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# Prove that: (i) cos 10° cos 30° cos 50° cos 70° = $\frac{3}{16}$ (ii) cos 40° cos 80° cos 160° = $-\frac{1}{8}$ (iii) sin 20° sin 40° sin 80° = $\frac{\sqrt{3}}{8}$ (iv) cos 20° cos 40° cos 80° = $\frac{1}{8}$ (v) tan 20° tan 40° tan 60° tan 80° = 3 (vi) tan 20° tan 30° tan 40° tan 80° = 1 (vii) sin 10° sin 50° sin 60° sin 70° = $\frac{\sqrt{3}}{16}$ (viii) sin 20° sin 40° sin 60° sin 80° = $\frac{3}{16}$

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Solution

## (i) $\mathrm{LHS}=\mathrm{cos}10°\mathrm{cos}30°\mathrm{cos}50°\mathrm{cos}70°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[2\mathrm{cos}10°\mathrm{cos}50°\right]\mathrm{cos}30°\mathrm{cos}70°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}\left(10°+50°\right)+\mathrm{cos}\left(10°-50°\right)\right]\mathrm{cos}30°\mathrm{cos}70°\left\{\because 2\mathrm{cos}A\mathrm{cos}B=\mathrm{cos}\left(A+B\right)-\mathrm{cos}\left(A-B\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}60°+\mathrm{cos}\left(-40°\right)\right]\mathrm{cos}30°\mathrm{cos}70°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\frac{1}{2}+\mathrm{cos}40°\right]\left(\frac{\sqrt{3}}{2}\right)×\mathrm{cos}70°$ $=\frac{\sqrt{3}}{4}\mathrm{cos}70°\left[\frac{1}{2}+\mathrm{cos}40°\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\mathrm{cos}70°+\frac{\sqrt{3}}{4}\left[\mathrm{cos}70°\mathrm{cos}40°\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\mathrm{cos}70°+\frac{\sqrt{3}}{8}\left[2\mathrm{cos}70°\mathrm{cos}40°\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\mathrm{cos}70°+\frac{\sqrt{3}}{8}\left[\mathrm{cos}\left(70°+40°\right)+\mathrm{cos}\left(70°-40°\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\mathrm{cos}70°+\frac{\sqrt{3}}{8}\left[\mathrm{cos}110°+\mathrm{cos}30°\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\mathrm{cos}70°+\frac{\sqrt{3}}{8}\left[\mathrm{cos}\left(180°-70°\right)+\frac{\sqrt{3}}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{2}\mathrm{cos}70°-\frac{\sqrt{3}}{8}\mathrm{cos}70°+\frac{3}{16}\left[\because \mathrm{cos}\left(180°-70°\right)=-\mathrm{cos}70°\right]\phantom{\rule{0ex}{0ex}}=\frac{3}{16}=\mathrm{RHS}$ (ii) $\mathrm{LHS}=\mathrm{cos}40°\mathrm{cos}80°\mathrm{cos}160°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[2\mathrm{cos}40°\mathrm{cos}80°\right]\mathrm{cos}160°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}\left(40°+80°\right)+\mathrm{cos}\left(40°-80°\right)\right]\mathrm{cos}160°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}120°+\mathrm{cos}\left(-40°\right)\right]\mathrm{cos}160°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{cos}\left(160°\right)\left[-\frac{1}{2}+\mathrm{cos}40°\right]\phantom{\rule{0ex}{0ex}}=-\frac{1}{4}\mathrm{cos}160°+\frac{1}{2}\mathrm{cos}160°\mathrm{cos}40°$ $=-\frac{1}{4}\mathrm{cos}160°+\frac{1}{4}\left[2\mathrm{cos}160°\mathrm{cos}40°\right]\phantom{\rule{0ex}{0ex}}=-\frac{1}{4}\mathrm{cos}160°+\frac{1}{4}\left[\mathrm{cos}\left(160°+40°\right)+\mathrm{cos}\left(160°-40°\right)\right]\phantom{\rule{0ex}{0ex}}=-\frac{1}{4}\mathrm{cos}160°+\frac{1}{4}\left[\mathrm{cos}200°+\mathrm{cos}120°\right]\phantom{\rule{0ex}{0ex}}=-\frac{1}{4}\mathrm{cos}160°+\frac{1}{4}\left[\mathrm{cos}\left(360°-160°\right)-\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}=-\frac{1}{4}\mathrm{cos}160°+\frac{1}{4}\mathrm{cos}160°-\frac{1}{8}\left[\because \mathrm{cos}\left(360°-160°\right)=\mathrm{cos}160°\right]\phantom{\rule{0ex}{0ex}}=-\frac{1}{8}=\mathrm{RHS}$ (iii) $\mathrm{LHS}=\mathrm{sin}20°\mathrm{sin}40°\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[2\mathrm{sin}20°\mathrm{sin}40°\right]\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}\left(20°-40°\right)-\mathrm{cos}\left(20°+40°\right)\right]\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}20°-\frac{1}{2}\right]\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{sin}80°\left[\mathrm{cos}20°-\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{sin}80°\mathrm{cos}20°-\frac{1}{4}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{sin}\left(90°-10°\right)\mathrm{cos}20°-\frac{1}{4}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{cos}10°\mathrm{cos}20°-\frac{1}{4}\mathrm{sin}80°$ $=\frac{1}{4}\left[2\mathrm{cos}10°c\mathrm{os}20°\right]-\frac{1}{4}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[\mathrm{cos}\left(10°+20°\right)+\mathrm{cos}\left(10°-20°\right)\right]-\frac{1}{4}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[\mathrm{cos}30°+\mathrm{cos}\left(-10°\right)\right]-\frac{1}{4}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[\mathrm{cos}30°+\mathrm{cos}\left(90°-80°\right)\right]-\frac{1}{4}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}+\frac{1}{4}\mathrm{sin}80°-\frac{1}{4}\mathrm{sin}80°\left[\because \mathrm{cos}\left(90°-80°\right)=\mathrm{sin}80°\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}=\mathrm{RHS}$ (iv) $\mathrm{LHS}=\mathrm{cos}20°\mathrm{cos}40°\mathrm{cos}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[2\mathrm{cos}20°\mathrm{cos}40°\right]\mathrm{cos}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}\left(20°+40°\right)+\mathrm{cos}\left(20°-40°\right)\right]\mathrm{cos}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}60°+\mathrm{cos}\left(-20°\right)\right]\mathrm{cos}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{cos}80°\left[\frac{1}{2}+\mathrm{cos}20°\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}cos80°+\frac{1}{2}\mathrm{cos}80°\mathrm{cos}20°$ $=\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[2\mathrm{cos}80°\mathrm{cos}20°\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[\mathrm{cos}\left(80°+20°\right)+\mathrm{cos}\left(80°-20°\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[\mathrm{cos}100°+\mathrm{cos}60°\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[\mathrm{cos}\left(180°-80°\right)+\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\mathrm{cos}80°-\frac{1}{4}\mathrm{cos}80°+\frac{1}{8}\left\{\because \mathrm{cos}\left(180°-80°\right)=-\mathrm{cos}80°\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{8}=\mathrm{RHS}$ (v) LHS = tan 20° tan 40° tan 60° tan 80° $=\mathrm{tan}60°\frac{\mathrm{sin}20°\mathrm{sin}40°\mathrm{sin}80°}{\mathrm{cos}20°\mathrm{cos}40°\mathrm{cos}80°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{2}\left[2\mathrm{sin}20°\mathrm{sin}40°\right]\mathrm{sin}80°}{\frac{1}{2}\left[2\mathrm{cos}20°\mathrm{cos}40°\right]\mathrm{cos}80°}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{2}\left[\mathrm{cos}\left(20°-40°\right)-\mathrm{cos}\left(20°+40°\right)\right]\mathrm{sin}80°}{\frac{1}{2}\left[\mathrm{cos}\left(20°+40°\right)+\mathrm{cos}\left(20°-40°\right)\right]\mathrm{cos}80°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{2}\left[\mathrm{cos}\left(-20°\right)-\mathrm{cos}60°\right]\mathrm{sin}80°}{\frac{1}{2}\left[\mathrm{cos}60°+\mathrm{cos}\left(-20°\right)\right]\mathrm{cos}80°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{2}\mathrm{sin}80°\left[\mathrm{cos}20°-\frac{1}{2}\right]}{\frac{1}{2}\mathrm{cos}80°\left[\frac{1}{2}+\mathrm{cos}20°\right]}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{2}\mathrm{sin}80°\mathrm{cos}20°-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{2}\mathrm{cos}80°\mathrm{cos}20°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{2}\mathrm{sin}\left(90°-10°\right)\mathrm{cos}20°-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{2}\mathrm{cos}80°\mathrm{cos}20°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{2}\mathrm{cos}10°\mathrm{cos}20°-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{2}\mathrm{cos}80°\mathrm{cos}20°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ $=\sqrt{3}×\frac{\frac{1}{4}\left[2\mathrm{cos}10°\mathrm{cos}20°\right]-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[2\mathrm{cos}80°\mathrm{cos}20°\right]}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{4}\left[\mathrm{cos}\left(10°+20°\right)+\mathrm{cos}\left(10°-20°\right)\right]-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[\mathrm{cos}\left(80°+20°\right)+\mathrm{cos}\left(80°-20°\right)\right]}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{4}\left[\mathrm{cos}30°+\mathrm{cos}\left(-10°\right)\right]-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[\mathrm{cos}100°+\mathrm{cos}60°\right]}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{4}\left[\mathrm{cos}30°+\mathrm{cos}\left(90°-80°\right)\right]-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[\mathrm{cos}\left(180°-80°\right)+\frac{1}{2}\right]}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{\sqrt{3}}{8}+\frac{1}{4}\mathrm{sin}80°-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°-\frac{1}{4}\mathrm{cos}80°+\frac{1}{8}}\left[\mathrm{cos}\left(90°-80°\right)=\mathrm{sin}80°,\mathrm{and}\mathrm{cos}\left(180°-80°\right)=-\mathrm{cos}\left(80°\right)\right]\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{\sqrt{3}}{8}}{\frac{1}{8}}\phantom{\rule{0ex}{0ex}}=3=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (vi) LHS = tan 20° tan 30° tan 60° tan 80° $=\mathrm{tan}30°\frac{\mathrm{sin}20°\mathrm{sin}40°\mathrm{sin}80°}{\mathrm{cos}20°\mathrm{cos}40°\mathrm{cos}80°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{3}}×\frac{\frac{1}{2}\left[2\mathrm{sin}20°\mathrm{sin}40°\right]\mathrm{sin}80°}{\frac{1}{2}\left[2\mathrm{cos}20°\mathrm{cos}40°\right]\mathrm{cos}80°}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{3}}×\frac{\frac{1}{2}\left[\mathrm{cos}\left(20°-40°\right)-\mathrm{cos}\left(20°+40°\right)\right]\mathrm{sin}80°}{\frac{1}{2}\left[\mathrm{cos}\left(20°+40°\right)+\mathrm{cos}\left(20°-40°\right)\right]\mathrm{cos}80°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{3}}×\frac{\frac{1}{2}\left[\mathrm{cos}20°-\frac{1}{2}\right]\mathrm{sin}80°}{\frac{1}{2}\left[\mathrm{cos}60°+\mathrm{cos}\left(-20°\right)\right]\mathrm{cos}80°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{3}}×\frac{\frac{1}{2}\mathrm{sin}80°\left[\mathrm{cos}20°-\frac{1}{2}\right]}{\frac{1}{2}\mathrm{cos}80°\left[\frac{1}{2}+\mathrm{cos}20°\right]}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{3}}×\frac{\frac{1}{2}\mathrm{sin}80°\mathrm{cos}20°-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{2}\mathrm{cos}80°\mathrm{cos}20°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{3}}×\frac{\frac{1}{2}\mathrm{sin}\left(90°-10°\right)\mathrm{cos}20°-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{2}\mathrm{cos}80°\mathrm{cos}20°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{3}}×\frac{\frac{1}{2}\mathrm{cos}10°\mathrm{cos}20°-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{2}\mathrm{cos}80°\mathrm{cos}20°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ $=\sqrt{3}×\frac{\frac{1}{4}\left[2\mathrm{cos}10°\mathrm{cos}20°\right]-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[2\mathrm{cos}80°\mathrm{cos}20°\right]}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{4}\left[\mathrm{cos}\left(10°+20°\right)+\mathrm{cos}\left(10°-20°\right)\right]-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[\mathrm{cos}\left(80°+20°\right)+\mathrm{cos}\left(80°-20°\right)\right]}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{4}\left[\mathrm{cos}30°+\mathrm{cos}\left(-10°\right)\right]-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[\mathrm{cos}100°+\mathrm{cos}60°\right]}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{1}{4}\left[\mathrm{cos}30°+\mathrm{cos}\left(90°-80°\right)\right]-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°+\frac{1}{4}\left[\mathrm{cos}\left(180°-80°\right)+\frac{1}{2}\right]}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{\sqrt{3}}{8}+\frac{1}{4}\mathrm{sin}80°-\frac{1}{4}\mathrm{sin}80°}{\frac{1}{4}\mathrm{cos}80°-\frac{1}{4}\mathrm{cos}\left(80°\right)+\frac{1}{8}}\left\{\because \mathrm{cos}\left(90°-80°\right)=\mathrm{sin}80°,\mathrm{cos}\left(180°-80°\right)=-\mathrm{cos}80°\right\}\phantom{\rule{0ex}{0ex}}=\sqrt{3}×\frac{\frac{\sqrt{3}}{8}}{\frac{1}{8}}\phantom{\rule{0ex}{0ex}}=3=\mathrm{RHS}$ (vii) $\mathrm{LHS}=\mathrm{sin}10°\mathrm{sin}50°\mathrm{sin}60°\mathrm{sin}70°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{sin}60°\left[2\mathrm{sin}10°\mathrm{sin}50°\right]\mathrm{sin}70°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\frac{\sqrt{3}}{2}\left[\mathrm{cos}\left(10°-50°\right)-\mathrm{cos}\left(10°+50°\right)\right]\mathrm{sin}70°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\left[\mathrm{cos}\left(-40°\right)-\frac{1}{2}\right]\mathrm{sin}70°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\mathrm{sin}70°\left[\mathrm{cos}40°-\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\mathrm{sin}70°\mathrm{cos}40°-\frac{\sqrt{3}}{8}\mathrm{sin}70°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\mathrm{sin}\left(90°-20°\right)\mathrm{cos}40°-\frac{\sqrt{3}}{8}\mathrm{sin}70°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\mathrm{cos}20°\mathrm{cos}40°-\frac{\sqrt{3}}{8}\mathrm{sin}70°$ $=\frac{\sqrt{3}}{8}\left[2\mathrm{cos}20°\mathrm{cos}40°\right]-\frac{\sqrt{3}}{8}\mathrm{sin}70°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\left[\mathrm{cos}\left(20°+40°\right)+\mathrm{cos}\left(20°-40°\right)\right]-\frac{\sqrt{3}}{8}\mathrm{sin}70°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\left[\mathrm{cos}60°+\mathrm{cos}\left(-20°\right)\right]-\frac{\sqrt{3}}{8}\mathrm{sin}70°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\left[\mathrm{cos}60°+\mathrm{cos}\left(90°-70°\right)\right]-\frac{\sqrt{3}}{8}\mathrm{sin}70°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{16}+\frac{\sqrt{3}}{8}\mathrm{sin}70°-\frac{\sqrt{3}}{8}\mathrm{sin}70°\left[\because \mathrm{cos}\left(90°-70°\right)=\mathrm{sin}70°\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{16}=\mathrm{RHS}$ (viii) $\mathrm{LHS}=\mathrm{sin}20°\mathrm{sin}40°\mathrm{sin}60°\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{sin}60°\left[2\mathrm{sin}20°\mathrm{sin}40°\right]\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\frac{\sqrt{3}}{2}\left[\mathrm{cos}\left(20°-40°\right)-\mathrm{cos}\left(20°+40°\right)\right]\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\left[\mathrm{cos}20°-\frac{1}{2}\right]\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\mathrm{sin}80°\left[\mathrm{cos}20°-\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\mathrm{sin}80°\mathrm{cos}20°-\frac{\sqrt{3}}{8}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\mathrm{sin}\left(90°-10°\right)\mathrm{cos}20°-\frac{\sqrt{3}}{8}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\mathrm{cos}10°\mathrm{cos}20°-\frac{\sqrt{3}}{8}\mathrm{sin}\left(80°\right)$ $=\frac{\sqrt{3}}{8}\left[2\mathrm{cos}10°\mathrm{cos}20°\right]-\frac{\sqrt{3}}{8}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\left[\mathrm{cos}\left(10°+20°\right)+\mathrm{cos}\left(10°-20°\right)\right]-\frac{\sqrt{3}}{8}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\left[\mathrm{cos}30°+\mathrm{cos}\left(-10°\right)\right]-\frac{\sqrt{3}}{8}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{8}\left[\mathrm{cos}30°+\mathrm{cos}\left(90°-80°\right)\right]-\frac{\sqrt{3}}{8}\mathrm{sin}80°\phantom{\rule{0ex}{0ex}}=\frac{3}{16}+\frac{\sqrt{3}}{8}\mathrm{sin}80°-\frac{\sqrt{3}}{8}\mathrm{sin}80°\left[\because \mathrm{cos}\left(90°-80°\right)=\mathrm{sin}80°\right]\phantom{\rule{0ex}{0ex}}=\frac{3}{16}=\mathrm{RHS}$

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