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Question

# Prove that: (i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A (ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A (iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos $\frac{A}{2}$ cos $\frac{3A}{2}$ sin 3A (iv) sin 3A + sin 2A − sin A = 4 sin A cos $\frac{A}{2}$ cos $\frac{3A}{2}$ (v) cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = −$\frac{3}{4}$ (vi) $\mathrm{sin}\frac{\mathrm{x}}{2}\mathrm{sin}\frac{7x}{2}+\mathrm{sin}\frac{3x}{2}\mathrm{sin}\frac{11x}{2}=\mathrm{sin}2x\mathrm{sin}5x.$ (vii) $\mathrm{cos}x\mathrm{cos}\frac{x}{2}-\mathrm{cos}3x\mathrm{cos}\frac{9x}{2}=\mathrm{sin}7x\mathrm{sin}8x$

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Solution

## (i) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}\mathrm{cos}3A+c\mathrm{os}5A+\mathrm{cos}7A+\mathrm{cos}15A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}\left(\frac{3A+5A}{2}\right)\mathrm{cos}\left(\frac{3A-5A}{2}\right)+2\mathrm{cos}\left(\frac{7A+15A}{2}\right)\mathrm{cos}\left(\frac{7A-15A}{2}\right)\left\{\because \mathrm{cos}A+\mathrm{cos}B=2\mathrm{cos}\left(\frac{A+B}{2}\right)\mathrm{cos}\left(\frac{A-B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}4A\mathrm{cos}\left(-A\right)+2\mathrm{cos}11A\mathrm{cos}\left(-4A\right)$ $=2\mathrm{cos}4A\mathrm{cos}A+2\mathrm{cos}11A\mathrm{cos}4A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}4A\left\{\mathrm{cos}A+\mathrm{cos}11A\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}4A×\left\{2\mathrm{cos}\left(\frac{A+11A}{2}\right)\mathrm{cos}\left(\frac{A-11A}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=4\mathrm{cos}4A\mathrm{cos}6A\mathrm{cos}\left(-5A\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=4\mathrm{cos}4A\mathrm{cos}5A\mathrm{cos}6A\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}$ (ii) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}\mathrm{cos}A+\mathrm{cos}3A+\mathrm{cos}5A+\mathrm{cos}7A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}\left(\frac{A+3A}{2}\right)\mathrm{cos}\left(\frac{A-3A}{2}\right)+2\mathrm{cos}\left(\frac{5A+7A}{2}\right)\mathrm{cos}\left(\frac{5A-7A}{2}\right)\left\{\because \mathrm{cos}A+\mathrm{cos}B=2\mathrm{cos}\left(\frac{A+B}{2}\right)\mathrm{cos}\left(\frac{A-B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}2A\mathrm{cos}\left(-A\right)+2\mathrm{cos}6A\mathrm{cos}\left(-A\right)$ $=2\mathrm{cos}2A\mathrm{cos}A+2\mathrm{cos}6A\mathrm{cos}A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}A\left(\mathrm{cos}2A+\mathrm{cos}6A\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}A×2\mathrm{cos}\left(\frac{2A+6A}{2}\right)\mathrm{cos}\left(\frac{2A-6A}{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=4\mathrm{cos}A\mathrm{cos}4A\mathrm{cos}\left(-2A\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=4\mathrm{cos}A\mathrm{cos}2A\mathrm{cos}4A\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}.$ (iii) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}\mathrm{sin}A+\mathrm{sin}2A+\mathrm{sin}4A+\mathrm{sin}5A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{A+2A}{2}\right)\mathrm{cos}\left(\frac{A-2A}{2}\right)+2\mathrm{sin}\left(\frac{4A+5A}{2}\right)\mathrm{cos}\left(\frac{4A-5A}{2}\right)\left\{\because \mathrm{sin}A+\mathrm{sin}B=2\mathrm{sin}\left(\frac{A+B}{2}\right)\mathrm{cos}\left(\frac{A-B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{3}{2}A\right)\mathrm{cos}\left(-\frac{A}{2}\right)+2\mathrm{sin}\left(\frac{9}{2}A\right)\mathrm{cos}\left(-\frac{A}{2}\right)\phantom{\rule{0ex}{0ex}}$ $=2\mathrm{sin}\left(\frac{3}{2}A\right)\mathrm{cos}\left(\frac{A}{2}\right)+2\mathrm{sin}\left(\frac{9}{2}A\right)\mathrm{cos}\left(\frac{A}{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}\left(\frac{A}{2}\right)\left\{\mathrm{sin}\frac{3}{2}A+\mathrm{sin}\frac{9}{2}A\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}\left(\frac{A}{2}\right)×2\mathrm{sin}\left(\frac{\frac{3}{2}A+\frac{9}{2}A}{2}\right)\mathrm{cos}\left(\frac{\frac{3}{2}A-\frac{9}{2}A}{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=4\mathrm{cos}\left(\frac{A}{2}\right)\mathrm{cos}3A\mathrm{cos}\left(-\frac{3}{2}A\right)\phantom{\rule{0ex}{0ex}}=4\mathrm{cos}\frac{A}{2}\mathrm{cos}\left(\frac{3}{2}A\right)\mathrm{cos}3A\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}$ (iv) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}=\mathrm{sin}3A+\mathrm{sin}2A-\mathrm{sin}A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{3A+2A}{2}\right)\mathrm{co}s\left(\frac{3A-2A}{2}\right)-\mathrm{sin}A\left\{\because \mathrm{sin}A+\mathrm{sin}B=2\mathrm{sin}\left(\frac{A+B}{2}\right)\mathrm{cos}\left(\frac{A-B}{2}\right)\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}\left(\frac{5}{2}A\right)\mathrm{cos}\left(\frac{A}{2}\right)-\mathrm{sin}A\phantom{\rule{0ex}{0ex}}$ $=2\mathrm{sin}\left(\frac{5}{2}A\right)\mathrm{co}s\left(\frac{A}{2}\right)-2\mathrm{sin}\frac{A}{2}\mathrm{cos}\frac{A}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}\left(\frac{A}{2}\right)\left\{\mathrm{sin}\frac{5}{2}A-\mathrm{sin}\frac{A}{2}\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cos}\left(\frac{A}{2}\right)×2\mathrm{sin}\left(\frac{\frac{5}{2}A-\frac{A}{2}}{2}\right)\mathrm{cos}\left(\frac{\frac{5}{2}A+\frac{A}{2}}{2}\right)\phantom{\rule{0ex}{0ex}}=4\mathrm{cos}\left(\frac{A}{2}\right)\mathrm{sin}A\mathrm{cos}\left(\frac{3}{2}A\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=4\mathrm{sin}A\mathrm{cos}\left(\frac{A}{2}\right)\mathrm{cos}\left(\frac{3}{2}A\right)\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}$ (v) $\mathrm{Consider}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}\mathrm{cos}20°\mathrm{cos}100°+\mathrm{cos}100°\mathrm{cos}140°-\mathrm{cos}140°\mathrm{cos}200°\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(2\mathrm{cos}20°\mathrm{cos}100°+2\mathrm{cos}100°\mathrm{cos}140°-2\mathrm{cos}140°\mathrm{cos}200°\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}\left(100°+20°\right)\mathrm{cos}\left(100°-20°\right)+\mathrm{cos}\left(140°+100°\right)\mathrm{cos}\left(140°-100°\right)-\mathrm{cos}\left(200°+140°\right)\mathrm{cos}\left(200°-140°\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}120°+\mathrm{cos}80°+\mathrm{cos}240°+\mathrm{cos}40°-\mathrm{cos}340°-\mathrm{cos}60°\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}120°+\mathrm{cos}240°-\mathrm{cos}60°+\mathrm{cos}80°+\mathrm{cos}40°-\mathrm{cos}340°\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\left(-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\right)+\mathrm{cos}80°+\mathrm{cos}40°-\mathrm{cos}340°\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[-\frac{3}{2}+\left\{2\mathrm{cos}\left(\frac{80°+40°}{2}\right)\mathrm{cos}\left(\frac{80°-40°}{2}\right)-\mathrm{cos}\left(360°-20°\right)\right\}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[-\frac{3}{2}+\left\{2\mathrm{cos}60°\mathrm{cos}20°-\mathrm{cos}20°\right\}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[-\frac{3}{2}+\mathrm{cos}20°-\mathrm{cos}20°\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[-\frac{3}{2}\right]\phantom{\rule{0ex}{0ex}}=-\frac{3}{4}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}$ (vi) $\mathrm{LHS}=\mathrm{sin}\frac{x}{2}\mathrm{sin}\frac{7x}{2}+\mathrm{sin}\frac{3x}{2}\mathrm{sin}\frac{11x}{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[2\mathrm{sin}\frac{x}{2}\mathrm{sin}\frac{7x}{2}+2\mathrm{sin}\frac{3x}{2}\mathrm{sin}\frac{11x}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}\left(\frac{7x}{2}-\frac{x}{2}\right)-\mathrm{cos}\left(\frac{7x}{2}+\frac{x}{2}\right)+\mathrm{cos}\left(\frac{11x}{2}-\frac{3x}{2}\right)-\mathrm{cos}\left(\frac{11x}{2}+\frac{3x}{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}3x-\mathrm{cos}4x+\mathrm{cos}4x-\mathrm{cos}7x\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}3x-\mathrm{cos}7x\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[-2\mathrm{sin}\left(\frac{3x+7x}{2}\right)\mathrm{sin}\left(\frac{3x-7x}{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[-2\mathrm{sin}\left(5x\right)\mathrm{sin}\left(-2x\right)\right]\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left(5x\right)\mathrm{sin}\left(2x\right)=\mathrm{RHS}$ Hence, LHS = RHS (vii) $\mathrm{LHS}=\mathrm{cos}x\mathrm{cos}\frac{x}{2}-\mathrm{cos}3x\mathrm{cos}\frac{9x}{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[2\mathrm{cos}x\mathrm{cos}\frac{x}{2}-2\mathrm{cos}3x\mathrm{cos}\frac{9x}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}\left(x+\frac{x}{2}\right)+\mathrm{cos}\left(x-\frac{x}{2}\right)-\mathrm{cos}\left(3x+\frac{9x}{2}\right)-\mathrm{cos}\left(3x-\frac{9x}{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}\frac{3x}{2}+\mathrm{cos}\frac{x}{2}-\mathrm{cos}\frac{15x}{2}-\mathrm{cos}\frac{3x}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\mathrm{cos}\frac{x}{2}-\mathrm{cos}\frac{15x}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[-2\mathrm{sin}\left(\frac{x+15x}{4}\right)\mathrm{sin}\left(\frac{x-15x}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[-2\mathrm{sin}\left(4x\right)\mathrm{sin}\left(-\frac{7x}{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left(4x\right)\mathrm{sin}\left(\frac{7x}{2}\right)=\mathrm{RHS}$ Hence, LHS = RHS Disclaimer: The given question is incorrect. The correct question should be $\mathrm{cos}x\mathrm{cos}\frac{x}{2}-\mathrm{cos}3x\mathrm{cos}\frac{9x}{2}=\mathrm{sin}4x\mathrm{sin}\frac{7x}{2}$.

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