Find an expression for #cos 3x# in terms of #cosx #?
2 Answers
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)#
# sin 3x = 3sinx -4sin^3x #
Explanation:
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
# (cos theta + isin theta)^n=cos ntheta + isin ntheta#
With
# (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 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
# 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 #
