How do you graph #theta=(5pi)/4#?

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Answer 1 · 2016-11-01T19:15:53

It is the line #y = x; x < 0#

This the 3rd quadrant, therefore, we must restrict x and y to be less than 0 , when we substitute #tan^-1(y/x)# for #theta#

#tan^-1(y/x) = (5pi)/4; x < 0 and y < 0#

Take the tangent of both sides:

#tan(tan^-1(y/x)) = tan((5pi)/4); x < 0 and y < 0#

The tangent "undoes" its inverse and substitute 1 for #tan((5pi)/4)#:

#y/x = 1; x < 0 and y < 0#

Multiply both sides by x:

#y = x; x < 0#

We dropped the restriction on y, because it is, now, a dependent variable.