Review the unit circle definition of the trigonometric functions.
Log in frank 8 years agoPosted 8 years ago. Direct link to frank's post “In the appendix, can some...” In the appendix, can someone explain to me WHY cotangent, tangent, secant, and cosecant are represented in that way? • (108 votes) Daivik 5 years agoPosted 5 years ago. Direct link to Daivik's post “tangent=sin(θ)/cos(θ)(tan...” tangent=sin(θ)/cos(θ)(tangent is the path from point(x,y) to the secant) (48 votes) nathanielbevins 8 years agoPosted 8 years ago. Direct link to nathanielbevins's post “I came across a problem. ...” I came across a problem. If theta represents an angle such that sin 2 theta = tan theta - cos 2 theta, then sin theta - cos theta=? The answer choices are neg root 2, 0, 1, 2 root 2 and can not be determined. I figured that theta must be 3pi/4. This makes the first equation true. But when I then plug in 3pi/4 for the second equation we come up with root 2, which isn't an option. • (29 votes) LEGO.M 8 years agoPosted 8 years ago. Direct link to LEGO.M's post “Thank you for this questi...” Thank you for this question, this is a great one I think. :D (46 votes) AJ 8 years agoPosted 8 years ago. Direct link to AJ's post “In the "Appendix: All tri...” In the "Appendix: All trig ratios in the unit circle" above, why is the Tangent at a negative slope? • (21 votes) Skywalker 7 years agoPosted 7 years ago. Direct link to Skywalker's post “Normally the unit circle ...” Normally the unit circle is pictured with the hypotenuse going from the origin to the point where the ray intersect the circle (meaning that the right-angle corner in on the X axis) whereas the one pictured here runs through point of intersection to the X and Y axis outside of the unit circle. Pictured this way, it allows for one to see all the trig function versus just the "Big 3" (sin, cos, and tan). That being said. The way this hypotenuse is oriented gives it a negative slope. (26 votes) gameboyjustin 7 years agoPosted 7 years ago. Direct link to gameboyjustin's post “Aside from rote memory, h...” Aside from rote memory, how would I be able to derive that sinθ equals to the Y coordinate or cosθ equals to the X coordinate? I feel like the only thing I would remember instinctively would be the mnemonic "SOH CAH TOA" ... would I be able to derive it from my understanding of that? • (11 votes) Ian 7 years agoPosted 7 years ago. Direct link to Ian's post “Yes. If you know SOHCAHTO...” Yes. If you know SOHCAHTOA and the unit circle, then you can derive it. The unit circle has a radius of one, so the hypotenuse is 1. Sine is opposite over hypotenuse. Since the hypotenuse is 1, sine on the unit circle is the opposite side. When you look at the unit circle, the opposite side is perpendicular to the x-axis. This means that, essentially, the opposite side is the height from, or the distance from, the x-axis, which is the y-value. (31 votes) kimberm1 4 years agoPosted 4 years ago. Direct link to kimberm1's post “I understand how to apply...” I understand how to apply the trig ratios in their purest form (absolute and a standard right angle triangle). I also understand the theory of using the unit circle to extend the domain of sin(theta) and cosine(theta) to all real values. With that said, how is a calculator able to take this basic principle and compute the logic that cos(89) = 0.017 and cos(91) = -0.017 (a reflection across the y-axis). What's going on under the hood? • (10 votes) Iron Programming 4 years agoPosted 4 years ago. Direct link to Iron Programming's post “Hi kimberm1,I would ass...” Hi kimberm1, I would assume that they created an algorithm that checks the angle just like a human would. Here is some PSUEDO CODE that might help you understand. And so on. What the computer would essentially do is follow an algorithm that checks what quadrant the point will be in based upon the angle. We can easily calculate that as seen above. Hope this helps, (16 votes) kenwishart1684 a year agoPosted a year ago. Direct link to kenwishart1684's post “if tan is opp/adj shouldn...” if tan is opp/adj shouldn't the answer be .87/-.5? • (5 votes) Bryland Rivera a year agoPosted a year ago. Direct link to Bryland Rivera's post “yeah just finish solving ...” yeah just finish solving .87/-.5 and you get the answer (9 votes) Auro soni 6 years agoPosted 6 years ago. Direct link to Auro soni's post “Why and how are trigonome...” Why and how are trigonometric ratios related to circles? • (4 votes) Sanjay Kulkacek 6 years agoPosted 6 years ago. Direct link to Sanjay Kulkacek's post “That's the subject of thi...” That's the subject of this video: https://www.khanacademy.org/math/algebra2/trig-functions/unit-circle-definition-of-trig-functions-alg2/v/unit-circle-definition-of-trig-functions-1, particularly around The cool thing about a unit circle (a circle with radius one), is that if you draw a right triangle where the hypotenuse of the triangle connects a point on the unit circle with the point (0, 0), the length of the hypotenuse is always going to be one. Because that's the radius of the circle! And the length of the other two sides are just going to be the x and y coordinates of the point on the unit circle. Take a little time drawing right triangles were the hypotenuse connects the middle of the unit circle to a point on the circle and see if you can convince yourself that this is the case. So you have a right triangle where one side length is the x-coordinate and one side length is the y-coordinate, and the hypotenuse is equal to 1. This means you can figure out the sine and cosine of the angle that the hypotenuse and the x-axis make: The cosine is the length of the adjacent side (which is the x-coordinate) divided by the length of the hypotenuse (which is 1). So the cosine is just the x-coordinate! Similarly, the sine is the length of the opposite side (which is the length of the y-coordinate) divided by the length of the hypotenuse (which is 1). So the sine is the y-coordinate. The trigonometric ratio defining the tangent (opposite over adjacent) may be less intuitive from the unit circle, but I hope you can see from the unit circle why the tangent of an angle is equal to the sine of the angle over the cosine. Hope this helps. (10 votes) Radha Krishna a year agoPosted a year ago. Direct link to Radha Krishna's post “what does actually *negat...” what does actually negative angle mean? • (4 votes) Siddharth a year agoPosted a year ago. Direct link to Siddharth's post “It means it is a clockwis...” It means it is a clockwise angle. (9 votes) Alianah ! <3 9 months agoPosted 9 months ago. Direct link to Alianah ! <3's post “I don't understand any of...” I don't understand any of this.... is my brain just to small ;-; • (7 votes) Kim Seidel 9 months agoPosted 9 months ago. Direct link to Kim Seidel's post “This is the review. Try ...” This is the review. Try starting at the beginning of the lesson. Make sure you understand each part as you work thru the content. Do the practice. Ask specific questions when you don't understand. Then, try the review again. (3 votes) gaumond.robert 8 years agoPosted 8 years ago. Direct link to gaumond.robert's post “In the Appendix (moving p...” In the Appendix (moving point example above) I have difficulty relating the tangent (in blue) to a line drawn from the center of the circle through the green dot. The tangent would only match if the x-intercept of the sine (red line) was equal on both sides of the red line. • (8 votes)Want to join the conversation?
cotangent=1/tan(θ)(cot α 1/tan)(cotangent is the path from the point (x,y) to the cosecant)
secant=1/cos(θ)(sec α 1/cos)
cosecant=1/sin(θ)(csc α 1/sin)
If you try theta= pi/4 or negative pi/4; the first equation holds in both situations and you get choice A and C respectively. Actually not just these two radians but a set of them are the solutions to the 1st equation; Check this:
First, I use m to represent theta, sqrt(2) as the square root of 2, for simplification: so we know sin(2m)=tan(m)-cos(2m) from the given facts, then we use the identities, we have left=sin(2m)=2*sinm*cosm=right=(sinm/cosm) - [1-2(sinm)^2],
then divide sinm on both sides (if sinm=0, the equation cannot be satisfied since left=0, right=0-1=-1, not equal to the left), then we have: 2cosm=1/cosm -1/sinm +2sinm;
then move the terms and change it into: 1/sinm -1/cosm = 2(sinm - cosm); simplify this as: (cosm-sinm)/(sinm*cosm)=2(sinm-cosm)
let x = sinm - cosm, which is the value we try to find; we have -x/(sinm*cosm)=2x
now it turns into a problem solving x: if x=0, it satisfies the equation and this means sinm=cosm; these angles can be the bisector of the 1st (I) and 3rd(III) quadrant; they can be pi/4 and 3pi/4 as you mentioned before and 2k*pi applied on them as well, under this condition, we choose 0 as the answer;
If x doesn't equal to 0, we divide x on both sides and we have: figure out the only situation to hold the equation true is that -1/sinm*cosm=2, that means (-1*2)/(sin2m)=2 and sin2m must be -1 to keep this true.
Under this situation, we know as long as m (the angle theta) satisfies sin2m=-1, it is the solution and now 2m= 3*pi/2 + 2*k*pi, k is an integer; so m=3*pi/4 + k*pi, k is an integer; we now see that the terminal side of m is the bisector of the 2nd (II) and 4th (IV) quadrant, if in the 2nd quadrant, sinm=sqrt(2)/2 and cosm= - sqrt(2)/2; we have sinm-cosm=sqrt(2); if in the 4th quadrant, sinm= - sqrt(2)/2 and cosm= sqrt(2)/2, we have sinm-cosm= - sqrt(2).
Here I think would be my answer to this question, if something is wrong, please let me know and we can figure out that~
Aka first calculate the absolute value of cos(theta), and then check to see what the value of theta is. IF (theta >= 0 AND theta < 90) THEN
ANSWER = cos(theta) // leave unchanged
IF (theta >= 90 AND theta < 180) THEN
ANSWER = cos(theta) * -1
Knowing the quadrant, we know know what sign the y and x values are.
- Convenient Colleague
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