Find an expression for #cos 3x# in terms of #cosx #?

It can be rewritten in terms of two addition identities:

#sin(u + v) = sinucosv + cosusinv#
#cos(u + v) = cosucosv - sinusinv#

#sin(3x) = sin(2x+x)#

#= sin2xcosx + cos2xsinx#

From the identities above, we have:

#sin(2x) = 2sinxcosx#
#cos(2x) = cos^2x - sin^2x#

Hence, we have:

#sin(3x) = (2sinxcosx)cosx + (cos^2x - sin^2x)sinx#

#= 2sinxcos^2x - sin^3x + sinxcos^2x#

#= 3sinxcos^2x - sin^3x#

#= 3sinx(1-sin^2x) - sin^3x#

#= color(blue)(3sinx - 4sin^3x)#

Another approach to the excellent answer from @Truong-Son N. , which is less work should you need higher powers/multiples is to use de Moivre's theorem, using complex numbers.

de Moivre's theorem states that for any complex number #z# in trigonometric form #z=cos theta + isin theta#, then

# (cos theta + isin theta)^n=cos ntheta + isin ntheta#

With #n=3# we have:

# (cos theta + isin theta)^3=cos 3theta + isin 3theta#

And expanding the LHS using the binomial theorem we have:

# (costheta)^3 + 3(costheta)^2(isintheta) + 3(costheta)(isintheta)^2 + (isintheta)^3 =cos 3theta + isin 3theta#

# :. cos^3theta + i3cos^2thetasintheta - 3costhetasin^2theta -isin^3theta =cos 3theta + isin 3theta#

# :. (cos^3theta - 3costhetasin^2theta)+ i(3cos^2thetasintheta -sin^3theta) =cos 3theta + isin 3theta#

If we equate imaginary components we get:

# sin 3theta = 3cos^2thetasintheta -sin^3theta #

As we want the expression in terms of #sin theta# we can replace #cos^2theta# using the fundamental identity #sin^2A+cos^2A-=1# to get:

# sin 3theta = 3(1-sin^2theta)sintheta -sin^3theta #
# " " = 3sintheta -3sin^3theta -sin^3theta #
# " " = 3sintheta -4sin^3theta #

Or,

# sin 3x = 3sinx -4sin^3x #

Incidentally, as an extension we also get an expression for #cos3x# for free! Equating real components we get:

# cos 3theta = cos^3theta - 3costhetasin^2theta#
# " "= cos^3theta - 3costheta(1-cos^2theta)#
# " "= cos^3theta - 3costheta+3cos^3theta#
# " "= 4cos^3theta - 3costheta#

Or,

# cos 3x = 4cos^3x - 3cosx #