# Cos x unit circle

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#1
I’m unsure why but I still have trouble understanding the cos x identities on the unit circle. Sin x is pretty straightforward.

Could someone help provide some context based on the unit circle for cos x? Or, provide a link which explains this?

Regards
Last edited by KingRich; 1 month ago
0
1 month ago
#2
Pick a point on the unit circle (x,y), then
x=cos(theta)
y=sin(theta)
As can be easily seen if you draw the right angled triangle (adjacent base is "x" or cos(theta), opposite vertical is y or sin(theta)). The hypotenuse is 1 (unit circle). So something like

The rest of the stuff follows (sign and symmetry in x/y axes). So cos() is symmetric about 0 and 180 degrees (x-axis) as the two points
(x,+/-y)
wil have the same value for cos(theta)=x. cos() is positive 0..90 and -90..0 as +/-90 is the y axis. cos() will therefore have the same sign when the angle is reflected in the x-axis and will flip signs when reflected in the y-axis.

Similarly for sin() being symmetric about 90 and 270 (y-axis) and positive 0..180 etc. sin() will therefore have the same sign when the angle is reflected in the y-axis and will flip signs when reflected in the x-axis.
Last edited by mqb2766; 1 month ago
0
#3
(Original post by mqb2766)
Pick a point on the unit circle (x,y), then
x=cos(theta)
y=sin(theta)
As can be easily seen if you draw the right angled triangle (adjacent base is "x" or cos(theta), opposite vertical is y or sin(theta)). The hypotenuse is 1 (unit circle). So something like

The rest of the stuff follows (sign and symmetry in x/y axes). So cos() is symmetric about 0 and 180 degrees (x-axis) as the two points
(x,+/-y)
wil have the same value for cos(theta)=x. cos() is positive 0..90 and -90..0 as +/-90 is the y axis. cos() will therefore have the same sign when the angle is reflected in the x-axis and will flip signs when reflected in the y-axis.

Similarly for sin() being symmetric about 90 and 270 (y-axis) and positive 0..180 etc. sin() will therefore have the same sign when the angle is reflected in the y-axis and will flip signs when reflected in the x-axis.
Thank you for this but I can’t really follow along with his video for cos and when it reaches 1, he says the width of the triangle will be 1.

I will just have to do my best to memorise the identities because this one is kind of kicking me in the butt. Partly, the visualisation part of it
Last edited by KingRich; 1 month ago
0
1 month ago
#4
(Original post by KingRich)
Thank you for this but I can’t really follow along with his video for cos and when it reaches 1, he says the width of the triangle will be 1.

I will just have to do my best to memorise the identities because this one is kind of kicking me in the butt. Partly, the visualisation part of it
This is similar
https://www.coursehero.com/study-gui...ine-functions/
For a point (x,y) on the unit circle, the hypotenuse is 1, so
cos(t) = x
sin(t) = y
where x is the "adjacent base" and y is the "opposite vertical".

A few identites that should be "easy"
cos(t) = cos(-t)
This is just a reflection in the x axis (0 or 180 degrees) and the x (base) will be the same.
cos(t) = -cos(180-t)
This is refleciton in the y-axis (90 or 270) and the x (base) will flips sign. You can relate both of these to the cos curve, as that is symmetric about the exteme values (0 or 180 degrees) and flips sign about the points it crosses 0 (so 90 or 270 degrees).

If there is an identity youre unsure about, which ones ... Learning the identities, looking them up in your formula sheet, ... is good but understanding the unit circle (cast, trig curves) is important.

Similarly for sin() as above.
Last edited by mqb2766; 1 month ago
0
#5
(Original post by mqb2766)
This is similar
https://www.coursehero.com/study-gui...ine-functions/
For a point (x,y) on the unit circle, the hypotenuse is 1, so
cos(t) = x
sin(t) = y
where x is the "adjacent base" and y is the "opposite vertical".

A few identites that should be "easy"
cos(t) = cos(-t)
This is just a reflection in the x axis (0 or 180 degrees) and the x (base) will be the same.
cos(t) = -cos(180-t)
This is refleciton in the y-axis (90 or 270) and the x (base) will flips sign. You can relate both of these to the cos curve, as that is symmetric about the exteme values (0 or 180 degrees) and flips sign about the points it crosses 0 (so 90 or 270 degrees).

If there is an identity youre unsure about, which ones ... Learning the identities, looking them up in your formula sheet, ... is good but understanding the unit circle (cast, trig curves) is important.

Similarly for sin() as above.
I agree, I want to be able to visualise on the unit circle and see where the identities come from. I could do this previously but some reason I’ve forgotten.

Tbh that’s what confuses me.
Cos x = cos-x
but then -cos x =cos (180-x)

if I’m looking at a quadrant circle then cos x being at 0 degrees, the cos-x would also be at 0 degrees too. At 180 degrees it would -cos x?
0
1 month ago
#6
(Original post by KingRich)
I agree, I want to be able to visualise on the unit circle and see where the identities come from. I could do this previously but some reason I’ve forgotten.

Tbh that’s what confuses me.
Cos x = cos-x
but then -cos x =cos (180-x)

if I’m looking at a quadrant circle then cos x being at 0 degrees, the cos-x would also be at 0 degrees too. At 180 degrees it would -cos x?
The cos(t) = cos(-t)
Assuming t = 30 degrees with values (x,y) both positive, then the reflection in the x axis (0 degrees) gives the point (x,-y) with angle t=-30 degrees. The "base" is the same so
cos(30) = cos(-30)

Now consider cos(150) with values (-x,y) where x and y are positive. This is a reflection in the y axis (90 (degrees) of (x,y) which corresponds to t=30. So
cos(150) = -cos(30) = -cos(180-150)

Sketch the points/triangles/reflections and if youre still unsure, upload the sketches.

Edit KingRich. If you draw the "usual" right angled triangle in quadrant 1, so t is acute and the point is (x,y) with both positive, then
* Reflecting it in the x axis gives (x,-y) so -t lies in quadrant 4 and
cos(-t) = cos(t) = x
sin(-t) = -sin(t) = -y
tan(-t) = -tan(t) = -y/x
* Reflecting it in the y axis gives (-x,y) so 180-t lies in quadrant 2 and
cos(180-t) = -cos(t) = -x
sin(180-t) = sin(t) = y
tan(180-t) = -tan(t) = -y/x
* Reflecting in both the x and y axes gives (-x,-y), so t-180 lies in quadrant 3 and
cos(t-180) = -cos(t) = -x
sin( t-180) = -sin(t) = -y
tan(t-180) = tan(t) = y/x
Those are your basic identities which are fairly obvious from the sin/cos/tan curves but just thinking about them in terms of reflecting the point/triangle in the axes is straight forward as well.

For me, the "unit" circle/point/triangle is more useful for thinking about transforming between trig functions/inverse functions. So if you knew
tan(t) = a/b
then
sin(t) = a/sqrt(a^2+b^2)
cos(t) = b/sqrt(a^2+b^2)
Similarly, something like finding the value of
cos(arctan(2/7))
seems "hard" but its just (similar to) a 7 : 2 : sqrt(53) triangle (its just pythagoras) and all the trig values like the above can be just "written down" without any identity chasing.
Last edited by mqb2766; 4 weeks ago
0
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