Question

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?

Answer 1

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

Answer 2

Okay

Answer 3

look for points that have an x coordinate of \(\LARGE \frac{\sqrt{3}}{2}\)

then look at the corresponding angles at those points

Answer 4

is there a formula though?

Answer 5

I mean something like sin(A-x) = ...?

Answer 6

you can use arccosine

Answer 7

you use the inverse trig function to undo the trig function to isolate the variable

Answer 8

Okay :)

Answer 9

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?

Answer 10

when you square both sides, you have to square the entire side

you cannot square a piece of the side

also

\[\Large (A+B)^2 \neq A^2 + B^2\]

Answer 11

Ah, dernit, I forgot that.
Um... is there another way I could solve this problem though? o-o

Answer 12

so you should have this instead

\[\Large \sin{\theta}-\sqrt{3}\cos{\theta}=\sqrt{3}\]

\[\Large \left(\sin{\theta}-\sqrt{3}\cos{\theta}\right)^2=(\sqrt{3})^2\]

\[\Large \left(\sin{\theta}-\sqrt{3}\cos{\theta}\right)\left(\sin{\theta}-\sqrt{3}\cos{\theta}\right)=(\sqrt{3})^2\]

Answer 13

Yes, but is there an easier way to solve the problem ^^;

Answer 14

let me think

Answer 15

Okay, thank you! :)

Answer 16

\[\Large \sin{\theta}-\sqrt{3}\cos{\theta}=\sqrt{3}\]

\[\Large \sin{\theta}=\sqrt{3}+\sqrt{3}\cos{\theta}\]

\[\Large \sin{\theta}=\sqrt{3}(1+\cos{\theta})\]

\[\Large (\sin{\theta})^2=\left[\sqrt{3}(1+\cos{\theta})\right]^2\]

\[\Large \sin^2{\theta}=3(1+\cos{\theta})^2\]

\[\Large \sin^2{\theta}=3(1+2\cos{\theta}+\cos^2{\theta})\]

\[\Large 1-\cos^2{\theta}=3(1+2\cos{\theta}+\cos^2{\theta})\]

\[\Large 1-z^2=3(1+2z+z^2) \ ... \ \text{ Let } z = \cos(\theta)\]

Now use the quadratic formula to solve for z

Answer 17

once you have your 2 solutions in terms of z, you'll use z = cos(theta) to find the solutions in terms of theta

Answer 18

O_O

Answer 19

A-alright...

Answer 20

Hey, I think I need awhile to go over this. You can go help someone else if you want

Answer 21

I don't wanna make you just wait here

Answer 22

ok tell me what you get for z

Answer 23

I have an insane amount of lag again lol, you really don't have to wait for me.

Answer 24

that's ok

Answer 25

@jim_thompson5910 I got \[4z^{2}+6z+2\]

Answer 26

now use the quadratic formula

in this case, a = 4, b = 6, c = 2

Answer 27

Okay

Answer 28

Well, I used this formula, is that okay?

Answer 29

sure factoring works too

Answer 30

Okay... I kinda forgot where to go from here. Is it\[\left(4z^{2}+4\right)\left(z+2\right)?\]

Answer 31

\[\Large 4z^2 + 6z + 2 \]

\[\Large 4z^2 + \color{red}{6z} + 2 \]

\[\Large 4z^2 + \color{red}{4z+2z} + 2 \]

now factor by grouping

Answer 32

Oh, okay! Thank you. :)

Answer 33

Uh... how exactly do I factor?
attempt: 4z(z+1)+2(z+1)?

Answer 34

then you factor out (z+1) to get (4z+2)(z+1)

Answer 35

Ah, I did do it right lol
So now I simplify?

Answer 36

4z+2=0
4z=-2
z=-2/4

z+1=0
z=-1

Answer 37

-2/4 reduces to -1/2

Answer 38

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

Answer 39

Okay I'm not sure why this is laggy again but
[\cos^{-1}{\left(-1,-\frac{1}{2}\right)}?\]

Answer 40

...why is the latex not working

Answer 41

you take the z solutions one at a time

Answer 42

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

Answer 43

Okay

Answer 44

if z = -1/2, then

z = cos(theta)
-1/2 = cos(theta)
cos(theta) = -1/2
theta = ??? or theta = ???

Answer 45

\[\frac{2\pi}{3}\]

Answer 46

what else

Answer 47

\[\cos^{-1}{\left(-1,-\frac{1}{2}\right)}?\] whoops missed a slashy! that's why it didn't work lol
hm... 240°?

Answer 48

again you do them one at a time

\[\Large \cos(\theta) = -\frac{1}{2}\]

\[\Large \theta = \arccos\left(-\frac{1}{2}\right)\]

Answer 49

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

Answer 50

Sorry, was just correcting my latex error

Answer 51

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.

Answer 52

Okay

Answer 53

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

Answer 54

:O how did you get that green checkmark lol

Answer 55

Yeah thank you for that, wolfram has been super laggy as well, idk why

Answer 56

I typed `\checkmark`

Answer 57

ah, thank you :)

Answer 58

if wolfram doesn't work, use

http://web2.0calc.com/

or google's calculator

Answer 59

okay thanks :)

Answer 60

@jim_thompson5910 2pi/3 gave me 0 http://prntscr.com/9auvep

Answer 61

same here, so \(\Large \theta = \frac{2\pi}{3}\) has been confirmed to be a true solution

now check \(\Large \theta = \frac{4\pi}{3}\)

Answer 62

and 4pi/3 gave me \[-\sqrt{3}\] http://prntscr.com/9auvl7

Answer 63

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"

Answer 64

OH right

Answer 65

4pi/3 is still wrong though http://prntscr.com/9auw7p

Answer 66

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

Answer 67

So if \(\Large \theta\) is restricted to the interval \(\Large 0 \le \theta < 2\pi\) then the only two solutions are \(\Large \theta = \pi, \theta = \frac{2\pi}{3}\)

Answer 68

so 2pi/3, pi... or in degree form: 180, 120

Answer 69

yep correct

Answer 70

alright thank you! :)