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) $\sin \frac{\mathrm{x}}{2}\sin \frac{7x}{2}+\sin \frac{3x}{2}\sin \frac{11x}{2}=\sin 2x\sin 5x.$
(vii) $\cos x\cos \frac{x}{2}-\cos 3x\cos \frac{9x}{2}=\sin 7x\sin 8x$

Answer 1

(i)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \cos 3A+c\mathrm{os}5A+\cos 7A+\cos 15A \\ =2\cos \left(\frac{3A+5A}{2}\right)\cos \left(\frac{3A-5A}{2}\right)+2\cos \left(\frac{7A+15A}{2}\right)\cos \left(\frac{7A-15A}{2}\right)\left\{\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\cos 4A\cos \left(-A\right)+2\cos 11A\cos \left(-4A\right) \end{gathered}$$

$$\begin{gathered} =2\cos 4A\cos A+2\cos 11A\cos 4A \\ =2\cos 4A\left\{\cos A+\cos 11A\right\} \\ =2\cos 4A\times \left\{2\cos \left(\frac{A+11A}{2}\right)\cos \left(\frac{A-11A}{2}\right)\right\} \\ =4\cos 4A\cos 6A\cos \left(-5A\right) \\ =4\cos 4A\cos 5A\cos 6A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

(ii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \cos A+\cos 3A+\cos 5A+\cos 7A \\ =2\cos \left(\frac{A+3A}{2}\right)\cos \left(\frac{A-3A}{2}\right)+2\cos \left(\frac{5A+7A}{2}\right)\cos \left(\frac{5A-7A}{2}\right)\left\{\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\cos 2A\cos \left(-A\right)+2\cos 6A\cos \left(-A\right) \end{gathered}$$

$$\begin{gathered} =2\cos 2A\cos A+2\cos 6A\cos A \\ =2\cos A(\cos 2A+\cos 6A) \\ =2\cos A\times 2\cos \left(\frac{2A+6A}{2}\right)\cos \left(\frac{2A-6A}{2}\right) \\ =4\cos A\cos 4A\cos \left(-2A\right) \\ =4\cos A\cos 2A\cos 4A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(iii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \sin A+\sin 2A+\sin 4A+\sin 5A \\ =2\sin \left(\frac{A+2A}{2}\right)\cos \left(\frac{A-2A}{2}\right)+2\sin \left(\frac{4A+5A}{2}\right)\cos \left(\frac{4A-5A}{2}\right)\left\{\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin \left(\frac{3}{2}A\right)\cos \left(-\frac{A}{2}\right)+2\sin \left(\frac{9}{2}A\right)\cos \left(-\frac{A}{2}\right) \end{gathered}$$

$$\begin{gathered} =2\sin \left(\frac{3}{2}A\right)\cos \left(\frac{A}{2}\right)+2\sin \left(\frac{9}{2}A\right)\cos \left(\frac{A}{2}\right) \\ =2\cos \left(\frac{A}{2}\right)\left\{\sin \frac{3}{2}A+\sin \frac{9}{2}A\right\} \\ =2\cos \left(\frac{A}{2}\right)\times 2\sin \left(\frac{{\displaystyle \frac{3}{2}}A+{\displaystyle \frac{9}{2}}A}{2}\right)\cos \left(\frac{{\displaystyle \frac{3}{2}}A-{\displaystyle \frac{9}{2}}A}{2}\right) \\ =4\cos \left(\frac{A}{2}\right)\cos 3A\cos \left(-\frac{3}{2}A\right) \\ =4\cos \frac{A}{2}\cos \left(\frac{3}{2}A\right)\cos 3A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

(iv)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ =\sin 3A+\sin 2A-\sin A \\ =2\sin \left(\frac{3A+2A}{2}\right)\mathrm{co}s\left(\frac{3A-2A}{2}\right)-\sin A\left\{\because \sin A+\sin B=2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin \left(\frac{5}{2}A\right)\cos \left(\frac{A}{2}\right)-\sin A \end{gathered}$$

$$\begin{gathered} =2\sin \left(\frac{5}{2}A\right)\mathrm{co}s\left(\frac{A}{2}\right)-2\sin \frac{A}{2}\cos \frac{A}{2} \\ =2\cos \left(\frac{A}{2}\right)\left\{\sin \frac{5}{2}A-\sin \frac{A}{2}\right\} \\ =2\cos \left(\frac{A}{2}\right)\times 2\sin \left(\frac{{\displaystyle \frac{5}{2}}A-{\displaystyle \frac{A}{2}}}{2}\right)\cos \left(\frac{{\displaystyle \frac{5}{2}}A+{\displaystyle \frac{A}{2}}}{2}\right) \\ =4\cos \left(\frac{A}{2}\right)\sin A\cos \left(\frac{3}{2}A\right) \\ =4\sin A\cos \left({\displaystyle \frac{A}{2}}\right)\cos \left(\frac{3}{2}A\right) \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

(v)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \cos 20°\cos 100°+\cos 100°\cos 140°-\cos 140°\cos 200° \\ =\frac{1}{2}(2\cos 20°\cos 100°+2\cos 100°\cos 140°-2\cos 140°\cos 200°) \\ =\frac{1}{2}\left[\cos \left(100°+20°\right)\cos \left(100°-20°\right)+\cos \left(140°+100°\right)\cos \left(140°-100°\right)-\cos \left(200°+140°\right)\cos \left(200°-140°\right)\right] \\ =\frac{1}{2}\left[\cos 120°+\cos 80°+\cos 240°+\cos 40°-\cos 340°-\cos 60°\right] \\ =\frac{1}{2}\left[\cos 120°+\cos 240°-\cos 60°+\cos 80°+\cos 40°-\cos 340°\right] \\ =\frac{1}{2}\left[\left(-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\right)+\cos 80°+\cos 40°-\cos 340°\right] \\ =\frac{1}{2}\left[-\frac{3}{2}+\left\{2\cos \left(\frac{80°+40°}{2}\right)\cos \left(\frac{80°-40°}{2}\right)-\cos \left(360°-20°\right)\right\}\right] \\ =\frac{1}{2}\left[-\frac{3}{2}+\left\{2\cos 60°\cos 20°-\cos 20°\right\}\right] \\ =\frac{1}{2}\left[-\frac{3}{2}+\cos 20°-\cos 20°\right] \\ =\frac{1}{2}\left[-\frac{3}{2}\right] \\ =-\frac{3}{4}=\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

(vi)
$$\begin{gathered} \mathrm{LHS}=\sin \frac{x}{2}\sin \frac{7x}{2}+\sin \frac{3x}{2}\sin \frac{11x}{2} \\ =\frac{1}{2}\left[2\sin \frac{x}{2}\sin \frac{7x}{2}+2\sin \frac{3x}{2}\sin \frac{11x}{2}\right] \\ =\frac{1}{2}\left[\cos \left(\frac{7x}{2}-\frac{x}{2}\right)-\cos \left(\frac{7x}{2}+\frac{x}{2}\right)+\cos \left(\frac{11x}{2}-\frac{3x}{2}\right)-\cos \left(\frac{11x}{2}+\frac{3x}{2}\right)\right] \\ =\frac{1}{2}\left[\cos 3x-\cos 4x+\cos 4x-\cos 7x\right] \\ =\frac{1}{2}\left[\cos 3x-\cos 7x\right] \\ =\frac{1}{2}\left[-2\sin \left(\frac{3x+7x}{2}\right)\sin \left(\frac{3x-7x}{2}\right)\right] \\ =\frac{1}{2}\left[-2\sin \left(5x\right)\sin \left(-2x\right)\right] \\ =\sin \left(5x\right)\sin \left(2x\right)=\mathrm{RHS} \end{gathered}$$

Hence, LHS = RHS

(vii)
$$\begin{gathered} \mathrm{LHS}=\cos x\cos \frac{x}{2}-\cos 3x\cos \frac{9x}{2} \\ =\frac{1}{2}\left[2\cos x\cos \frac{x}{2}-2\cos 3x\cos \frac{9x}{2}\right] \\ =\frac{1}{2}\left[\cos \left(x+\frac{x}{2}\right)+\cos \left(x-\frac{x}{2}\right)-\cos \left(3x+\frac{9x}{2}\right)-\cos \left(3x-\frac{9x}{2}\right)\right] \\ =\frac{1}{2}\left[\cos \frac{3x}{2}+\cos \frac{x}{2}-\cos \frac{15x}{2}-\cos \frac{3x}{2}\right] \\ =\frac{1}{2}\left[\cos \frac{x}{2}-\cos \frac{15x}{2}\right] \\ =\frac{1}{2}\left[-2\sin \left(\frac{x+15x}{4}\right)\sin \left(\frac{x-15x}{4}\right)\right] \\ =\frac{1}{2}\left[-2\sin \left(4x\right)\sin \left(-\frac{7x}{2}\right)\right] \\ =\sin \left(4x\right)\sin \left(\frac{7x}{2}\right)=\mathrm{RHS} \end{gathered}$$

Hence, LHS = RHS

Disclaimer: The given question is incorrect. The correct question should be $\cos x\cos \frac{x}{2}-\cos 3x\cos \frac{9x}{2}=\sin 4x\sin \frac{7x}{2}$.