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
