Trigo Equations/Identities Problems

**gp3** · November 28th 2009, 10:14 PM

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:

**bigwave** · November 28th 2009, 10:43 PM

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

do you understand the answers from the quadradric

**gp3** · November 28th 2009, 11:06 PM

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.

**bigwave** · November 28th 2009, 11:29 PM

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

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

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

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

**gp3** · November 29th 2009, 12:03 AM

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!

**bigwave** · November 29th 2009, 07:43 PM

$2\pi$ interval is $0 \le \theta \le2\pi$

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

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

$u^2 + u - 2 = 0$

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

it is easier to use $sin\theta$ rather than $csc\theta$ since we can then recognise the commonly known $\theta$
so $csc\theta = \frac{1}{sin}$ 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???

**gp3** · November 29th 2009, 10:02 PM

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

**gp3** · November 29th 2009, 10:23 PM

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?

**bigwave** · November 30th 2009, 03:26 PM

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
$csc^2\theta$ with the identity $\frac{1}{sin^2\theta}$

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

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

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

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

**gp3** · November 30th 2009, 06:23 PM

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

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

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