Question

$$\begin{gathered} (1+\tan \beta \tan 2\beta )\sin 2\beta =\tan 2\beta \\ (1+\tan 2{\beta }^2)\sin 2\beta \\ sec2{\beta }^2\times {\sin }^2\beta \\ \frac{1}{\cos 2\beta }\times {\sin }^2\beta \\ {\tan }^{2\beta } \\ \sin 2\alpha +\sin 4\alpha -\sin 6\alpha =4\sin \alpha \sin 2\alpha \sin 3\alpha \end{gathered}$$
solve this

Answer 1

Dear student
$$\begin{gathered} \mathrm{To}\mathrm{prove}:\sin 2\mathrm{\alpha }+\sin 4\mathrm{\alpha }-\sin 6\mathrm{\alpha }=4\mathrm{sin\alpha }\sin 2\mathrm{\alpha sin}3\mathrm{\alpha } \\ \mathrm{Consider},\mathrm{LHS} \\ \sin 4\mathrm{\alpha }+\sin 2\mathrm{\alpha }-\sin 6\mathrm{\alpha } \\ =2\sin \left(\frac{4\mathrm{\alpha }+2\mathrm{\alpha }}{2}\right)\cos \left(\frac{4\mathrm{\alpha }-2\mathrm{\alpha }}{2}\right)-\sin 6\mathrm{\alpha } \\ =2\sin 3\mathrm{\alpha }\mathrm{cos\alpha }-\sin 6\mathrm{\alpha } \\ =2\sin 3\mathrm{\alpha }\mathrm{cos\alpha }-2\sin 3\mathrm{\alpha }\cos 3\mathrm{\alpha } \\ =2\sin 3\mathrm{\alpha }\left[\mathrm{cos\alpha }-\cos 3\mathrm{\alpha }\right] \\ =2\sin 3\mathrm{\alpha }\left[-2\sin \left(\frac{\mathrm{\alpha }+3\mathrm{\alpha }}{2}\right)\sin \left(\frac{\mathrm{\alpha }-3\mathrm{\alpha }}{2}\right)\right] \\ =2\sin 3\mathrm{\alpha }\left[-2\sin 2\mathrm{\alpha }\sin \left(-\mathrm{\alpha }\right)\right] \\ =2\sin 3\mathrm{\alpha }\left[-2\sin 2\mathrm{\alpha }-\mathrm{sin\alpha }\right] \\ =4\mathrm{sin\alpha }\sin 2\mathrm{\alpha }\sin 3\mathrm{\alpha } \\ =\mathrm{RHS} \\ \mathrm{identitites}\mathrm{used}: \\ 1)\sin \left(-\mathrm{x}\right)=-\mathrm{sinx} \\ 2)\sin 2\mathrm{x}=2\mathrm{sinxcosx} \\ 3)\mathrm{sinA}+\mathrm{sinB}=2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \\ 4)\mathrm{cosA}-\mathrm{cosB}=-2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \end{gathered}$$
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