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As θ increases from 0 to π/2 (90o) sin θ increases until it reaches a maximum value of 1 when θ = π/2.

As θ increases from π/2 (90o) to 3π/2 (270o) sin θ decreases until it reaches a minimum value of -1 when θ = 3π/2.

As θ increases from 3π/2 (270o) to 3π (360o) sin θ increases until it reaches the value of 0 it had when θ = 0.

From this point, as θ increases (by the same amounts) sin θ repeats this same cyclic behaviour.

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Q: Does the value of sin theta increases as theta increases?
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How do solve sin 2 theta - square root of 3 theta- equals 0?

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