Question

Prove That $\mathrm{Sin}50°-\mathrm{Sin}70°+\mathrm{Sin}10°=0.$

Answer 1

As we have $\mathrm{Sin}50°-\mathrm{Sin}70°+\mathrm{Sin}10°=0.$

Taking LHS

$$\begin{gathered} \sin 50°–\sin 70°+\sin 10° \\ \Rightarrow \sin (60°–10°)–\sin (60°+10°)+\sin 10° \\ \mathrm{We}\mathrm{know}\mathrm{that}\sin (\mathrm{A}°–\mathrm{B}°)–\sin (\mathrm{A}°+\mathrm{B}°)=–2\cos \mathrm{A}\sin \mathrm{B} \\ \Rightarrow –2\sin 10°\cos 60°+\sin 10° \\ \Rightarrow 0 \\ \mathrm{Therefore},\sin 50°–\sin 70°+\sin 10°=0. \end{gathered}$$

Hence proved.