Let A_{1} be the area of the region bounded b | vedclass

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(A) The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$ in the first quadrant.$A_{1}$ is the area bounded by $y = \cos x$,$y = \

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$A_{1}: A_{2} = 1: \sqrt{2}$ and $A_{1} + A_{2} = 1$

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(A) The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$ in the first quadrant.$A_{1}$ is the area bounded by $y = \cos x$,$y = \sin x$ and the $y$-axis from $x = 0$ to $x = \frac{\pi}{4}$:$A_{1} = \int_{0}^{\pi/4} (\cos x - \sin x) dx = [\sin x + \cos x]_{0}^{\pi/4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.$A_{2}$ is the area bounded by $y = \sin x$ from $x = 0$ to $x = \frac{\pi}{4}$ and $y = \cos x$ from $x = \frac{\pi}{4}$ to $x = \frac{\pi}{2}$:$A_{2} = \int_{0}^{\pi/4} \sin x dx + \int_{\pi/4}^{\pi/2} \cos x dx = [-\cos x]_{0}^{\pi/4} + [\sin x]_{\pi/4}^{\pi/2}$$A_{2} = (-\frac{1}{\sqrt{2}} - (-1)) + (1 - \frac{1}{\sqrt{2}}) = 1 - \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2} = \sqrt{2}(\sqrt{2} - 1)$.Now,$A_{1} : A_{2} = (\sqrt{2} - 1) : \sqrt{2}(\sqrt{2} - 1) = 1 : \sqrt{2}$.And $A_{1} + A_{2} = (\sqrt{2} - 1) + (2 - \sqrt{2}) = 1$.

Let $A_{1}$ be the area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and the $y$-axis in the first quadrant. Also,let $A_{2}$ be the area of the region bounded by the curves $y = \sin x$,$y = \cos x$,the $x$-axis and $x = \frac{\pi}{2}$ in the first quadrant. Then ..... .

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