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