Question

find the distance between the points p(cosΦ, sinΦ) ;q(sinΦ,-cosΦ)​

Answer 1

Answer:

ANSWER=2

Step-by-step explanation:

pq=$\sqrt{(sinθ-cosθ)^2+ (-cosθ-sinθ)^2}$

    =sin^2θ+cos^2θ-2sinθcosθ +cos^2θ+sin^2θ-2cosθsinθ

     =1+1

     =2

Answer 2

$\color{red}\large\underline{\underline{Question:}}$

$\mathrm{\mapsto Find\:the\:distance\:between\:the\:points:}$

$\mathtt{\left(cos\:\theta\:,\:sin\:\theta\right)\:;\:</p><p>\left(sin\:\theta\:,\:cos\:\theta\right)}$

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$\color{purple}\large\underline{\underline{To\:Find:}}$

$\mathtt{The\:distance\:between\:the\:points:}$

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$\color{purple}\large\underline{\underline{We\:Know:}}$

• $\color{blue}{\underline\mathtt{Distance\:formula}}$

$\mathtt{\sqrt{\left(x_{2} - x_{1}\right)^{2} + \left(y_{2} - y_{1}\right)^{2}}}$

• $\mathtt{cos^{2}\theta + sin^{2}\theta}$
• $\mathtt{(a + b)^{2} = a^{2} + 2ab + b^{2}}$
• $\mathtt{(a - b)^{2} = a^{2} - 2ab + b^{2}}$

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$\color{lime}\large\underline{\underline{Given:}}$

• $x_{1} = cos\:\theta$
• $x_{2} = sin\:\theta$
• $y_{2} = - cos\:\theta$
• $y_{1} = sin\:\theta$

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$\color{orange}\large\underline{\underline{Solution:}}$

$\textsf{Applying the Distance Formula , we get}$

$\mathtt{\sqrt{\left(x_{2} - x_{1}\right)^{2} + \left(y_{2} - y_{1}\right)^{2}}}$

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

$\Rightarrow \sqrt{\left(sin^{2}\theta + 2·sin\theta·cos\theta + cos^{2}\theta\right) + \left(- cos^{2}\theta - 2·sin\:\theta·\cos\:theta + sin^{2}\theta\right)}$

$\Rightarrow \sqrt{\left(sin^{2}\theta \cancel{+ 2·sin\theta·cos\theta} + cos^{2}\theta\right) + \left(- cos^{2}\theta \cancel{ - 2·sin\:\theta·\cos\:theta} + sin^{2}\theta\right)}$

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

$\Rightarrow \sqrt{1 + 1}$

$\Rightarrow \sqrt{2}$

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$\mathtt{\therefore \left(cos\:\theta\:,\:sin\:\theta\right)\:;\:</p><p>\left(sin\:\theta\:,\:cos\:\theta\right) = \sqrt{2}}$