How do you find the exact value of $sin 690$ $sin690$ ?

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Answer 1 · 2018-07-26T03:13:07

$- \frac{1}{2}$ $-1/2$ .

${sin 690}^{\circ} = sin (720^{\circ} - 30^{\circ})$ $sin690^@=sin(720^@-30^@)$ ,

$= sin (4 π - \frac{π}{6})$ $=sin(4pi-pi/6)$ .

But, $sin (4 π - θ) = - sin θ$ $sin(4pi-theta)=-sintheta$ .

$:. sin690^@=-sin30^@=-1/2$ .

How do you find the exact value of sin690sin690sin690?