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) $\sin \frac{5\mathrm{\pi }}{18}-\cos \frac{4\mathrm{\pi }}{9}=\sqrt{3}\sin \frac{\mathrm{\pi }}{9}$

(vi) $\cos \frac{\mathrm{\pi }}{12}-\sin \frac{\mathrm{\pi }}{12}=\frac{1}{\sqrt{2}}$

(vii) sin 80° − cos 70° = cos 50°

(viii) sin 51° + cos 81° = cos 21°

Answer 1

(i)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \cos 55°+\cos 65°+\cos 175° \\ =2\cos \left(\frac{55°+65°}{2}\right)\cos \left(\frac{55°-65°}{2}\right)+\cos 175°\left\{\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\cos 60°\cos \left(-5°\right)+\cos 175° \\ =2\times \frac{1}{2}\cos 5°+\cos 175° \\ =\cos 5°+\cos 175° \\ =2\cos \left(\frac{5°+175°}{2}\right)\cos \left(\frac{5°-175°}{2}\right) \\ =2\cos 90°\cos 85° \\ =0 \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$


(ii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \sin 50°-\sin 70°+\sin 10° \\ =2\sin \left(\frac{50°-70°}{2}\right)\cos \left(\frac{50°+70°}{2}\right)+\sin 10°\left\{\because \mathrm{si}nA-\sin B=2\mathrm{si}n\left(\frac{A-B}{2}\right)c\mathrm{os}\left({\displaystyle \frac{A+B}{2}}\right)\right\} \\ =2\sin \left(-10°\right)\mathrm{co}s60°+\sin 10° \\ =2\times \frac{1}{2}\sin \left(-10°\right)+\sin 10° \\ =-\sin 10°+\sin 10° \\ =0 \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(iii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \cos 80°+\cos 40°-\cos 20° \\ =2\mathrm{co}s\left(\frac{80°+40°}{2}\right)\cos \left(\frac{80°-40°}{2}\right)-\cos 20°\left\{\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\cos 60°\cos 20°-\cos 20° \\ =2\times \frac{1}{2}\cos 20°-\cos 20° \\ =\cos 20°-\cos 20° \\ =0 \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(iv)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \cos 20°+\cos 100°+\cos 140° \\ =2\cos \left(\frac{20°+100°}{2}\right)\cos \left(\frac{20°-100°}{2}\right)+\cos 140°\left\{\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\cos 60°\cos \left(-40°\right)+\cos 140° \\ =2\times \frac{1}{2}\cos 40°+\cos 140° \\ =\cos 40°+\cos 140° \\ =2\cos \left(\frac{40°+140°}{2}\right)\cos \left(\frac{40°-140°}{2}\right) \\ =2\cos 90°\cos 50° \\ =0 \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$


(v)

$$\begin{gathered} \mathrm{LHS}=\sin \left(\frac{5\mathrm{\pi }}{18}\right)-\cos \frac{4\mathrm{\pi }}{9} \\ =\sin \left(\frac{5\mathrm{\pi }}{18}\right)-\cos \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{18}\right) \\ =\sin \left(\frac{5\mathrm{\pi }}{18}\right)-\sin \left(\frac{\mathrm{\pi }}{18}\right) \\ =2\sin \left(\frac{\frac{5\mathrm{\pi }}{18}-\frac{\mathrm{\pi }}{18}}{2}\right)\cos \left(\frac{\frac{5\mathrm{\pi }}{18}+\frac{\mathrm{\pi }}{18}}{2}\right)\left[\because \sin A-\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\right] \\ =2\sin \left(\frac{\mathrm{\pi }}{9}\right)\cos \frac{\mathrm{\pi }}{6} \\ =2\sin \left(\frac{\mathrm{\pi }}{9}\right)\cos \frac{\mathrm{\pi }}{6} \\ =2\times \frac{\sqrt{3}}{2}\sin \left(\frac{\mathrm{\pi }}{9}\right) \\ =\sqrt{3}\sin \left(\frac{\mathrm{\pi }}{9}\right)=\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(vi)

$$\begin{gathered} \mathrm{LHS}=\cos \frac{\mathrm{\pi }}{12}-\sin \frac{\mathrm{\pi }}{12} \\ =\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-{\displaystyle \frac{5\mathrm{\pi }}{12}}\right)-\sin \frac{\mathrm{\pi }}{12} \\ =\sin \left(\frac{5\mathrm{\pi }}{12}\right)-\sin \frac{\mathrm{\pi }}{12} \\ =2\sin \left(\frac{{\displaystyle \frac{5\mathrm{\pi }}{12}}-{\displaystyle \frac{\mathrm{\pi }}{12}}}{2}\right)\cos \left(\frac{{\displaystyle \frac{5\mathrm{\pi }}{12}}+{\displaystyle \frac{\mathrm{\pi }}{12}}}{2}\right)\left\{\because \sin A-\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\right\} \\ =2\sin \left(\frac{\mathrm{\pi }}{6}\right)\cos \left(\frac{\mathrm{\pi }}{4}\right) \\ =2\times \frac{1}{2}\times \frac{1}{\sqrt{2}} \\ =\frac{1}{\sqrt{2}} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$


(vii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ s\mathrm{in}80°-\cos 70° \\ =\sin 80°-\cos \left(90°-20°\right) \\ =\sin 80°-\sin 20° \\ =2\sin \left(\frac{80°-20°}{2}\right)\cos \left(\frac{80°+20°}{2}\right)\left\{\because \sin A-\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\right\} \\ =2\sin 30°\cos 50° \\ =2\times \frac{1}{2}\cos 50° \\ =\cos 50° \\ =\mathrm{RH}S \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(viii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \sin 51°+\cos 81° \\ =\sin 51°+\cos \left(90°-9°\right) \\ =\sin 51°+\sin 9° \\ =2\sin \left(\frac{51°+9°}{2}\right)\cos \left(\frac{51°-9°}{2}\right)\left\{\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin 30°\cos 21° \\ =2\times \frac{1}{2}\cos \left(21°\right) \\ =\cos \left(21°\right) \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$