Question

prove lhs equals rhs :cos theta *cos theta/2-cos 3theta*cos 9theta/2=sin 7theta/2*sin 4theta

Answer 1

$$\begin{gathered} \mathrm{Dear}\mathrm{Student}, \\ \mathrm{Prove}\mathrm{that}:\mathrm{cos\theta cos}\frac{\mathrm{\theta }}{2}-\cos 3\mathrm{\theta cos}\frac{9\mathrm{\theta }}{2}=\sin 4\mathrm{\theta sin}\frac{7\mathrm{\theta }}{2} \\ \mathrm{LHS}: \\ \mathrm{cos\theta cos}\frac{\mathrm{\theta }}{2}-\cos 3\mathrm{\theta cos}\frac{9\mathrm{\theta }}{2} \\ =\left[\frac{\cos (\mathrm{\theta }+{\displaystyle \frac{\mathrm{\theta }}{2}})+\cos (\mathrm{\theta }-{\displaystyle \frac{\mathrm{\theta }}{2}})}{2}\right]-\left[\frac{\cos ({\displaystyle \frac{9\mathrm{\theta }}{2}}+3\mathrm{\theta })+\cos ({\displaystyle \frac{9\mathrm{\theta }}{2}}-3\mathrm{\theta })}{2}\right]\{\mathrm{Because},\mathrm{cosxcosy}=\frac{\cos (\mathrm{x}+\mathrm{y})+\cos (\mathrm{x}-\mathrm{y})}{2}\} \\ =\frac{1}{2}\times [\cos \frac{3\mathrm{\theta }}{2}+\cos \frac{\mathrm{\theta }}{2}-(\cos \frac{15\mathrm{\theta }}{2}+\cos \frac{3\mathrm{\theta }}{2}\left)\right] \\ =\frac{1}{2}\times [\cos \frac{3\mathrm{\theta }}{2}+\cos \frac{\mathrm{\theta }}{2}-\cos \frac{15\mathrm{\theta }}{2}-\cos \frac{3\mathrm{\theta }}{2}] \\ =\frac{1}{2}\times [\cos \frac{\mathrm{\theta }}{2}-\cos \frac{15\mathrm{\theta }}{2}] \\ =-\frac{1}{2}\times [\cos \frac{15\mathrm{\theta }}{2}-\cos \frac{\mathrm{\theta }}{2}] \\ =-\frac{1}{2}\times [-2\sin (\frac{\frac{15\mathrm{\theta }}{2}+\frac{\mathrm{\theta }}{2}}{2}\left)\sin \right(\frac{\frac{15\mathrm{\theta }}{2}-\frac{\mathrm{\theta }}{2}}{2}) \\ =\sin (\frac{8\mathrm{\theta }}{2}\left)\sin \right(\frac{7\mathrm{\theta }}{2}) \\ \mathbf{=}\mathbf{sin}\mathbf{4}\mathbf{\theta sin}\frac{\mathbf{7}\mathbf{\theta }}{\mathbf{2}}\mathbf{=}\mathbf{RHS}\mathbf{,}\mathbf{}\mathbf{Hence}\mathbf{}\mathbf{Proved} \\ \mathrm{Regards}. \end{gathered}$$