Curve Skething -- Inverse trig (1 Viewer)

[abc123doremi]

we recently had 4u half yearlies and they asked us to draw:
cos (x+y) = 0
for curve sketching

i'm sure it must be a line or an array of lines, but anyone know how to solve it in such a way?

graphmatica gives me a line y= -x + pi/2 but then again it's not good for implicit fuctions either

any suggestions?

 

[m.jakaran]

It is an array of lines, as if the substitution, x+y=a is taken, then all we need to do is find the values for which cos(a)=0 and then sketch the lines, this goes back to three unit general solutions.

cos(a)=0 for n(pi/2) and n(3pi/2) hence we must draw
.
.
.
x+y=(-pi/2)

x+y=(pi/2)
.
.
.

by simply changing the subject, we obtain
.
.
.
y=(-pi/2)-x

y=(pi/2)-x
.
.
.
however your graphing program only did this for the pi/2 solution

 

[abc123doremi]

yeah i got that out
but i added 2k(pi) to account for all the other possible solutions
is that wrong?
coz i somewhat remember during inverse in three unit that that's what you gotta do
so there should be an infinite array of lines across the whole cartesian plane?

 

[m.jakaran]

Sorry, I just learned inverse trig last week, the correct answer is 2npi or minus(pi/2), and yes they should be an infinite amount of lines across the plane.

 

[youngminii]

cos(x + y) = cosxcosy - sinxsiny = 0
cosxcosy = sinxsiny
cosx = sinxtany
tany = 1/tanx
y = inversetan(1/tanx)
Which should be pretty simple if considering inversetan(tanx) = x, so I THINK inversetan(1/tanx) = 1/x
So it'll be a hyperbola? Can't be bothered thinking further.

Edit: Okay after reading the answers above, maybe I'm completely wrong.
Edit #2: inversetan(cotx) is right. It just doesn't = 1/x.
It's a series of zig zags with domain -pi/2 < x < pi/2 with a period of pi. It literally looks like this:

|\|\|\|\|\|\|\|\|\

 

[abc123doremi]

right thanks