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Question

# Prove that: (i) cos 55° + cos 65° + cos 175° = 0 (ii) sin 50° − sin 70° + sin 10° = 0 (iii) cos 80° + cos 40° − cos 20° = 0 (iv) cos 20° + cos 100° + cos 140° = 0 (v) $\mathrm{sin}\frac{5\mathrm{\pi }}{18}-\mathrm{cos}\frac{4\mathrm{\pi }}{9}=\sqrt{3}\mathrm{sin}\frac{\mathrm{\pi }}{9}$ (vi) $\mathrm{cos}\frac{\mathrm{\pi }}{12}-\mathrm{sin}\frac{\mathrm{\pi }}{12}=\frac{1}{\sqrt{2}}$ (vii) sin 80° − cos 70° = cos 50° (viii) sin 51° + cos 81° = cos 21°

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Solution

## (i) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}\mathrm{cos}55°+\mathrm{cos}65°+\mathrm{cos}175°\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}\left(\frac{55°+65°}{2}\right)\mathrm{cos}\left(\frac{55°-65°}{2}\right)+\mathrm{cos}175°\left\{\because \mathrm{cos}A+\mathrm{cos}B=2\mathrm{cos}\left(\frac{A+B}{2}\right)\mathrm{cos}\left(\frac{A-B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}60°\mathrm{cos}\left(-5°\right)+\mathrm{cos}175°\phantom{\rule{0ex}{0ex}}=2×\frac{1}{2}\mathrm{cos}5°+\mathrm{cos}175°\phantom{\rule{0ex}{0ex}}=\mathrm{cos}5°+\mathrm{cos}175°\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}\left(\frac{5°+175°}{2}\right)\mathrm{cos}\left(\frac{5°-175°}{2}\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}90°\mathrm{cos}85°\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}.$ (ii) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}\mathrm{sin}50°-\mathrm{sin}70°+\mathrm{sin}10°\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{50°-70°}{2}\right)\mathrm{cos}\left(\frac{50°+70°}{2}\right)+\mathrm{sin}10°\left\{\because \mathrm{si}nA-\mathrm{sin}B=2\mathrm{si}n\left(\frac{A-B}{2}\right)c\mathrm{os}\left(\frac{A+B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(-10°\right)\mathrm{co}s60°+\mathrm{sin}10°\phantom{\rule{0ex}{0ex}}=2×\frac{1}{2}\mathrm{sin}\left(-10°\right)+\mathrm{sin}10°\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}10°+\mathrm{sin}10°\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}.$ (iii) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}\mathrm{cos}80°+\mathrm{cos}40°-\mathrm{cos}20°\phantom{\rule{0ex}{0ex}}=2\mathrm{co}s\left(\frac{80°+40°}{2}\right)\mathrm{cos}\left(\frac{80°-40°}{2}\right)-\mathrm{cos}20°\left\{\because \mathrm{cos}A+\mathrm{cos}B=2\mathrm{cos}\left(\frac{A+B}{2}\right)\mathrm{cos}\left(\frac{A-B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}60°\mathrm{cos}20°-\mathrm{cos}20°\phantom{\rule{0ex}{0ex}}=2×\frac{1}{2}\mathrm{cos}20°-\mathrm{cos}20°\phantom{\rule{0ex}{0ex}}=\mathrm{cos}20°-\mathrm{cos}20°\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}.$ (iv) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}\mathrm{cos}20°+\mathrm{cos}100°+\mathrm{cos}140°\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}\left(\frac{20°+100°}{2}\right)\mathrm{cos}\left(\frac{20°-100°}{2}\right)+\mathrm{cos}140°\left\{\because \mathrm{cos}A+\mathrm{cos}B=2\mathrm{cos}\left(\frac{A+B}{2}\right)\mathrm{cos}\left(\frac{A-B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}60°\mathrm{cos}\left(-40°\right)+\mathrm{cos}140°\phantom{\rule{0ex}{0ex}}=2×\frac{1}{2}\mathrm{cos}40°+\mathrm{cos}140°\phantom{\rule{0ex}{0ex}}=\mathrm{cos}40°+\mathrm{cos}140°\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}\left(\frac{40°+140°}{2}\right)\mathrm{cos}\left(\frac{40°-140°}{2}\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}90°\mathrm{cos}50°\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}.$ (v) $\mathrm{LHS}=\mathrm{sin}\left(\frac{5\mathrm{\pi }}{18}\right)-\mathrm{cos}\frac{4\mathrm{\pi }}{9}\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left(\frac{5\mathrm{\pi }}{18}\right)-\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{18}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left(\frac{5\mathrm{\pi }}{18}\right)-\mathrm{sin}\left(\frac{\mathrm{\pi }}{18}\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{\frac{5\mathrm{\pi }}{18}-\frac{\mathrm{\pi }}{18}}{2}\right)\mathrm{cos}\left(\frac{\frac{5\mathrm{\pi }}{18}+\frac{\mathrm{\pi }}{18}}{2}\right)\left[\because \mathrm{sin}A-\mathrm{sin}B=2\mathrm{sin}\left(\frac{A-B}{2}\right)\mathrm{cos}\left(\frac{A+B}{2}\right)\right]\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{\mathrm{\pi }}{9}\right)\mathrm{cos}\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{\mathrm{\pi }}{9}\right)\mathrm{cos}\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}=2×\frac{\sqrt{3}}{2}\mathrm{sin}\left(\frac{\mathrm{\pi }}{9}\right)\phantom{\rule{0ex}{0ex}}=\sqrt{3}\mathrm{sin}\left(\frac{\mathrm{\pi }}{9}\right)=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}.$ (vi) $\mathrm{LHS}=\mathrm{cos}\frac{\mathrm{\pi }}{12}-\mathrm{sin}\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-\frac{5\mathrm{\pi }}{12}\right)-\mathrm{sin}\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left(\frac{5\mathrm{\pi }}{12}\right)-\mathrm{sin}\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{\frac{5\mathrm{\pi }}{12}-\frac{\mathrm{\pi }}{12}}{2}\right)\mathrm{cos}\left(\frac{\frac{5\mathrm{\pi }}{12}+\frac{\mathrm{\pi }}{12}}{2}\right)\left\{\because \mathrm{sin}A-\mathrm{sin}B=2\mathrm{sin}\left(\frac{A-B}{2}\right)\mathrm{cos}\left(\frac{A+B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{\mathrm{\pi }}{6}\right)\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}=2×\frac{1}{2}×\frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}.$ (vii) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}s\mathrm{in}80°-\mathrm{cos}70°\phantom{\rule{0ex}{0ex}}=\mathrm{sin}80°-\mathrm{cos}\left(90°-20°\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}80°-\mathrm{sin}20°\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{80°-20°}{2}\right)\mathrm{cos}\left(\frac{80°+20°}{2}\right)\left\{\because \mathrm{sin}A-\mathrm{sin}B=2\mathrm{sin}\left(\frac{A-B}{2}\right)\mathrm{cos}\left(\frac{A+B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}30°\mathrm{cos}50°\phantom{\rule{0ex}{0ex}}=2×\frac{1}{2}\mathrm{cos}50°\phantom{\rule{0ex}{0ex}}=\mathrm{cos}50°\phantom{\rule{0ex}{0ex}}=\mathrm{RH}S\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}.$ (viii) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}\mathrm{sin}51°+\mathrm{cos}81°\phantom{\rule{0ex}{0ex}}=\mathrm{sin}51°+\mathrm{cos}\left(90°-9°\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}51°+\mathrm{sin}9°\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{51°+9°}{2}\right)\mathrm{cos}\left(\frac{51°-9°}{2}\right)\left\{\because \mathrm{sin}A+\mathrm{sin}B=2\mathrm{sin}\left(\frac{A+B}{2}\right)\mathrm{cos}\left(\frac{A-B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}30°\mathrm{cos}21°\phantom{\rule{0ex}{0ex}}=2×\frac{1}{2}\mathrm{cos}\left(21°\right)\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left(21°\right)\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}.$

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