Question Number 211496 by RojaTaniya last updated on 11/Sep/24
Answered by som(math1967) last updated on 11/Sep/24
$$\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{sin}\mathrm{20}{sin}\mathrm{40}+\mathrm{2}{sin}\mathrm{20}{sin}\mathrm{60}+\mathrm{2}{sin}\mathrm{20}{sin}\mathrm{80}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{20}−{cos}\mathrm{60}+{cos}\mathrm{40}−{cos}\mathrm{80}+{cos}\mathrm{60}−{cos}\mathrm{100}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{20}+{cos}\mathrm{40}−{cos}\mathrm{80}+{cos}\mathrm{80}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{cos}\mathrm{30}{cos}\mathrm{10} \\ $$$$={sin}\mathrm{60}{sin}\mathrm{80} \\ $$$$={sin}\left(\mathrm{180}−\mathrm{60}\right){sin}\left(\mathrm{180}−\mathrm{80}\right) \\ $$$$={sin}\mathrm{120}{sin}\mathrm{100} \\ $$$$ \\ $$