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

Hint: Sine function is periodic with the period of $2n\pi $and so we will convert the given degrees of angle of sine in the form of the $2n\pi $finding the correlation then will identify the location of the angle in the quadrant then will apply All STC rule for the resultant required value.

Complete step-by-step solution:
Take the given expression: $\sin 690^\circ $
The above expression can be re-written: $\sin (720^\circ - 30^\circ )$
Again, the above expression can be written in the form of “pi”
$\sin \left( {4\pi - \dfrac{\pi }{6}} \right)$
Since, the sine function is periodic with the period of $2n\pi $
Hence, the above expression becomes $\sin \left( { - \dfrac{\pi }{6}} \right)$or $ - \sin \left( {\dfrac{\pi }{6}} \right)$
Place the value using the trigonometric table for the above expression-
$ - \sin \left( {\dfrac{\pi }{6}} \right) = - \dfrac{1}{2}$
Simplifying the fraction,
$ - \sin \left( {\dfrac{\pi }{6}} \right) = - 0.5$
Hence, the resultant required value for the $\sin 690^\circ $is $( - 0.5)$
This is the required solution.

Hence, the resultant required value for the $\sin 690^\circ $is $( - 0.5)$

Additional Information: Remember the All STC rule, it is also known as ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ($0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ($90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ($180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ($270^\circ {\text{ to 360}}^\circ $ ).

Note: Be careful in identifying any degree of measure in the one of the four quadrants and also refer to sine and cosine even and odd functions. Remember the trigonometric table for the reference values for different angles for sine, cosine and tangent functions for direct substitution for the accurate and the efficient solution.