Question

Answer the question in the attached

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Answer 1

Answer

$\sf{\sqrt{2}\:units}$

Step-by-step explanation

$\sf{\mathsf{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}$

$\bf{\mathbf{The\:distance\:between\:}\left(\cos \left(\theta\right),\:\sin \left(\theta\right)\right)\mathbf{\:and\:}\left(\sin \left(\theta\right),\:-\cos \left(\theta\right)\right)\mathbf{\:is\:}}$$\sf{=\sqrt{\left(\sin \left(\theta\right)-\cos \left(\theta\right)\right)^2+\left(-\cos \left(\theta\right)-\sin \left(\theta\right)\right)^2}}$

$\sf{=\sqrt{\sin ^2\left(\theta\right)-2\sin \left(\theta\right)\cos \left(\theta\right)+\cos ^2\left(\theta\right)+\left(-\cos \left(\theta\right)-\sin \left(\theta\right)\right)^2}}$

$\sf{=\sqrt{\sin ^2\left(\theta\right)-2\sin \left(\theta\right)\cos \left(\theta\right)+\cos ^2\left(\theta\right)+\cos ^2\left(\theta\right)+2\cos \left(\theta\right)\sin \left(\theta\right)+\sin ^2\left(\theta\right)}}$

$\sf{=\sqrt{2\sin ^2\left(\theta\right)+2\cos ^2\left(\theta\right)}}$

$\sf{=\sqrt{2(\sin ^2\left(\theta\right)+\cos ^2\left(\theta\right))}\:\:\:....\bf{sin^2\theta+cos^2\theta=1}}$

$\sf{=\sqrt{2(1)}}$

$\sf{=\sqrt{2}\:units}$

Answer 2

GIVEN :–

• Two points are (cosθ , - sinθ) and (sinθ , - cosθ).

TO FIND :–

• Distance between given points = ?

SOLUTION :–

• We know that distance between points (x₁ , y₁) and (x₂ , y₂) is –

$\\ \large\implies{ \boxed{\bf Distance = \sqrt{{( x_2-x_1)}^{2} + {(y_2-y_1)}^{2}}}}$

• Here –

$\\ \: \: \: \bf{\huge{.}} \: \: \: x_1 = \cos( \theta)$

$\\ \: \: \: \bf{\huge{.}} \: \: \: x_2 = \sin( \theta)$

$\\ \: \: \: \bf{\huge{.}} \: \: \: y_1 = \sin( \theta)$

$\\ \: \: \: \bf{\huge{.}} \: \: \: y_2 = -\cos( \theta)$

• Now put the values –

$\\\implies\bf Distance = \sqrt{{( \sin \theta - \cos \theta)}^{2} + {( - \cos\theta- \sin \theta)}^{2}}$

$\\\implies\bf Distance = \sqrt{{( \sin \theta - \cos \theta)}^{2} + {( \cos\theta + \sin \theta)}^{2}}$

$\\\implies\bf Distance = \sqrt{{( \sin^{2} \theta + \cos^{2} \theta - 2\sin \theta\cos\theta)}+ ( \sin^{2} \theta + \cos^{2} \theta + 2\sin \theta\cos\theta)}$

$\\\implies\bf Distance = \sqrt{{\sin^{2} \theta + \cos^{2} \theta - 2\sin \theta\cos\theta}+\sin^{2} \theta + \cos^{2} \theta +2\sin \theta\cos\theta}$

$\\\implies\bf Distance = \sqrt{\sin^{2} \theta + \cos^{2} \theta +\sin^{2} \theta + \cos^{2} \theta}$

$\\\implies\bf Distance = \sqrt{2\sin^{2} \theta + 2\cos^{2} \theta}$

$\\\implies\bf Distance = \sqrt{2(\sin^{2} \theta + \cos^{2} \theta)}$

$\\\implies\bf Distance = \sqrt{2(1)}$

$\\\implies\large{\boxed{\bf Distance = \sqrt{2}}}$

• Hence , The distance is √2.