Help explaining how to find the domain of tangent functions
so the domain of a tangent is all real number except odd multiples of \[\frac{ \pi }{ 2b }+\frac{ c }{ b }\]
The characteristic of a tangent function is \[f(\theta)=atan(b \theta-c)+d\]
so with this how would you do that with \[f(\theta)=\tan(1/2 \pi \theta)-2\]
would pi be b?
b is the "stuff" in front of theta I can't tell if it's 1/(2pi) or pi/2 . but whatever it is, that is b once you know b, you use it in the very first post
@phi sorry its tan(1/2)pi theta)
so c would be 1/2 then so \[\frac{ \pi }{ 2\pi }+\frac{ \frac{ 1 }{ 2 } }{ \pi }\]
if it's \[ \tan\left(\frac{1}{2} \pi \theta \right) -2\] that is the same as \[ \tan\left(\frac{\pi}{2} \theta \right) -2\]
Why would that be exactly I am confused :/
and also if it is pi/2 then what would be c? 0?
the site is being a pain. anyway, we match up \[ \tan\left(\frac{\pi}{2} \theta \right) -2 \] with \[ f(\theta)=\tan(b \theta-c)+d \] there is no "c" (c is 0) b is pi/2 so we use those in \[ \frac{ \pi }{ 2b }+\frac{ c }{ b } \] we need 2*b, which is 2*pi/2 = pi and put that in your formula: \[ \frac{ \pi }{ \pi }+\frac{ 0 }{ b } \] it looks like we get 1
so it is all real numbers except odd multiples of 1?
the domain of a tangent is all real number except odd multiples of 1 I guess that would be all numbers except 1,3,5,... (the odd numbers)
so except odd integers
yes
Thank you so much for your help. I kinda have just been trying to solve this for like 3 hours now s: so this should help me solve the other ones. I cannot thank you enough
so one more thing I was doing the next one which is \[f(\theta)=\tan(\frac{ 1 }{ 3 }\pi \theta)+1 \] Which would be \[\tan(\frac{ \pi \theta }{ 3 })+1\] so if you put this into the equation so this would be 3/2+0/pi @phi would the final answer be All real numbers except for the odd multiples of 3/2?
I believe this is right so :P you explaining did help me alot ahaha
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