How do you convert (-sqrt(6), -sqrt(2))(−√6,−√2) to polar form?
1 Answer
Mar 1, 2016
Explanation:
Use the formulae that links Cartesian to Polar coordinates.
• r^2 = x^2 + y^2∙r2=x2+y2
• theta = tan^-1(y/x)∙θ=tan−1(yx) here
x = -sqrt6 " and " y = -sqrt2 x=−√6 and y=−√2
r^2 = (-sqrt6)^2 + (-sqrt2)^2 = 6 + 2 = 8 r2=(−√6)2+(−√2)2=6+2=8
r^2 = 8 rArr r = sqrt8 = 2sqrt2 r2=8⇒r=√8=2√2 The point
(-sqrt6,-sqrt2)" is in 3rd quadrant" .(−√6,−√2) is in 3rd quadrant. care must be taken to ensure that
theta" is in 3rd quadrant"θ is in 3rd quadrant
theta = tan^-1((-sqrt2)/(-sqrt6)) = 30^@ orpi/6" radians "θ=tan−1(−√2−√6)=30∘orπ6 radians This is the 'related' angle in the 1st quadrant , require 3rd.
rArr theta = (180+30)^@ = 210^@ or (7pi)/6" radians "⇒θ=(180+30)∘=210∘or7π6 radians
