Question

Show that:
(i) $sin{50}^{\circ }cos{85}^{\circ }=\frac{1-\sqrt{2}sin{35}^{\circ }}{2\sqrt{2}}$
(ii) $sin{25}^{\circ }cos{115}^{\circ }=\frac{1}{2}(sin{140}^{\circ }-1)$

Answer 1

(i) $sin{50}^{\circ }cos{85}^{\circ }=\frac{1-\sqrt{2}sin{35}^{\circ }}{2\sqrt{2}}$
LHS$=sin{50}^{\circ }cos{85}^{\circ }=\frac{2sin{50}^{\circ }cos{85}^{\circ }}{2}\because 2sinAcosB=sin(A+B)+sin(A-B)\Rightarrow \frac{2sin{50}^{\circ }cos{85}^{\circ }}{2}=\frac{1}{2}\left[sin\right({50}^{\circ }+{85}^{\circ })+sin({50}^{\circ }-{85}^{\circ }\left)\right]=\frac{1}{2}[sin{135}^{\circ }+sin(-{35}^{\circ }\left)\right]=\frac{1}{2}\left[sin\right({90}^{\circ }+{45}^{\circ })-sin{35}^{\circ }][\because sin(-\theta )=-sin\theta ]=\frac{1}{2}[cos{45}^{\circ }-sin{35}^{\circ }][\because sin({90}^{\circ }+\theta )=cos\theta ]$
Now,
$cos{45}^{\circ }=\frac{1}{\sqrt{2}}=\frac{1}{2}[\frac{1}{\sqrt{2}}-sin{35}^{\circ }]=\frac{1-\sqrt{2}sin{35}^{\circ }}{2\sqrt{2}}$

(ii) $sin{25}^{\circ }cos{115}^{\circ }=\frac{1}{2}(sin{140}^{\circ }-1)$
LHS$=sin{25}^{\circ }cos{115}^{\circ }=\frac{2sin{25}^{\circ }cos{115}^{\circ }}{2}$
We know that,
$2sinAcosB=sin(A+B)+sin(A-B)=\frac{1}{2}\left[sin\right({25}^{\circ }+{115}^{\circ })+sin({25}^{\circ }-{115}^{\circ }\left)\right]=\frac{1}{2}[sin{140}^{\circ }+sin(-{90}^{\circ }\left)\right]\left[As\right(sin(-\theta )=-sin\theta \left)\right]\Rightarrow \frac{1}{2}\left[sin\right({90}^{\circ }+{50}^{\circ })-sin{90}^{\circ }]\Rightarrow \frac{1}{2}\left[sin\right({140}^{\circ }-1\left)\right](Assin{90}^{\circ }=1)$
Hence proved.