sin 10°sin50° + sin50°sin250°+ sin250°sin10°=-3/4
Prove That,sin 10°sin50° + sin50°sin250°+ sin250°sin10°=-3/4
Solution:
sin 10°sin50° + sin50°sin250° + sin250°sin10°
= $\frac{1}{2}$ (sin 10°sin50° + sin50°sin250° + sin250°sin10°)
= $\frac{1}{2}$(2sin 10°sin50° + 2sin50°sin250° + 2sin250°sin10°)
= $\frac{1}{2}$ {cos (10°-50°) – cos(10° + 50°) + cos(50° – 250°)$- cos(50° + 250°) + cos(250° – 10°) – cos(250° + 10°)}
= $\frac{1}{2}$(cos40° -cos60° + cos200° – cos300° + cos240° – cos260°)
= $\frac{1}{2}$ (cos40°+cos200°-cos260°-cos300°+cos240°-cos60°)
= $\frac{1}{2}$ { cos 40°+cos(180° + 20°) – cos(180° + 80°) – cos(360° – 60°) + cos(180° + 60°) – cos60°}
= $\frac{1}{2}$(cos 40°-cos20°+cos80°-cos60°-cos60°-cos60°)
= $\frac{1}{2}$(cos40° -cos20°+sin10°– $\frac{1}{2}$ -$\frac{1}{2}$ – $\frac{1}{2}$)
=$\frac{1}{2}$(2 $\sin \frac{40 + 20}{2}sin\frac{20 – 40}{2}$ +sin10° – $\frac{3}{2}$ )
= $\frac{1}{2}${ 2 sin30°sin(-10°)+sin10°-$\frac{3}{2}$}
= $\frac{1}{2}$(-2× $\frac{1}{2}$ ×sin10°+sin10°-$\frac{3}{2}$)
= $\frac{1}{2}$ (-sin10°+sin10°-$\frac{3}{2}$)
= $- \frac{3}{4}$ [Proved]