Find 2 solutions for cot(θ)=-√3 what is the easiest way to find this?
Im stuck at cot (θ)=1/-√3
click on that link
wwowwww thx!
Thank Mathway.com it helps me with some of my problems :)
Unit circle is every trig student's friend.
unit circle is boss
amazing u need to upgrade for it to show steps?
if your willing to pay for it then yeah lol
can u show me the unit circle way?
cheat sheet at paul's notes has excellent unit circle. i am not at home or i would attach it
Here's one...
yes normally with sin and cos i can find the values but not with others
lets rearrange he problem a little first: \[\cot(\theta) = -\sqrt{3} \iff \tan(\theta) = -\frac{1}{\sqrt{3}}\iff \frac{\sin(\theta)}{\cos(\theta)} = -\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\]
so we are looking for an angle that satisfies one of these two possibilities: \[\sin (\theta) = \frac{1}{2}, \cos(\theta) = -\frac{\sqrt{3}}{2}\] or \[\sin(\theta) = -\frac{1}{2}, \cos(\theta) = \frac{\sqrt{3}}{2}\]
cos(theta)=1/-√3 is 1/tan=cot(theta) right?
Remember, sin is the y-value and cos is the x-value... find ( sqrt(3)/2, -1/2 ) or ( -sqrt(3)/2, 1/2 ) as points on the circle...
Or you could just remember a 1,2 sqrt 3 triangle....
waahh confused i meant cot(theta)=-√3 not cos
that is correct chelsea
yes in other words im lookin for what value 1/tan=-sqrt3
that right?
yes :)
-√3divided by 1 or other way around?
this aint makin nooo sensee
take a look at my second post to help you out.
i did confusing though it only makes sense to me if i say to myself what value x/y will give me -sqrt3?
right, the post under that says what values of x (which is cos) and y (which is sin) will give you that.
Chelsea - That's the way I think about it too...I first take what I am looking for (in your case cot) then convert to sin and cos.. So I know I need cos/sin which I think convert to x and y...since x = cos and y = sin So I need x/y = -sqrt(3) so my x will be sqrt(3)/2 and my y will be 1/2...and one will be negative I find that on the unit circle and I'm done! :)
oh so those are 2 different coordinates but how did u narrow it down to just those two?
did u mentally think of those 2 fractions ?
Those fractions come from the 30-60-90 triangle rule. The sides of a 30-60-90 triangle are always in the ratio of 1:sqrt(3):2, respectively. So when it goes to the unit circle, all 30-60-90 points will have coordinate involving sqr(3)/2 and 1/2. For instance, look at pi/6. This is 30 degrees so it's coordinates will be (sqrt(3)/2, 1/2)
If you have a problem involving sqrt(2), you're going to use the fractions sqrt(2)/2 and 1/2
ooooh that makes alot more sense up untill now i havent used any triangle methods reason why i didnt recognize those fractions.
thanks bender
my pleasure :)
Join our real-time social learning platform and learn together with your friends!