now let's turn this into a right triangle the hypotenuse of the right triangle is r and this is x and y and here we have the angle theta sine theta is equal to the opposite side divided by the hypotenuse so it's y divided by r now keep in mind when dealing with the unit circle r is one so therefore the equation becomes sine theta is equal to one but when you're not dealing with a unit circle sine theta is actually y over r it's equal to the opposite side divided by the hypotenuse cosine theta according to sohcahtoa cosine theta is equal to the adjacent side divided by the hypotenuse the adjacent side is x hypotenuse is r tangent theta is equal to the opposite side divided by the adjacent side in torah so it's y divided by x cosecant is the reciprocal of sine so if sine is y over r cosecant is r over y and secant is the reciprocal of cosine so that's r over x and cotangent is x divided by y so those are some values that you need to know now let's review this sign of the trigonometric functions so here we have quadrant one quadrant two quadrant three and quadrant four so if you recall sine is positive in quadrants one and two sine is associated with the y value y is positive above the x-axis sine is negative in quadrants three and four cosine is positive in quadrants one and four x is positive on the right side x is negative on the left so cosine is negative on quadrants two and three tangent which is y over x sine over cosine tangent is positive in quadrants one and three tangent is negative and two and four so these are some things that you want to commit to memory so in quadrant one everything is positive in quadrant two sine is positive in quadrant three tangent is positive in quadrant four cosine is positive perhaps you heard of the expression all students take calculus all means that in quadrant one all signs sine cosine and tangent all of them are positive students sign is positive as for students take t is positive calculus c is positive in quadrant four so all students stay calculus something that you may find useful now let's say we're given a point p which is negative 5 comma 12 and it's on the terminal side of theta using this find the values of the six trigonometric functions so if you're given a point what you want to do is plot the point and also make a triangle at the same time so x is negative five that means we've got to travel five units to the left y is positive so you have to travel up 12 units and then draw the hypotenuse which is r so we know this is the 5 12 13 triangle and here is the angle theta that's the reference angle by the way the actual angle is here relative to the positive x-axis now let's go ahead and find sine theta we need to find all six trigonometric functions so sine theta which is y over r also it's opposite divided by hypotenuse where hypotenuse are sine theta is going to be positive 12 over 13. cosine theta is equal to the adjacent side divided by the hypotenuse so it's negative 5 over 13. tangent theta is equal to opposite divided by adjacent so that's going to be 12 over negative 5. now once you have these three you could find cosecant you just got to flip this fraction so cosecant is going to be 13 over 12 secant is the reciprocal of this one so that's a negative 13 over 5 and cotangent is the reciprocal of this value so that's negative 5 over 12. and so that's how you could find the six trigonometric functions if you're given the point let's try another problem so let's say if we're given the point negative 8 negative 15 find the value of the six trigonometric functions so let's plot it let's travel eight units to the left and down 15 units because it's negative 15. x is negative in quadrants two and three but y is negative and three and four so the triangle has to be in quadrant three now this is the 8 15 17 triangle and here's the angle theta between the terminal side and the x-axis so sine theta is going to be opposite which is negative 15 divided by the hypotenuse and negative eight is the adjacent side relative to theta so sine theta is negative 15 over 17. cosine theta according to sohcahtoa is going to be adjacent which is negative 8 divided by the hypotenuse and tangent theta is equal to the opposite side divided by the adjacent side so the two negative signs will cancel and we're just going to get 15 over 8. so keep in mind in quadrant 3 tangent is positive but sine and cosine or negative cosecant to find that all we need to do is flip this value so cosecant is going to be 17 over 15 but with a negative sign and secant is the reciprocal of cosine so we just got to flip this one so it's negative 17 over eight and finally cotangent which is one over tangent that's going to be eight over fifteen and so that's it for this example let's try one more problem so let's say we have the point positive two negative four so pause the video and work on this one so let's begin by plotting the point so x is positive two so we got to go 2 units towards the right y is negative 4 so we have to go down 4 units now this is not a special triangle so we need to use the pythagorean theorem in order to find r the hypotenuse a squared plus b squared is equal to c squared a is two in this example b is negative four and let's find c the hypotenuse is always positive by the way 2 squared is 4 negative 4 squared is positive 16 and 4 plus 16 is 20. so we got to take the square root of both sides and 20 we can write it as 4 times 5. and the square root of four is two so the hypotenuse is two square root five now that we have the hypotenuse we could find everything else so 2 is on the adjacent side relative to theta negative 4 is on the opposite side and 2 square root 5 is the hypotenuse sine theta which is uh y divided by r is equal to the opposite side divided by the hypotenuse so it's negative four divided by two square root five which reduces to negative two over square root five and if we rationalize it by multiplying the top and bottom by square root 5 it becomes negative 2 square root 5 over 5. so that's the value for sine theta cosine theta is equal to the adjacent side which is 2 divided by the hypotenuse so that simplifies to 1 over square root 5 because these cancel and then if we rationalize it it becomes square root five over five so that's the value of cosine next we have tangent which is opposite divided by adjacent so that's negative two now let's find the value of cosecant cosecant is one over sine but what you want to do is you want to flip this portion of sine before you rationalize it so cosecant is going to be negative square root 5 over 2 and now for secant we want to flip this portion before we rationalize it so if cosine is 1 over square root of five secant is square root of five over one or simply just square root five so that's the value of secant and here's the value for cosecant now tangent is negative two so cotangent is going to be negative one over two you just gotta flip it and so that's it for this example you