Question

Prove that:
$\cos 4\theta -\cos 4\alpha =8\left(\cos \theta -\cos \alpha \right)\left(\cos \theta +\cos \alpha \right)\left(\cos \theta -\sin \alpha \right)\left(\cos \theta +\sin \alpha \right)$

Answer 1

$$\begin{gathered} \mathrm{RHS}=8\left(\mathrm{cos\theta }-\mathrm{cos\alpha }\right)\left(\mathrm{cos\theta }+\mathrm{cos\alpha }\right)\left(\mathrm{cos\theta }-\mathrm{sin\alpha }\right)\left(\mathrm{cos\theta }+\mathrm{sin\alpha }\right) \\ =8\left({\cos }^2\mathrm{\theta }-{\cos }^2\mathrm{\alpha }\right)\left({\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\alpha }\right) \\ =8\left({\cos }^4\mathrm{\theta }-{\cos }^2\mathrm{\theta }\times {\sin }^2\mathrm{\alpha }-{\cos }^2\mathrm{\alpha }\times {\cos }^2\mathrm{\theta }+{\cos }^2\mathrm{\alpha }\times {\sin }^2\mathrm{\alpha }\right) \\ =8\left\{{\cos }^4\mathrm{\theta }-{\cos }^2\mathrm{\theta }\left({\sin }^2\mathrm{\alpha }+{\cos }^2\mathrm{\alpha }\right)+{\cos }^2\mathrm{\alpha }\times {\sin }^2\mathrm{\alpha }\right\} \\ =8\left\{{\cos }^4\mathrm{\theta }-{\cos }^2\mathrm{\theta }+{\cos }^2\mathrm{\alpha }\times \left(1-{\cos }^2\mathrm{\alpha }\right)\right\} \\ =8\left\{{\cos }^4\mathrm{\theta }-{\cos }^2\mathrm{\theta }+{\cos }^2\mathrm{\alpha }-{\cos }^4\mathrm{\alpha }\right\} \\ =8\left\{{\cos }^2\mathrm{\theta }\left({\cos }^2\mathrm{\theta }-1\right)+{\cos }^2\mathrm{\alpha }\times \left(1-{\cos }^2\mathrm{\alpha }\right)\right\} \end{gathered}$$

$$\begin{gathered} =8\left\{\frac{1}{2}{\cos }^2\mathrm{\theta }\left(2{\cos }^2\mathrm{\theta }-2\right)+\frac{1}{2}{\cos }^2\mathrm{\alpha }\times \left(2-2{\cos }^2\mathrm{\alpha }\right)\right\} \\ =8\left\{\frac{1}{2}{\cos }^2\mathrm{\theta }\left(2{\cos }^2\mathrm{\theta }-1-1\right)-\frac{1}{2}{\cos }^2\mathrm{\alpha }\times \left(2{\cos }^2\mathrm{\alpha }-1-1\right)\right\} \\ =8\left\{\frac{1}{2}{\cos }^2\mathrm{\theta }\left(\cos 2\mathrm{\theta }-1\right)-\frac{1}{2}{\cos }^2\mathrm{\alpha }\times \left(\cos 2\mathrm{\alpha }-1\right)\right\}\left(\because \cos 2\mathrm{\alpha }=2{\cos }^2\mathrm{\alpha }-1\right) \\ =8\left[\frac{1}{4}\left\{2{\cos }^2\mathrm{\theta }\left(\cos 2\mathrm{\theta }-1\right)-2{\cos }^2\mathrm{\alpha }\times \left(\cos 2\mathrm{\alpha }-1\right)\right\}\right] \\ =8\left[\frac{1}{4}\left\{\left(1+\cos 2\mathrm{\theta }\right)\left(\cos 2\mathrm{\theta }-1\right)-\left(1+\cos 2\mathrm{\alpha }\right)\left(\cos 2\mathrm{\alpha }-1\right)\right\}\right] \end{gathered}$$

$$\begin{gathered} =8\left[\frac{1}{4}\left\{{\cos }^{2}2\mathrm{\theta }-1-{\cos }^{2}2\mathrm{\alpha }+1\right\}\right] \\ =8\left[\frac{1}{8}\left\{2{\cos }^{2}2\mathrm{\theta }-2{\cos }^{2}2\mathrm{\alpha }\right\}\right] \\ =\left[\left\{\left(1+\cos 4\mathrm{\theta }\right)-\left(1+\cos 4\mathrm{\alpha }\right)\right\}\right] \\ =\left[1+\cos 4\mathrm{\theta }-1-\cos 4\mathrm{\alpha }\right] \\ =\cos 4\mathrm{\theta }-\cos 4\mathrm{\alpha }=\mathrm{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$