kittiwitti1
  • kittiwitti1
http://prntscr.com/9atnn7 I know there's a formula for this, but I seriously forgot. Can't find it, either... can anyone jog my memory?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jim_thompson5910
  • jim_thompson5910
I would use the unit circle https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/2000px-Unit_circle_angles_color.svg.png
kittiwitti1
  • kittiwitti1
Okay
jim_thompson5910
  • jim_thompson5910
look for points that have an x coordinate of \(\LARGE \frac{\sqrt{3}}{2}\) then look at the corresponding angles at those points

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kittiwitti1
  • kittiwitti1
is there a formula though?
kittiwitti1
  • kittiwitti1
I mean something like sin(A-x) = ...?
jim_thompson5910
  • jim_thompson5910
you can use arccosine
jim_thompson5910
  • jim_thompson5910
you use the inverse trig function to undo the trig function to isolate the variable
kittiwitti1
  • kittiwitti1
Okay :)
kittiwitti1
  • kittiwitti1
Hey @jim_thompson5910 I wanted to ask another question if that's okay...\[\sin{\theta}-\sqrt{3}\cos{\theta}=\sqrt{3}\]I went about it like this: \[\sin^{2}{\theta}-(\sqrt{3}\cos{\theta})^{2}=(\sqrt{3})^{2}\]\[\sin^{2}{\theta}-\left(3\cos^{2}{\theta}\right)=\left(\sqrt{3}\right)^{2}\]\[1-\cos^{2}{\theta}-3\cos^{2}{\theta}=3\]\[-4cos^{2}{\theta}=4\]\[cos^{2}{\theta}=-1\]How is this so far?
jim_thompson5910
  • jim_thompson5910
when you square both sides, you have to square the entire side you cannot square a piece of the side also \[\Large (A+B)^2 \neq A^2 + B^2\]
kittiwitti1
  • kittiwitti1
Ah, dernit, I forgot that. Um... is there another way I could solve this problem though? o-o
jim_thompson5910
  • jim_thompson5910
so you should have this instead \[\Large \sin{\theta}-\sqrt{3}\cos{\theta}=\sqrt{3}\] \[\Large \left(\sin{\theta}-\sqrt{3}\cos{\theta}\right)^2=(\sqrt{3})^2\] \[\Large \left(\sin{\theta}-\sqrt{3}\cos{\theta}\right)\left(\sin{\theta}-\sqrt{3}\cos{\theta}\right)=(\sqrt{3})^2\]
kittiwitti1
  • kittiwitti1
Yes, but is there an easier way to solve the problem ^^;
jim_thompson5910
  • jim_thompson5910
let me think
kittiwitti1
  • kittiwitti1
Okay, thank you! :)
jim_thompson5910
  • jim_thompson5910
\[\Large \sin{\theta}-\sqrt{3}\cos{\theta}=\sqrt{3}\] \[\Large \sin{\theta}=\sqrt{3}+\sqrt{3}\cos{\theta}\] \[\Large \sin{\theta}=\sqrt{3}(1+\cos{\theta})\] \[\Large (\sin{\theta})^2=\left[\sqrt{3}(1+\cos{\theta})\right]^2\] \[\Large \sin^2{\theta}=3(1+\cos{\theta})^2\] \[\Large \sin^2{\theta}=3(1+2\cos{\theta}+\cos^2{\theta})\] \[\Large 1-\cos^2{\theta}=3(1+2\cos{\theta}+\cos^2{\theta})\] \[\Large 1-z^2=3(1+2z+z^2) \ ... \ \text{ Let } z = \cos(\theta)\] Now use the quadratic formula to solve for z
jim_thompson5910
  • jim_thompson5910
once you have your 2 solutions in terms of z, you'll use z = cos(theta) to find the solutions in terms of theta
kittiwitti1
  • kittiwitti1
O_O
kittiwitti1
  • kittiwitti1
A-alright...
kittiwitti1
  • kittiwitti1
Hey, I think I need awhile to go over this. You can go help someone else if you want
kittiwitti1
  • kittiwitti1
I don't wanna make you just wait here
jim_thompson5910
  • jim_thompson5910
ok tell me what you get for z
kittiwitti1
  • kittiwitti1
I have an insane amount of lag again lol, you really don't have to wait for me.
jim_thompson5910
  • jim_thompson5910
that's ok
kittiwitti1
  • kittiwitti1
@jim_thompson5910 I got \[4z^{2}+6z+2\]
jim_thompson5910
  • jim_thompson5910
now use the quadratic formula in this case, a = 4, b = 6, c = 2
kittiwitti1
  • kittiwitti1
Okay
kittiwitti1
  • kittiwitti1
Well, I used this formula, is that okay?|dw:1449372482375:dw|
jim_thompson5910
  • jim_thompson5910
sure factoring works too
kittiwitti1
  • kittiwitti1
Okay... I kinda forgot where to go from here. Is it\[\left(4z^{2}+4\right)\left(z+2\right)?\]
jim_thompson5910
  • jim_thompson5910
\[\Large 4z^2 + 6z + 2 \] \[\Large 4z^2 + \color{red}{6z} + 2 \] \[\Large 4z^2 + \color{red}{4z+2z} + 2 \] now factor by grouping
kittiwitti1
  • kittiwitti1
Oh, okay! Thank you. :)
kittiwitti1
  • kittiwitti1
Uh... how exactly do I factor? attempt: 4z(z+1)+2(z+1)?
jim_thompson5910
  • jim_thompson5910
then you factor out (z+1) to get (4z+2)(z+1)
kittiwitti1
  • kittiwitti1
Ah, I did do it right lol So now I simplify?
kittiwitti1
  • kittiwitti1
4z+2=0 4z=-2 z=-2/4 z+1=0 z=-1
jim_thompson5910
  • jim_thompson5910
-2/4 reduces to -1/2
jim_thompson5910
  • jim_thompson5910
the two solutions in terms of z are z = -1 or z = -1/2 now use z = cos(theta) to find the solutions in terms of theta
kittiwitti1
  • kittiwitti1
Okay I'm not sure why this is laggy again but [\cos^{-1}{\left(-1,-\frac{1}{2}\right)}?\]
kittiwitti1
  • kittiwitti1
...why is the latex not working
jim_thompson5910
  • jim_thompson5910
you take the z solutions one at a time
jim_thompson5910
  • jim_thompson5910
if z = -1, then... z = cos(theta) -1 = cos(theta) cos(theta) = -1 theta = pi ... use the unit circle (link below) https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/2000px-Unit_circle_angles_color.svg.png
kittiwitti1
  • kittiwitti1
Okay
jim_thompson5910
  • jim_thompson5910
if z = -1/2, then z = cos(theta) -1/2 = cos(theta) cos(theta) = -1/2 theta = ??? or theta = ???
kittiwitti1
  • kittiwitti1
\[\frac{2\pi}{3}\]
jim_thompson5910
  • jim_thompson5910
what else
kittiwitti1
  • kittiwitti1
\[\cos^{-1}{\left(-1,-\frac{1}{2}\right)}?\] whoops missed a slashy! that's why it didn't work lol hm... 240°?
jim_thompson5910
  • jim_thompson5910
again you do them one at a time \[\Large \cos(\theta) = -\frac{1}{2}\] \[\Large \theta = \arccos\left(-\frac{1}{2}\right)\]
jim_thompson5910
  • jim_thompson5910
I prefer the unit circle method. There are 2 points with x coordinate of -1/2 they correspond to the angles of 2pi/3 and 4pi/3
kittiwitti1
  • kittiwitti1
Sorry, was just correcting my latex error
jim_thompson5910
  • jim_thompson5910
The three *possible* solutions are theta = pi theta = 2pi/3 theta = 4pi/3 You have to check each solution one at a time. Plug each one into the original equation \[\Large \sin{\theta}-\sqrt{3}\cos{\theta}=\sqrt{3}\] and make sure you get a true equation (when you simplify both sides). If you get a true equation, then that possible solution is indeed a true solution. If you get a false equation, then that solution is considered extraneous and not a real solution at all.
kittiwitti1
  • kittiwitti1
Okay
jim_thompson5910
  • jim_thompson5910
For example Checking the solution \(\Large \theta = \pi\) \[\Large \sin(\theta)-\sqrt{3}\cos(\theta)=\sqrt{3}\] \[\Large \sin(\pi)-\sqrt{3}\cos(\pi)=\sqrt{3}\] \[\Large 0-\sqrt{3}(-1)=\sqrt{3}\] \[\Large \sqrt{3}=\sqrt{3} \ \ \color{green}{\checkmark}\] So \(\Large \theta = \pi\) has been confirmed to be a true solution
kittiwitti1
  • kittiwitti1
:O how did you get that green checkmark lol
kittiwitti1
  • kittiwitti1
Yeah thank you for that, wolfram has been super laggy as well, idk why
jim_thompson5910
  • jim_thompson5910
I typed `\checkmark`
kittiwitti1
  • kittiwitti1
ah, thank you :)
jim_thompson5910
  • jim_thompson5910
if wolfram doesn't work, use http://web2.0calc.com/ or google's calculator
kittiwitti1
  • kittiwitti1
okay thanks :)
kittiwitti1
  • kittiwitti1
@jim_thompson5910 2pi/3 gave me 0 http://prntscr.com/9auvep
jim_thompson5910
  • jim_thompson5910
same here, so \(\Large \theta = \frac{2\pi}{3}\) has been confirmed to be a true solution now check \(\Large \theta = \frac{4\pi}{3}\)
kittiwitti1
  • kittiwitti1
and 4pi/3 gave me \[-\sqrt{3}\] http://prntscr.com/9auvl7
jim_thompson5910
  • jim_thompson5910
Checking the solution \(\Large \theta = \frac{2\pi}{3}\) \[\Large \sin(\theta)-\sqrt{3}\cos(\theta)=\sqrt{3}\] \[\Large \sin\left(\frac{2\pi}{3}\right)-\sqrt{3}\cos\left(\frac{2\pi}{3}\right)=\sqrt{3}\] \[\Large \frac{\sqrt{3}}{2}-\sqrt{3}*\frac{-1}{2}=\sqrt{3}\] \[\Large \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\sqrt{3}\] \[\Large \frac{\sqrt{3}+\sqrt{3}}{2}=\sqrt{3}\] \[\Large \frac{2\sqrt{3}}{2}=\sqrt{3}\] \[\Large \sqrt{3}=\sqrt{3} \ \ \color{green}{\checkmark}\] So \(\Large \theta = \frac{2\pi}{3}\) has been confirmed to be a true solution I think you meant to have a "minus" not a "plus"
kittiwitti1
  • kittiwitti1
OH right
kittiwitti1
  • kittiwitti1
4pi/3 is still wrong though http://prntscr.com/9auw7p
jim_thompson5910
  • jim_thompson5910
Checking the solution \(\Large \theta = \frac{4\pi}{3}\) \[\Large \sin(\theta)-\sqrt{3}\cos(\theta)=\sqrt{3}\] \[\Large \sin\left(\frac{4\pi}{3}\right)-\sqrt{3}\cos\left(\frac{4\pi}{3}\right)=\sqrt{3}\] \[\Large -\frac{\sqrt{3}}{2}-\sqrt{3}*\frac{-1}{2}=\sqrt{3}\] \[\Large -\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\sqrt{3}\] \[\Large \frac{-\sqrt{3}+\sqrt{3}}{2}=\sqrt{3}\] \[\Large \frac{0\sqrt{3}}{2}=\sqrt{3}\] \[\Large 0=\sqrt{3} \ \ \color{red}{X ... \text{FALSE}}\] So \(\Large \theta = \frac{4\pi}{3}\) is NOT true solution. It is extraneous I agree with what you got
jim_thompson5910
  • jim_thompson5910
So if \(\Large \theta\) is restricted to the interval \(\Large 0 \le \theta < 2\pi\) then the only two solutions are \(\Large \theta = \pi, \theta = \frac{2\pi}{3}\)
kittiwitti1
  • kittiwitti1
so 2pi/3, pi... or in degree form: 180, 120
jim_thompson5910
  • jim_thompson5910
yep correct
kittiwitti1
  • kittiwitti1
alright thank you! :)

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