# Trigo Equations/Identities Problems

• November 28th 2009, 10:14 PM
gp3
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!!! (Angry)

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

http://img341.imageshack.us/img341/2276/scan0001.gif
Ans: http://img163.imageshack.us/img163/9...an0002copy.jpg

http://img341.imageshack.us/img341/8...an0001copy.gif
Ans: http://img694.imageshack.us/img694/6...02copycopy.jpg
• November 28th 2009, 10:43 PM
bigwave
$cot^2\theta + csc\theta = 1$
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
$\frac{\pi}{2}, \frac{11\pi}{6}, \frac{7\pi}{6}$

• November 28th 2009, 11:06 PM
gp3
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. (Headbang)
• November 28th 2009, 11:29 PM
bigwave
$(csc\theta - 1)(csc\theta +2) = 0$
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
• November 29th 2009, 12:03 AM
gp3
sorry man, i dont really understand: 'only place on the http://www.mathhelpforum.com/math-he...cdff3fc7-1.gif interval'

how did you factor to get http://www.mathhelpforum.com/math-he...7dd64c44-1.gif ?

I use the quadratic formula and only get http://www.mathhelpforum.com/math-he...74d72168-1.gif.

• November 29th 2009, 07:43 PM
bigwave
$2\pi$ 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???

• November 29th 2009, 10:02 PM
gp3
thanks buddy! I'll try to see if i can solve it first. (Bow)
• November 29th 2009, 10:23 PM
gp3
alright i got the first ans. http://www.mathhelpforum.com/math-he...a002d0ad-1.gif

but how do you get the other 2 answers since http://www.mathhelpforum.com/math-he...3323ac03-1.gif 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? (Wondering)(Wondering)
• November 30th 2009, 03:26 PM
bigwave
Quote:

Originally Posted by gp3
alright i got the first ans. http://www.mathhelpforum.com/math-he...a002d0ad-1.gif

but how do you get the other 2 answers since http://www.mathhelpforum.com/math-he...3323ac03-1.gif 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? (Wondering)(Wondering)

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
• November 30th 2009, 06:23 PM
gp3
hey really big thanks man! I understand how to solve it now! (Nod)(Nod)

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