Question

Show that :
(i) $\sin 50°\cos 85°=\frac{1-\sqrt{2}\sin 35°}{2\sqrt{2}}$

(ii) $\sin 25°\cos 115°=\frac{1}{2}\left(\sin 140°-1\right)$

Answer 1

(i)

$$\begin{gathered} \mathrm{LHS}=2\sin 50°\cos 85° \\ =\frac{\sin \left(50°+85°\right)+\sin \left(50°-85°\right)}{2}\left[\because \sin A\cos B=\frac{1}{2}\left\{\sin (A+B)+\sin (A-B)\right\}\right] \\ =\frac{\sin 135°+\sin \left(-35°\right)}{2} \\ =\frac{\sin 135°-\sin 35°}{2} \\ =\frac{\cos 45°-\sin 35°}{2}\left[\because \sin \left(90°+45°\right)=\cos 45°\right] \\ =\frac{1}{2}\left(\frac{1}{\sqrt{2}}-\sin 35°\right) \\ =\frac{1}{2}\left[\frac{1-\sqrt{2}\sin 35°}{\sqrt{2}}\right] \\ =\frac{1-\sqrt{2}\sin 35°}{2\sqrt{2}} \\ \mathrm{RHS}=\frac{1-\sqrt{2}\sin 35°}{2\sqrt{2}} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

(ii)

$$\begin{gathered} \mathrm{LHS}=2\sin 25°\cos 115° \\ =\frac{\sin \left(25°+115°\right)+\sin \left(25°-115°\right)}{2}\left[\because \sin A\cos B=\frac{1}{2}\left\{\sin (A+B)+\sin (A-B)\right\}\right] \\ =\frac{\sin 140°+\sin \left(-90°\right)}{2} \\ =\frac{\sin 140°-\sin \left(90°\right)}{2} \\ =\frac{\sin 140°-1}{2} \\ \mathrm{RHS}=\frac{\sin 140°-1}{2} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$