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December 18, 2014

December 18, 2014

Posted by **jacob** on Monday, April 30, 2007 at 5:42pm.

the first one is: Show that cos(x/2) sin(3x/2) = ½(sinx + sin2x)

i know that you are supposed to substitute all those trig function things in it but i kind of forgot how to

the only that i can see substituting in is the double angle one for sin2x

could anyone walk me through the process maybe?

What a very nasty trig identity.

I am sure this is not the most efficient way, but the neat thing about identities, if you do legitimate steps, usually you end up showing LS = RS.

LS= cos(x/2)sin(3x/2)

=cosA(sin(x+x/2))

=cosA(sinxcos(x/2) + cosxsin(x/2))

=cosA(2sinAcosAcosA + (cos^2 A - sin^2 A)sinA)

= 2(sin^2 A)(cos^3 A + sinA(cos^3 A) - (sin^3 A)(cosA)

=(sinA(cosA)[(3cos^2 A) - (sin^2 A)]

let that one sit for a while

R.S.

= 1/2(2sinAcosA + 2sinxcosx_

=sinAcosA + sinxcosx

=sinAcosA + (2sinAcosA)(cosx)

=sinAcosA(1 + 2cosx)

=sinAcosA(1 + cosx + cosx)

=sinAcosA(sin^2A + cos^2A + 2cos^2A -1 + 1 - 2sin^2A)

= sinAcosA(3cos^2A - sin^2A)

= L.S. !!!!!!!!!

please, somebody come up with a better way.

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