# Thread: Sorry....I've got one more thing I need help with....

1. ## Sorry....I've got one more thing I need help with....

So, not like anyone cares, but this was on my math test today and I didn't know how to solve it. I'd wait and see what I had done wrong when my test comes back, but my teacher NEVER gives our tests back, nor does she go over anything.......... My teacher sucks. You can tell I'm just a beginner when it comes to trig.....ANYWAY, here's the problem:

$\displaystyle tan2(x+41)=1$

by the way, the 41 is really 41degrees.... I don't know how to put a degree sign in it....

It seems so simple....why can't I figure it out?

2. ## Ok

Originally Posted by blair_alane
So, not like anyone cares, but this was on my math test today and I didn't know how to solve it. I'd wait and see what I had done wrong when my test comes back, but my teacher NEVER gives our tests back, nor does she go over anything.......... My teacher sucks. You can tell I'm just a beginner when it comes to trig.....ANYWAY, here's the problem:

$\displaystyle tan2(x+41)=1$

by the way, the 41 is really 41degrees.... I don't know how to put a degree sign in it....

It seems so simple....why can't I figure it out?
assuming this is $\displaystyle tan^{2}(x+41)=1$ then $\displaystyle tan(x+41)=\sqrt{1}=1\Rightarrow{x+41=arctan(1)=\fr ac{\pi}{4}=45^{\circ}}$$\displaystyle \Rightarrow{x=4^{\circ}}\Rightarrow{tan(4+41)^2=ta n(45)^2=1^2=1} 3. Originally Posted by blair_alane I'd wait and see what I had done wrong when my test comes back, but my teacher NEVER gives our tests back :/ that would really frustrate me. Originally Posted by blair_alane by the way, the 41 is really 41degrees.... I don't know how to put a degree sign in it.... My laptop doesn't work with this, so I can't test it, but I think if you hold down alt and type 167 or 168, then release alt, that it should put that in (I think one is the degree symbol, and the other is an upside down question mark. I used to know them all back when I gamed heavily, but that was years ago.) Originally Posted by Mathstud28 assuming this is \displaystyle tan^{2}(x+41)=1 then \displaystyle tan(x+41)=\sqrt{1}=1\Rightarrow{x+41=arctan(1)=\fr ac{\pi}{4}=45^{\circ}}$$\displaystyle \Rightarrow{x=4^{\circ}}\Rightarrow{tan(4+41)^2=ta n(45)^2=1^2=1}$
Good job, but I think your work would be more legible (and therefore easier to appreciate ) if it were broken down into multiple lines. Effective use of white space greatly increases readability, there are actually many reasons for this, but it's a bit of a tangent that I'll avoid.

4. ## Haha

Originally Posted by angel.white
:/ that would really frustrate me.
My laptop doesn't work with this, so I can't test it, but I think if you hold down alt and type 167 or 168, then release alt, that it should put that in (I think one is the degree symbol, and the other is an upside down question mark. I used to know them all back when I gamed heavily, but that was years ago.)

Good job, but I think your work would be more legible (and therefore easier to appreciate ) if it were broken down into multiple lines. Effective use of white space greatly increases readability, there are actually many reasons for this, but it's a bit of a tangent that I'll avoid.
You are very right... it

is just a lot of the time

I am in a rush so I do

that as quick as possible... thanks

for the suggestion thought

haha just kidding thanks for the advice!!

5. Originally Posted by Mathstud28
assuming this is $\displaystyle tan^{2}(x+41)=1$ then $\displaystyle tan(x+41)=\sqrt{1}=1\Rightarrow{x+41=arctan(1)=\fr ac{\pi}{4}=45^{\circ}}$$\displaystyle \Rightarrow{x=4^{\circ}}\Rightarrow{tan(4+41)^2=ta n(45)^2=1^2=1}$
Wow....you make it look so easy..... Thanks SO MUCH!!!!! I get it now!

Teehee! I SO didn't get that on my test! oh well! Thanks again! I owe you!

-BAM

6. Originally Posted by angel.white
:/ that would really frustrate me.
It's VERY frustrating.... I mean, does she REALLY want to keep our tests?! hehe.... well, you have to have a bad teacher sometime..... I could go on about my teacher, but I wont 'cause I doubt anyone cares!

Originally Posted by angel.white
My laptop doesn't work with this, so I can't test it, but I think if you hold down alt and type 167 or 168, then release alt, that it should put that in (I think one is the degree symbol, and the other is an upside down question mark. I used to know them all back when I gamed heavily, but that was years ago.)
I'll have to try this at home! I'm at school and, let's face it, my school computers suck! heehee!

Thanks!

BAM

7. Originally Posted by Mathstud28
assuming this is $\displaystyle tan^{2}(x+41)=1$ then $\displaystyle tan(x+41)=\sqrt{1}=1\Rightarrow{x+41=arctan(1)= \frac{\pi}{4}=45^{\circ}}$ $\displaystyle \Rightarrow{x=4^{\circ}}\Rightarrow{tan(4+41)^2= tan(45)^2=1^2=1}$
All your calculations are OK - but (there is allways a "but") I'm missing a second solution:

$\displaystyle \tan^2(x+41^\circ) = 1~\implies~|\tan(x+41^\circ)|=1$ $\displaystyle ~\implies~\tan(x+41^\circ)=1~ \vee~\tan(x+41^\circ)=-1$

You have solved the first equation. If you solve the second equation you'll get:

$\displaystyle x+41^\circ = 135^\circ~\implies~x = 94^\circ$

To be exact the solution of this equation consists of an unlimited amount of values.

8. Originally Posted by earboth
All your calculations are OK - but (there is allways a "but") I'm missing a second solution:

$\displaystyle \tan^2(x+41^\circ) = 1~\implies~|\tan(x+41^\circ)|=1$ $\displaystyle ~\implies~\tan(x+41^\circ)=1~ \vee~\tan(x+41^\circ)=-1$

You have solved the first equation. If you solve the second equation you'll get:

$\displaystyle x+41^\circ = 135^\circ~\implies~x = 94^\circ$

To be exact the solution of this equation consists of an unlimited amount of values.
wow..... you're right..... thanks a bunch! I NEVER would have thought about the "but"!

9. Originally Posted by earboth
All your calculations are OK - but (there is allways a "but") I'm missing a second solution:

$\displaystyle \tan^2(x+41^\circ) = 1~\implies~|\tan(x+41^\circ)|=1$ $\displaystyle ~\implies~\tan(x+41^\circ)=1~ \vee~\tan(x+41^\circ)=-1$

You have solved the first equation. If you solve the second equation you'll get:

$\displaystyle x+41^\circ = 135^\circ~\implies~x = 94^\circ$

To be exact the solution of this equation consists of an unlimited amount of values.
Yeah I thought about that but I assumed that this was either implied or that she was disregarding this since she said there is only one answer