Question

Prove that:
(i) sin 38° + sin 22° = sin 82°
(ii) cos 100° + cos 20° = cos 40°
(iii) sin 50° + sin 10° = cos 20°
(iv) sin 23° + sin 37° = cos 7°
(v) sin 105° + cos 105° = cos 45°
(vi) sin 40° + sin 20° = cos 10°

Answer 1

(i)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ s\mathrm{in}38°+\sin 22° \\ =2\sin \left(\frac{38°+22°}{2}\right)\cos \left(\frac{38°-22°}{2}\right)\left\{\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin 30°\cos 8° \\ =2\times \frac{1}{2}\cos (90°-8°) \\ =\sin 82° \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(ii)

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

(iii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \sin 50°+\sin 10° \\ =2\sin \left(\frac{50°+10°}{2}\right)\cos \left(\frac{50°-10°}{2}\right)\left\{\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin 30°\cos 20° \\ =2\times \frac{1}{2}\cos 20° \\ =\cos 20° \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$


(iv)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ s\mathrm{in}23°+\sin 37° \\ =2\sin \left(\frac{23°+37°}{2}\right)\cos \left(\frac{23°-37°}{2}\right)\left\{\because \sin A+\sin B=2\mathrm{si}n\left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin 30°\cos \left(-7°\right) \\ =2\sin 30°\cos 7° \\ =2\times \frac{1}{2}\cos 7° \\ =\cos 7° \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(v)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ s\mathrm{in}105°+\cos 105° \\ =\sin 105°+\cos \left(90°+15°\right) \\ =\sin 105°-\sin 15° \\ =2\mathrm{si}n\left(\frac{105°-15°}{2}\right)\cos \left(\frac{105°+15°}{2}\right)\left\{\because \sin A+\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\right\} \\ =2\sin 45°\cos 60° \\ =2\sin \left(90°-45°\right)\cos 60° \\ =2\times \frac{1}{2}\cos \left(45°\right) \\ =\cos 45° \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(vi)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \sin 40°+\sin 20° \\ =2\sin \left(\frac{40°+20°}{2}\right)\cos \left(\frac{40°-20°}{2}\right)\left\{\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin 30°\cos 10° \\ =2\times \frac{1}{2}\cos 10° \\ =\cos \left(10°\right) \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$