# Thread: Not quite getting this trig stuff

1. ## Not quite getting this trig stuff

hi guys.

currently studying some maths at university, and they assume a whole bunch of high school maths knowledge. problem is, i was at high school 7 years ago, i've forgotten it all.

anyway....

I'm having some trouble working these kind of problems out...

tan x = 1

i would love to reach for my calculator and get 0.785398, but unfortunately the kind of answer i need is pi/4 ....

I've been staring at unit circles and triangles and stuff trying to work them out, but i'm just not getting it. how do i answer these kind of problems without reaching for the calculator?

Thanks in advance for any help, exam is tomorrow, and boy do i need it!!

2. $\tan x=1\iff x=\arctan1,\,\forall x\in\left.\right]-\frac\pi2,\frac\pi2\left[\right.$

Then $x=\frac\pi4$

3. could you dumb that down....a fair bit if possible. i really don't get this stuff at all.

4. Originally Posted by bruxism
could you dumb that down....a fair bit if possible. i really don't get this stuff at all.
here are some stuff you have to memorize anyway, so let's get them out of the way

You MUST know that:

$\tan x = \frac {\sin x}{\cos x}$

$\sec x = \frac {1}{\cos x}$ ........note that csc(x) and sec(x) are counter-intuitive, the s for secant goes with the c for cosine and the c for cosecant goes with the s for sine

$\csc x = \frac {1}{\sin x}$

$\cot x = \frac {1}{\tan x} = \frac {\cos x}{\sin x}$

Now, see the table below. it is a lot easier to memorize than the tables most professors give their class.

here's how to fill it out.

STEP 1:
label the first row and first column as you see them. it's easy to remember because you count up with the angles, 30, 45, 60 ...

STEP 2:
for sin(x), you count up, you write starting in the slot beside sin(x), 1 , 2 , 3

for cosine, count down, fill in 3, 2, 1 (i highlighted these numbers so you can pin-point them)

STEP 3:
divide all angles by 2

STEP 4:
take the square root of the numerators of all the fractions

STEP 5: Pat yourself on the back to congratulate yourself on remembering that, and whisper "Thank God Jhevon showed this to me, Jhevon is the man!"

now you know the main angles you need to know for the trig functions. any other angle you need to know by heart, you can use this table with your knowledge of the formulas above (which you have to memorize anyway) and your knowledge of reference angles (which you have to know anyway).

now, recall that: $\tan x = \frac {\sin x}{\cos x}$

so, $\tan x = 1$ means $\frac {\sin x }{\cos x} = 1$

$\Rightarrow \sin x = \cos x$

but from your memory of the table, this happens when $x = \frac {\pi}{4}$

and so, we're done

any questions?

5. that's made things a lot clearer!! i've just got one more question for now....

lets say i'm faced with tan x = 1/sqrt 3

i reverse engineered that question, but hopefully the answer is pi/6.

how would you come to an answer like that?

6. Originally Posted by bruxism
that's made things a lot clearer!! i've just got one more question for now....

lets say i'm faced with tan x = 1/sqrt 3

i reverse engineered that question, but hopefully the answer is pi/6.

how would you come to an answer like that?
that's correct

again, you would officially write:

$\tan x = \frac {1}{\sqrt {3}}$

$\Rightarrow x = \arctan \left( \frac {1}{\sqrt {3}} \right)$

$\Rightarrow x = \frac {\pi}{6}$

and everyone would marvel at your genius.

but what you actually thought was...

ok, i am so dead, this question is so hard! wait, calm down. tangent is sine over cosine right? yeah. i have 1 in the top and square root of 3 in the bottom. so i need the cosine to be in the bottom for tangent, so i think i'll try to make the cosine be something with square root of 3 in it.

aha! i remember Jhevon showed me that table. $\cos \left( \frac {\pi}{6} \right) = \frac {\sqrt {3}}{2}$. Yes! that might be it! now, let's just hope the sine part behaves.

i remember from the table that $\sin \left( \frac {\pi}{6} \right) = \frac {1}{2}$

ok, so let's reverse engineer this baby!

so then, $\tan \left( \frac {\pi}{6} \right) = \frac { \sin \left( \frac {\pi}{6} \right)}{ \cos \left( \frac {\pi}{6} \right)} = \frac {\frac {1}{2}}{ \frac {\sqrt {3}}{2}} = \frac {1}{2} \cdot \frac {2}{\sqrt {3}}$

the two's cancel, and i'm left with $\frac {1}{\sqrt {3}}$

aha! so $x = \frac {\pi}{6}$ was the answer.

i am the man, wasn't worried for one second. trig ain't that hard. that professor has got to get up pretty early in the morning to put one over on me. he better come good for the final and not insult me with such easy questions!

the moral of the story? stay of the coffee and see a therapist about that voice in your head

7. thanks, you really are the man. Entertaining and educational!! unbelievably i've actually signed up at uni to become a MATHS TEACHER!! i loved maths in school and enjoy teaching so thought this might be the go. I forgot how hard it can be though!

The exam tomorrow...if anyone cares, is on the cartesian plane, methods of proof, and complex numbers.

Most of it is converting polar coordinates to rectangular coordinates and back, and converting complex numbers to polar form and exponential forms. I'm gonna have to study my little butt off, but i think i'll get there.

Thanks for the help!!!