Question
What is the distance between the points $\text{A}(\sin\theta-\cos\theta,0)$ and $\text{B}(0,\sin\theta+\cos\theta)?$
✓
Answer
Distance between $\text{A}(\sin\theta-\cos\theta,0)$ and $\text{B}(0,\sin\theta+\cos\theta)$
$=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(0-\sin\theta+\cos\theta)^2+(\sin\theta+\cos\theta-0)^2}$
$=\sqrt{(-\sin\theta+\cos\theta)^2+(\sin\theta+\cos\theta)^2}$
$=\sqrt{\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta+\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta}$
$=\sqrt{2\sin^2\theta+2\cos^2\theta}=\sqrt{2(\sin^2\theta+\cos^2\theta)}$
$=\sqrt{2\times1}=\sqrt{2}$
$(\because\ \sin^2\theta+\cos^2\theta=1)$
$=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(0-\sin\theta+\cos\theta)^2+(\sin\theta+\cos\theta-0)^2}$
$=\sqrt{(-\sin\theta+\cos\theta)^2+(\sin\theta+\cos\theta)^2}$
$=\sqrt{\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta+\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta}$
$=\sqrt{2\sin^2\theta+2\cos^2\theta}=\sqrt{2(\sin^2\theta+\cos^2\theta)}$
$=\sqrt{2\times1}=\sqrt{2}$
$(\because\ \sin^2\theta+\cos^2\theta=1)$
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