Let $$A_1$$ be the area of the region bounded by the curves $$y = \sin x$$, $$y = \cos x$$ and $$y$$-axis in the first quadrant. Also, let $$A_2$$ be the area of the region bounded by the curves $$y = \sin x$$, $$y = \cos x$$, $$x$$-axis and $$x = \frac{\pi}{2}$$ in the first quadrant. Then,

In the first quadrant, $$y = \sin x$$ and $$y = \cos x$$ intersect at $$x = \dfrac{\pi}{4}$$, where both equal $$\dfrac{\sqrt{2}}{2}$$. For $$0 \leq x \leq \dfrac{\pi}{4}$$, $$\cos x \geq \sin x$$.

$$A_1$$ is the area bounded by $$y = \sin x$$, $$y = \cos x$$, and the $$y$$-axis in the first quadrant. This is the region between the two curves from $$x = 0$$ to $$x = \dfrac{\pi}{4}$$:

$$A_1 = \displaystyle\int_0^{\pi/4}(\cos x - \sin x)\,dx = [\sin x + \cos x]_0^{\pi/4} = \left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}\right) - (0 + 1) = \sqrt{2} - 1$$.

$$A_2$$ is the area bounded by $$y = \sin x$$, $$y = \cos x$$, the $$x$$-axis, and $$x = \dfrac{\pi}{2}$$ in the first quadrant. This region lies below both curves and above the $$x$$-axis, specifically: from $$x = 0$$ to $$\dfrac{\pi}{4}$$, the lower boundary is $$\sin x$$ (and above it is $$\cos x$$, so the region below $$\sin x$$ up to the $$x$$-axis contributes); from $$\dfrac{\pi}{4}$$ to $$\dfrac{\pi}{2}$$, the lower boundary is $$\cos x$$.

$$A_2 = \displaystyle\int_0^{\pi/4}\sin x\,dx + \displaystyle\int_{\pi/4}^{\pi/2}\cos x\,dx = [-\cos x]_0^{\pi/4} + [\sin x]_{\pi/4}^{\pi/2}$$

$$= \left(-\dfrac{\sqrt{2}}{2} + 1\right) + \left(1 - \dfrac{\sqrt{2}}{2}\right) = 2 - \sqrt{2}$$.

Now $$\dfrac{A_1}{A_2} = \dfrac{\sqrt{2} - 1}{2 - \sqrt{2}} = \dfrac{\sqrt{2} - 1}{\sqrt{2}(\sqrt{2} - 1)} = \dfrac{1}{\sqrt{2}}$$, so $$A_1 : A_2 = 1 : \sqrt{2}$$.

Also, $$A_1 + A_2 = (\sqrt{2} - 1) + (2 - \sqrt{2}) = 1$$.

Therefore $$A_1 : A_2 = 1 : \sqrt{2}$$ and $$A_1 + A_2 = 1$$.