prove that cos theta × cos theta/2 - cos 3 theta × cos 9 theta/2 = sin 7 theta × sin 8 theta

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{LHS}=\mathrm{cos\theta cos}\left(\frac{\mathrm{\theta }}{2}\right)-\cos 3\mathrm{\theta cos}\left(\frac{9\mathrm{\theta }}{2}\right) \\ =\frac{1}{2}\left[\cos \left(\mathrm{\theta }-\frac{\mathrm{\theta }}{2}\right)+\cos \left(\mathrm{\theta }+\frac{\mathrm{\theta }}{2}\right)\right]-\frac{1}{2}\left[\cos \left(3\mathrm{\theta }-\frac{9\mathrm{\theta }}{2}\right)+\cos \left(3\mathrm{\theta }+\frac{9\mathrm{\theta }}{2}\right)\right] \\ =\frac{1}{2}\left[\cos \left(\frac{\mathrm{\theta }}{2}\right)+\cos \left(\frac{3\mathrm{\theta }}{2}\right)\right]-\frac{1}{2}\left[\cos \left(\frac{-3\mathrm{\theta }}{2}\right)+\cos \left(\frac{15\mathrm{\theta }}{2}\right)\right] \\ =\frac{1}{2}\left[\cos \left(\frac{\mathrm{\theta }}{2}\right)+\cos \left(\frac{3\mathrm{\theta }}{2}\right)-\cos \left(\frac{3\mathrm{\theta }}{2}\right)-\cos \left(\frac{15\mathrm{\theta }}{2}\right)\right] \\ =\frac{1}{2}\left[\cos \left(\frac{\mathrm{\theta }}{2}\right)-\cos \left(\frac{15\mathrm{\theta }}{2}\right)\right] \\ =\frac{1}{2}\left[2\sin \left(\frac{\frac{\mathrm{\theta }}{2}+\frac{15\mathrm{\theta }}{2}}{2}\right)\sin \left(\frac{\frac{\mathrm{\theta }}{2}-\frac{15\mathrm{\theta }}{2}}{2}\right)\right] \\ =\sin \left(\frac{16\mathrm{\theta }}{4}\right)\sin \left(\frac{-14\mathrm{\theta }}{4}\right) \\ =-\sin 4\mathrm{\theta sin}\left(\frac{7\mathrm{\theta }}{2}\right) \\ \ne \mathrm{RHS} \end{gathered}$$

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