1. Find the intersection of $y=\sin x$ and $y=\cos x$ in the first quadrant.

We solve $\sin x=\cos x$ $\tan x=1$ In the first quadrant, $x=\frac{\pi}{4}, \qquad y=\frac{1}{\sqrt2}.$

2. Area $A_1$

The region bounded by $y=\sin x$, $y=\cos x$, and the $y$-axis in the first quadrant lies between $x=0$ and $x=\frac{\pi}{4}$.

On $\left[0,\frac{\pi}{4}\right]$, we have $\cos x \ge \sin x.$ So, $A_1=\int_0^{\pi/4}(\cos x-\sin x)\,dx.$

Now, $A_1=\left[\sin x+\cos x\right]_0^{\pi/4}$ $=\left(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\right)-(0+1)$ $=\sqrt2-1.$

So, $A_1=\sqrt2-1.$

3. Area $A_2$

The region bounded by $y=\sin x$, $y=\cos x$, $x$-axis, and $x=\frac{\pi}{2}$ in the first quadrant is the region under the lower of the two curves from their intersection $x=\frac{\pi}{4}$ to $x=\frac{\pi}{2}$, with the $x$-axis as lower boundary.

• On $\left[\frac{\pi}{4},\frac{\pi}{2}\right]$, we have $\sin x \ge \cos x$.
• Hence the lower curve is $y=\cos x$.

Therefore, $A_2=\int_{\pi/4}^{\pi/2}\cos x\,dx.$

Now, $A_2=\left[\sin x\right]_{\pi/4}^{\pi/2}$ $=1-\frac{1}{\sqrt2}.$

Rationalizing/comparing, $1-\frac{1}{\sqrt2}=\frac{\sqrt2-1}{\sqrt2}.$

So, $A_2=1-\frac{1}{\sqrt2}.$

4. Compare $A_1$ and $A_2$

We have $A_1=\sqrt2-1, \qquad A_2=1-\frac{1}{\sqrt2}=\frac{\sqrt2-1}{\sqrt2}.$

Thus, $\frac{A_1}{A_2}=\frac{\sqrt2-1}{(\sqrt2-1)/\sqrt2}=\sqrt2.$ So, $A_1:A_2=\sqrt2:1.$ Equivalently, $A_1:A_2 \ne 1:\sqrt2.$

Now sum: $A_1+A_2=(\sqrt2-1)+\left(1-\frac{1}{\sqrt2}\right)=\sqrt2-\frac{1}{\sqrt2}=\frac{1}{\sqrt2}.$ Wait, simplify carefully: $\sqrt2-\frac{1}{\sqrt2}=\frac{2-1}{\sqrt2}=\frac{1}{\sqrt2}.$ So the sum is not $1$.

This indicates the interpretation of $A_2$ must be reconsidered.

5. Correct geometric interpretation of $A_2$

The region bounded by $y=\sin x$, $y=\cos x$, $x$-axis, and $x=\frac{\pi}{2}$ is the closed region enclosed by all these boundaries:

• from $(\pi/4,1/\sqrt2)$ to $(\pi/2,1)$ along $y=\sin x$,
• from $(\pi/2,1)$ vertically not possible to x-axis, so this cannot form the boundary alone.

Hence the intended region is the area enclosed by:

• $y=\cos x$ from $x=\pi/4$ to $x=\pi/2$,
• the $x$-axis from $x=0$ to $x=\pi/2$,
• $y=\sin x$ from $x=0$ to $x=\pi/4$, which is the region below both curves and above the $x$-axis from $x=0$ to $x=\pi/2$.

So, $A_2=\int_0^{\pi/4}\sin x\,dx+\int_{\pi/4}^{\pi/2}\cos x\,dx.$

Now compute: $\int_0^{\pi/4}\sin x\,dx=[-\cos x]_0^{\pi/4}=1-\frac{1}{\sqrt2},$ $\int_{\pi/4}^{\pi/2}\cos x\,dx=[\sin x]_{\pi/4}^{\pi/2}=1-\frac{1}{\sqrt2}.$

Hence, $A_2=2\left(1-\frac{1}{\sqrt2}\right)=2-\sqrt2.$

6. Now compare again

We have $A_1=\sqrt2-1, \qquad A_2=2-\sqrt2.$

Ratio: $\frac{A_1}{A_2}=\frac{\sqrt2-1}{2-\sqrt2}.$ Since $2-\sqrt2=\sqrt2(\sqrt2-1),$ we get $\frac{A_1}{A_2}=\frac{\sqrt2-1}{\sqrt2(\sqrt2-1)}=\frac{1}{\sqrt2}.$ Thus, $A_1:A_2=1:\sqrt2.$

Sum: $A_1+A_2=(\sqrt2-1)+(2-\sqrt2)=1.$

7. Check options

• A: $A_1:A_2=1:\sqrt2$ and $A_1+A_2=1$ ✔️
• B: false
• C: false
• D: false

Therefore, the correct option is A.