If R^2=3, then R=+- sqrt(3)
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If R^2=3, then R=+- sqrt(3)

[From: ] [author: ] [Date: 11-12-29] [Hit: ]
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RcosΘ=1 and RsinΘ=-sqrt(2)

Need to find R for Rsin(x+Θ)=1

R^2[sin^2(Θ)]+R^2[cos^2(Θ)]=R^2(1)=1^2… so that R=sqrt(3)

Isn't R here supposed to be -sqrt(3) and +sqrt(3) ?

Do we get rid of the negative square root of 3 because R is a radius? Why?

Thank You.

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I think you're right to get rid of the negative value.
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