Answer: \(sin 0^{\circ} = 0\)
Step-by-step explanation:
Since, \(sin x = cos (90-x)\) ( In first quadrant )
And, \(- sin x = cos (90+x)\) ( In second quadrant )
Since, we have to find the value of \(sin 0^{\circ}\)
And, we have \(cos 90^{\circ} = 0\)
⇒ \(cos (90-0)^{\circ} = 0\) or \(cos (90+0)^{\circ} = 0\)
⇒ \(sin 0^{\circ} = 0\) or \(-sin 0^{\circ}= 0\) ⇒ \(sin 0^{\circ}= 0\)
Thus, \(sin 0^{\circ} = 0\)
