Verified answer

[tex]\displaystyle \sf \left[cos(\theta)+i.sen(\theta)\right]^n = cos(n.\theta)+i.sen(n.\theta) \\\\\ \text{ou seja} : \\\\ \left[cis(\theta)\right]^{n} = cis(n.\theta) \\\\ \text{Note que } :\\\\ \boxed{\sf 2\cdot cos(n.\theta) = \left[cis(\theta)\right]^{n}+\left[cis(\theta)\right]^{-n} \ } \\\\\ \underline{\text{Fa\c camos n = 3}} : \\\\ 2\cdot cos(3.\theta) = \left[cis(\theta)\right]^{3}+\left[cis(\theta)\right]^{-3} \\\\ Da{\'i}}:[/tex]

[tex]\displaystyle \sf \left[cis(\theta)\right]^{3} = \left[cos(\theta)+i.sen(\theta)\right]^3 \\\\ \left[cis(\theta)\right]^{-3} = [cos(\theta)-i.sen(\theta)]^{3} \\\\ \underline{\text{Desenvolvendo}} : \\\\ \left[cos(\theta)+i.sen(\theta)\right]^3 = cos^3(\theta)-i.sen^3(\theta)+3cos^2(\theta)sen(\theta).i-3cos(\theta).sen^2(\theta) \\\\ \underline{\left[cos(\theta)-i.sen(\theta)\right]^3=cos^3(\theta) +i.sen^3(\theta) -3cos^2(\theta )sen(\theta).i -3cos(\theta)sen^2(\theta) \ \ +}[/tex]

[tex]\displaystyle \sf \left[cis(\theta)\right]^{3} +\left[cis(\theta)\right]^{-3} = 2cos^3(\theta) -2.3cos(\theta)sen^2(\theta) \\\\ 2\cdot cos(3\theta) = 2cos^3(\theta) -2.3cos(\theta)sen^2(\theta) \\\\\\ \Large\boxed{\sf \ cos(3.\theta) = cos^3(\theta)-3cos(\theta)sen^2(\theta) \ }\checkmark \\\\ C.Q.D[/tex]