Question

$\sin(8\pi \div 3) \times \cos(23\pi \div 6 ) + \cos(13\pi \div 6) \times \sin(35\pi \div 2) = \frac{1}{2}$
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Answer 1

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sin(8π÷3)×cos(23π÷6)+cos(13π÷6)×sin(35π÷2)=21

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sin 8pi/3 cos 23 pi/6 + cos 13 pi/3 sin 35pi/6

=sin(3pi-pi/3)cos(4pi-pi/6) + cos(4pi + pi/3) sin(6pi-pi/6)

=sin(pi/3). cos(pi/6)+ cos(pi/3) [-sin(pi/6)]

=sin(pi/3).cos(pi/6)-cos(pi/3)sin(pi/6)

=sin(pi/3 - pi/6)[sin(a-b)=sin a cos b - cos a sin b]

=sin(pi/6)

=1/2