Prove that: cos 2 ⁡ θ sin 4 ⁡ θ = 1 32 ( cos ⁡ 6 θ − cos ⁡ 2 θ + 2 − 2 cos ⁡ 4 θ )

Question

Prove that:      cos    2    ⁡  θ      sin    4    ⁡  θ  =      1    32    (  cos  ⁡  6  θ  −  cos  ⁡  2  θ  +  2  −  2  cos  ⁡  4  θ  )

Madisyn Avery · Accepted Answer

Use the linearisation and the duplication formulae:

\begin{aligned} \cos^{2} θ \sin^{4} θ & = \cos^{2} θ \sin^{2} θ \sin^{2} θ = \frac{1}{4} \sin^{2} 2 θ \sin^{2} θ \\ = \frac{1}{4} \frac{1 - \cos 4 θ}{2} \frac{1 - \cos 2 θ}{2} = \frac{1}{32} (2 \cos 4 θ \cos 2 θ - 2 \cos 4 θ - 2 \cos 2 θ + 2) \\ = \frac{1}{32} (\cos 6 θ + \cos 2 θ - 2 \cos 4 θ - 2 \cos 2 θ + 2) \\ = \frac{1}{32} (\cos 6 θ - 2 \cos 4 θ - \cos 2 θ + 2) \end{aligned}

If you are allowed to use complex numbers, this is much easier: set

u = e^{i θ}

. Then we have, by Euler's formulae:

\bar{u} = e^{- i θ}, \cos θ = \frac{u + \bar{u}}{2}, \sin θ = \frac{u - \bar{u}}{2 i},

whence (note

u \bar{u} = 1

)

\begin{aligned} \cos^{2} θ \sin^{4} θ & = \frac{(u + \bar{u})^{2}}{4} \frac{(u - \bar{u})^{4}}{16} = \frac{1}{64} (u^{2} - {\bar{u}}^{2})^{2} (u - \bar{u})^{2} \\ = \frac{1}{64} (u^{4} - 2 + {\bar{u}}^{4}) (u^{2} - 2 + {\bar{u}}^{2}) \\ = \frac{1}{64} (u^{6} - 2 u^{4} + u^{2} - 2 u^{2} + 4 - 2 {\bar{u}}^{2} + {\bar{u}}^{2} - 2 {\bar{u}}^{4} + {\bar{u}}^{6}) \\ = \frac{1}{64} (u^{6} + {\bar{u}}^{6} - 2 (u^{4} + {\bar{u}}^{4}) - (u^{2} + {\bar{u}}^{2}) + 4) \\ = \frac{1}{32} (\cos 6 θ - 2 \cos 4 θ - \cos 2 θ + 2) . \end{aligned}

Prove that: cos 2 </msup> ⁡ θ sin 4