 12, 2009 at 10:14pmradical

- thinking needed -
**bobpursley**, Thursday, February 12, 2009 at 10:31pmmultiply 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) -
**Reiny**, Thursday, February 12, 2009 at 10:38pmI 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) -
**bobpursley**, Thursday, February 12, 2009 at 10:48pmThanks, Reiny.

- Pre-Cal (verifying trig identities) -
**John**, Monday, February 8, 2010 at 9:59amI think I have the same book as you Crystal... I have no clue how to solve this thing... :p

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