MCQ
The distance between the points $(\cos\theta,\sin\theta)$ and $(\sin\theta,-\cos\theta)$ is:
- A$\sqrt{3}$
- ✓$\sqrt{2}$
- C$2$
- D$1$
✓
Answer
Correct option: B.
$\sqrt{2}$
We have to find the distance between $\text{A}(\cos\theta,\sin\theta)$ and $\text{B}(\sin\theta,-\cos\theta).$
In general, the distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by,
$\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
So,
$\text{AB}=\sqrt{(\sin\theta-\cos\theta)^2+(-\cos\theta-\sin\theta)^2}$
$=\sqrt{2(\sin^2\theta+\cos^2\theta)}$
But according to the trigonometric identity,
$\sin^2\theta+\cos^2\theta=1$
Therefore,
$\text{AB}=\sqrt{2}$
In general, the distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by,
$\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
So,
$\text{AB}=\sqrt{(\sin\theta-\cos\theta)^2+(-\cos\theta-\sin\theta)^2}$
$=\sqrt{2(\sin^2\theta+\cos^2\theta)}$
But according to the trigonometric identity,
$\sin^2\theta+\cos^2\theta=1$
Therefore,
$\text{AB}=\sqrt{2}$
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