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Mathematics 11 Online
OpenStudy (anonymous):

Rewrite with only sin x and cos x. cos 3x

OpenStudy (anonymous):

I got to cos(2x+x)= cos2xcosx-sin2xsinx

OpenStudy (campbell_st):

thats a great start... so now you need to look at cos(2x) and sin(2x) sin(2x) = 2sin(x)cos(x) and cos(2x) = cos^2(x) - sin^2(x) or one of the other variations. and substitute and simplify

OpenStudy (anonymous):

(cos^2x-sin^2x)(cosx)-(2sinxcosx)(sinx) ?

OpenStudy (campbell_st):

well done... now its just a case of distributing... for the answer..

OpenStudy (anonymous):

(cos^3x-sin^2xcosx)(2sin^2xcosx) ?

OpenStudy (campbell_st):

almost you need a - sign and then collect like terms \[\cos^3(x) - \sin^2(x) \cos(x) - 2\sin^2(x) \cos(x)\] so just collect the like terms

OpenStudy (anonymous):

cos^3x-3sin^2xcosx ?

OpenStudy (campbell_st):

yes, that's it... well done

OpenStudy (anonymous):

But that's not one of my choices. cos x - 4 cos x sin2x -sin3x + 2 sin x cos x -sin2x + 2 sin x cos x 2 sin2x cos x - 2 sin x cos x

OpenStudy (campbell_st):

ok... so look at \[\cos^3(x) = \cos^2(x) \times \cos(x)\] and \[\cos^2(x) = 1 - \sin^2(x)\] so you have \[(1 -\sin^2(x))\cos(x) - 3\sin^2(x)\cos(x)\] and then simplify

OpenStudy (anonymous):

So the answer is cosx-4sin^2xcosx ! Thank you so much!

OpenStudy (campbell_st):

yes, thats the answer thats ok... sorry for the long approach... the solution may have been quicker by choosing a different substitution for cos(2x) but glad to help

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