Which of the following relations is correct for $$0<\theta<90^{\circ}$$

Let $$\theta = 30^\circ$$
Then, sin $$\theta$$ = sin $$30^\circ = \dfrac{1}{2}$$
sin $$2\theta$$ = sin $$60^\circ = \dfrac{\sqrt{3}}{2}$$

Comparing, both, 
$$\dfrac{1}{2} < \dfrac{\sqrt{3}}{2}$$

Let $$\theta = 45^\circ$$
Then, sin $$\theta$$ = sin $$45^\circ = \dfrac{1}{\sqrt{2}}$$
sin $$2\theta$$ = sin $$90^\circ = 1$$

Comparing both,
$$\dfrac{1}{\sqrt{2}} < 1$$

Hence, we can say that sin $$\theta$$ < sin $$2\theta$$ for 0 < $$\theta$$ < 90$$^\circ$$