# Thread: Simplifying Fractions and Solving Trig. Equations

1. ## [Help needed] Simplifying a fraction & Solving 2 Trig. Equations

Alright...I need some help with an Algebra problem and 2 Trig. problems...

"Express in Simplest form:" X/4 -4/x / 1 -4/x

I wrote: (x/4 - 4/x / 1- 4/x) x
x^2/4x - 4 / 1-4

...I'm stuck from this point on.

"Solve the following Trig. equation."

5 cos^2 (theta) -1 = 3 (1 - cos^2 (theta))

I wrote: 5 cos^2 (theta) -1= 3 (1 - cos^2 (theta)
5 cos^2 (theta) -1= 3 - 3 cos^2 (theta)
5 cos^2 (theta)-1 = cos^2 (theta)

...and I'm stuck.

Sin^2 (theta) - 2 Sin (theta) = 3
...I wrote, WTF? ...

Seriously...Is there some type of "trick" to solving trig. equations? I seem to be the only one in my class to be "not-getting" this! The teacher will not take the time to explain trig. equation-solving to me...because she says "you should know how to do these"... If you could please give reasoning in words to your work on solving these 2 trig. equations I would appreciate it!

P.S. Thank you to all on this forum who have helped me so far in other problems, I appreciate it.

2. x/4 -4/x / 1 -4/x

Please try to bracket your terms to make it less ambiguous for us so we can actually decipher your problem. Is this your question: $\displaystyle \frac{\frac{x}{4} - \frac{4}{x}}{1 - \frac{4}{x}}$ ?

-----------------------------------
Hmm your algebra is a bit off:
$\displaystyle 5\cos^{2} \theta - 1 = 3 - 3\cos^{2} \theta$
$\displaystyle 5\cos^{2} \theta + 3\cos^{2} \theta = 3 + 1$ (move $\displaystyle 3\cos^{2}\theta$ to the left and -1 to the right)
$\displaystyle 8\cos^{2} \theta = 4$ (Simplify)

Keep going.

-----------------------------------
$\displaystyle \sin^2 \theta - 2 \sin \theta = 3$

For this question, you're going to have to factor as if you were with a trinomial:
$\displaystyle \sin^{2} \theta - 2\sin \theta - 3 = 0$

Imagine $\displaystyle x = \sin \theta$: $\displaystyle x^{2} - 2x - 3 = 0$

Factor as you would with that. Then solve for theta.

3. Originally Posted by o_O
Please try to bracket your terms to make it less ambiguous for us so we can actually decipher your problem. Is this your question: ?
Sorry about the bracket thing... Yes, that is my problem. Could you tell me how you got the problem to look so neat? Are you using some type of software or...do you just use the brackets?

4. What I'm using is LaTex. Here's a tutorial on how to use it in your posts if you're interested: Latex Tutorial

As for the problem, you can get rid of the denominators by multiplying both top and bottom by 4x:
$\displaystyle \frac{\frac{x}{4} - \frac{4}{x}}{1 - \frac{4}{x}} \cdot {\color{blue} \frac{4x}{4x}}$

See what you can do and post back if you have any troubles.

5. Thank you very much! I factored both trig. equations and I "got them" now.

For the simplification of the fraction, I got [x/4] for an answer.
Is this correct? I hope so...

Thank you again for your help!

6. Not quite:
$\displaystyle \frac{\frac{x}{4} - \frac{4}{x}}{1 - \frac{4}{x}} \cdot {\color{blue} \frac{4x}{4x}} = \frac{x^{2} - 16}{4x - 16}$

Factor top and bottom and see what cancels

7. Hello, MrOats!

Since you don't use parentheses, I'll have to guess what you meant . . .

Express in simplest form: .$\displaystyle \frac{\dfrac{x}{4} -\dfrac{4}{x}}{1 -\dfrac{4}{x }}$
Multiply top and bottom by $\displaystyle 4x\!:$

. . $\displaystyle \frac{4x\left(\dfrac{x}{4}-\dfrac{4}{x}\right)}{4x\left(1 - \dfrac{4}{x}\right)} \;=\;\frac{x^2-16}{4x-16} \;=\;\frac{(x-4)(x+4)}{4(x-4)} \;=\;\frac{x+4}{4}$

Solve: .$\displaystyle 5\cos^2\!\theta -1 \:= \:3 (1 - \cos^2\!\theta)$

I wrote: .$\displaystyle 5\cos^2\!\theta -1\:= \:3 - 3\cos^2\!\theta$
. . . . . . .$\displaystyle 5\cos^2\!\theta-1 \:= \:\cos^2\!\theta\quad\hdots$ What?
Start again . . .

. . $\displaystyle 5\cos^2\!\theta - 1 \:=\:3 - 3\cos^2\!\theta \qquad\Rightarrow\qquad 8\cos^2\!\theta \:=\:4$

. . $\displaystyle \cos^2\!\theta \:=\:\frac{1}{2}\qquad\Rightarrow\qquad \cos\theta \:=\:\pm\frac{1}{\sqrt{2}}$

Therefore: .$\displaystyle \theta \:=\:\frac{\pi}{4} + \frac{\pi}{2}n$

$\displaystyle \sin^2\!\theta - 2\sin\theta \:=\:3$

We have: .$\displaystyle \sin^2\!\theta - 2\sin\theta - 3 \:=\:0$

Factor: .$\displaystyle (\sin\theta - 3)( \sin\theta + 1)\:=\:0$

. . $\displaystyle \sin\theta-3\:=\:0\quad\Rightarrow\quad \sin\theta \:=\:3\quad\hdots$ no real solutions

. . $\displaystyle \sin\theta + 1\:=\:0\quad\Rightarrow\quad \sin\theta \:=\:-1\quad\Rightarrow\quad \theta \;=\:\frac{3\pi}{2} + 2\pi n$

8. To: o_O , THANK YOU.

To: Soroban, Thank you as well!