# Thread: Trigo Equations/Identities Problems

1. ## Trigo Equations/Identities Problems

Hi guys,

I've been studying the notes that my teacher gave me but i still can't solve these few questions!!!

Can someone help me? I have a test coming up next week! Thanks a lot.

Ans:

Ans:

2. ## it becomes a quadratic

$\displaystyle cot^2\theta + csc\theta = 1$
use the identity $\displaystyle cot^2\theta = csc^2\theta - 1$
so,
$\displaystyle csc^2\theta - 1 + csc\theta = 1$
set it to zero
$\displaystyle csc^2\theta + csc\theta -2 = 0$
this factors
$\displaystyle (csc\theta - 1)(csc\theta +2) = 0$
from this you will get
$\displaystyle \frac{\pi}{2}, \frac{11\pi}{6}, \frac{7\pi}{6}$

3. now one thing I don't really understand is why is it that some of these questions only have 1 answers while some have more than 5!

I know you have to look at which quadrant they are in and the range, is that true? for eg. in the 2nd quad, you have to take 180deg + 'basic angle'. I'm not really clear on this.

4. ok starting from the factored quadradic which is
$\displaystyle (csc\theta - 1)(csc\theta +2) = 0$
if you take $\displaystyle (csc\theta - 1)$ so that it becomes zero the whole equation goes to zero so if $\displaystyle csc\theta = 1$ then that will be some our answer

$\displaystyle csc\theta = \frac{1}{sin\theta}$

so $\displaystyle \frac{1}{sin\theta} = 1$or$\displaystyle sin\theta = 1$ so $\displaystyle sin\theta$ is only 1 at $\displaystyle \frac{\pi}{2}$ that is only place on the $\displaystyle 2\pi$ interval that it can

we do the same with $\displaystyle (csc^2\theta + 2)$ except the results will show up in 2 places on the $\displaystyle 2\pi$ interval

can you see this....
sorry i am slow on this not real fast with the latex
I will help you more if you need.

5. sorry man, i dont really understand: 'only place on the interval'

and means the first quadrant?

how did you factor to get ?

I use the quadratic formula and only get .

really thanks for your help!

6. $\displaystyle 2\pi$ interval is $\displaystyle 0 \le \theta \le2\pi$

$\displaystyle \frac{2}{\pi}$ is within $\displaystyle 0 \le \theta \le2\pi$ we are not concerned about what quadrant it is in.

$\displaystyle csc^2\theta + csc\theta -2 = 0$
using
$\displaystyle u$ substitution where $\displaystyle u = csc\theta$
you have

$\displaystyle u^2 + u - 2 = 0$

which is factored as
$\displaystyle (u - 1)(u +2)$

it is easier to use $\displaystyle sin\theta$ rather than $\displaystyle csc\theta$ since we can then recognise the commonly known $\displaystyle \theta$
so $\displaystyle csc\theta = \frac{1}{sin}$ so substituting it back into the factors we can solve
see the chart below(which i stole from another thread to see how these were derived..

are we getting the picture???

7. thanks buddy! I'll try to see if i can solve it first.

8. alright i got the first ans.

but how do you get the other 2 answers since gives you a negative answer (-2), and not in the range, I thought we must leave it out, as in we cant get an answer from it?

9. Originally Posted by gp3
alright i got the first ans.

but how do you get the other 2 answers since gives you a negative answer (-2), and not in the range, I thought we must leave it out, as in we cant get an answer from it?
ok, again, it will be easier to replace
$\displaystyle csc^2\theta$ with the identity $\displaystyle \frac{1}{sin^2\theta}$

then we can work with $\displaystyle \frac{1}{sin^2\theta} = -2$

multiply both sides by $\displaystyle sin^2\theta$ to get $\displaystyle 1 = -2sin\theta$

now divide both sides by $\displaystyle -2$ to get $\displaystyle -\frac{1}{2} = sin\theta$
this has 2 answers $\displaystyle \frac{7\pi}{6}$and $\displaystyle \frac{11\pi}{6}$
they are in the III and IV quadrants respectfully

do you still need help with the next one you originally posted

10. hey really big thanks man! I understand how to solve it now!

and i got how to solve the other one as well. Thanks!