Find Trig Ratios Given Angles in Standard Form

追加: 2026-05-12

[00:00:00]all right here we're going to talk about ratios of angles in standard position now what does that mean from scratch here we're going to be talking about the cartesian coordinate plane right that's just the x comma y points that you're used to let's say we're given that x comma y right given an ordered pair you can actually create an angle angles in standard position start at the origin and the initial side is the positive x-axis always now the ray that goes through that given point actually creates an angle and notice it goes through the point and the angle let's call it theta and so there we have it now if we drop an altitude drop an altitude to the x-axis we can form what's called the reference triangle so this reference triangle can be sort of generalized to find the six trig ratios now the length right here for the base is x the same length as the coordinate and y is that distance there but these are actually coordinates now there's also a distance from zero to r we'll call or zero to the point and we'll call that r and so right away we could see r squared equals x squared plus y squared right that's the pythagorean theorem but more importantly here we can redefine the trig ratios using those coordinates now before we know sine was defined to be opposite over hypotenuse in this case we could say sine is equal to the ratio y over r cosine before was defined to be adjacent over hypotenuse so the adjacent here is x so we defined that as x over r and then tangent before opposite over adjacent we define tangent to be y over x now the difference here is that x and y are actually coordinates coordinates of an ordered pair all right now the the other three are just the reciprocals so cotangent is defined to be x over y secant is defined to be r over x and cosecant is defined to be r over y all right so we have a brand new definition in terms of coordinates now let's do an example okay so you can see here they're giving us a coordinate negative 3 comma 4 and we want to find all six trig ratios okay so we start by drawing a picture here's a cartesian coordinate plane the x-axis is the y-axis so what we're going to do is plot the point negative 3 comma 4.

[00:02:54]that's right about there negative 3 comma 4.

[00:02:58]now notice that lands in q2 so if we write the angle here in standard position the initial side is the positive x-axis now we draw a ray through that point and that's my terminal side right the terminal side of the angle here where the angle is this guy and you can see it's sort of obtuse right that's the angle in question all right so we're going to drop our altitude always drop your altitude to the x-axis and create this reference triangle okay that's a right triangle where this isn't really a distance what we're looking at is the coordinates so the coordinate here is x is negative 3 and the coordinate here for y is 4.

[00:03:47]now the goal to get our six trig ratio is we're going to need this distance from the origin to that point we'll call that r to get that distance you just use the pythagorean theorem and say r squared equals negative 3 squared leg squared plus leg squared and notice when you square negative 3 it's positive so you get 9 plus 16 is 25 right so r squared equals 25 taking the square root don't forget the plus or minus but it turns out the r is always positive so we're going to choose positive 5 for that and then i can go back to my problem here and i know this here is going to be 5 right so i have a negative 3 a 4 and a 5.

[00:04:32]using those coordinates we can get all six trig ratios sine of theta defined to be y over r is four-fifths cosine of theta defined to be x over r is negative three-fifths all right notice the ratio is negative and that's new for today trig ratios can be negative all right tangent of theta defined to be y over x so in this case 4 over negative 3.

[00:05:04]yeah now we usually don't leave negatives in the denominator we'll put that in front of the fraction bar and those are the three basic trig ratios where in quadrant two sine is positive cosine is negative and tangent is negative right now the other three are just the reciprocals so cotangent sorry about that cotangent would be negative three-fourths secant of theta would be negative five thirds and cosecant of theta would be five fourths right those are all six trig ratios in quadrant two okay here's another one they gave us an ordered pair negative square root of five comma negative two we'll begin by drawing that in the card or plotting that point in the cartesian coordinate plane okay so that's right about here negative square root of 5 comma negative 2 i guess not drawn to scale right we begin by drawing our angle in standard position so our terminal side goes through that point and this is the angle in question all right once we have the angle written we can then drop our altitude to the x-axis always to the x-axis to form our reference triangle here all right so this coordinate is negative square root of 5 right that's the x value and the y coordinate is negative 2.

[00:06:41]all right now to get our six trig ratios we need to figure out the r in this case to do that we'll use the pythagorean theorem and we'll say r squared equals leg squared so negative 2 squared plus the other leg squared negative square root of 5 squared r squared equals negative 2 squared is 4.

[00:07:02]and negative square root of 5 squared is 5. so r squared equals 9. now when i take the square root i'll get plus or minus but r is always positive so r equals 3 in this case once i know this is a 3 i have all my values of my reference triangle and so i can get my my 6 trig ratios all right let's do it sine of theta defined to be y over r in this case negative two over three cosine of theta defined to be x over r in this case negative root five over three tangent of theta defined to be y over x opposite over adjacent right so negative 2 over negative square root of 5 which is 2 over square root of 5. i'll leave that unrationalized notice that tangent ratio is positive so we're over here terminating in q3 and q3 sine is negative cosine is negative and tangent is actually positive all right so the other three are just the reciprocals we have cosine of theta is root 5 or i'm sorry cotangent of theta is root 5 over 2.

[00:08:28]secant of theta would be the reciprocal of cosine so negative 3 over root 5 and cosecant of theta is the reciprocal of sine so negative 3 over two and there's the six trig ratios given that ordered pair