Question

prove that
sin8π/3 × cos23π/6 + cos13π/3 × sin35π/6 =1/2

Answer 1

this can be represented as
$\sin(2\pi + \frac{2\pi}{3} ) \times \cos(4\pi - \frac{\pi}{6} ) + \\ \cos(4\pi + \frac{\pi}{3} ) \times \sin(6\pi - \frac{\pi}{6} )$
$\sin( \frac{2\pi}{3} ) \times \cos( \frac{\pi}{6} ) - \cos( \frac{\pi}{3} ) \\ \times \sin( \frac{\pi}{6} )$
$\frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} - \frac{1}{2} \times \frac{ 1 }{2}$
$\frac{3}{4} - \frac{1}{4}$
$\frac{3 - 1}{4} \\ = \frac{2}{4} \\ = \frac{1}{2}$
hence proved

Answer 2

Answer:

\sin(2\pi +   \frac{2\pi}{3} )  \times  \cos(4\pi -  \frac{\pi}{6} )  +  \\  \cos(4\pi +  \frac{\pi}{3} )  \times  \sin(6\pi -  \frac{\pi}{6} )  

\sin( \frac{2\pi}{3} )  \times  \cos( \frac{\pi}{6} )   -   \cos( \frac{\pi}{3} )  \\  \times  \sin( \frac{\pi}{6} )  

 \frac{ \sqrt{3} }{2}  \times  \frac{ \sqrt{3} }{2}   -   \frac{1}{2}  \times  \frac{ 1 }{2}  

\frac{3}{4}   -   \frac{1}{4}  

\frac{3 - 1}{4}  \\ =   \frac{2}{4}  \\  =  \frac{1}{2}

hence proved