Quick Review Evaluate Six Trigonometric Functions Given Angle

so when we want to evaluate what we need to understand is like what these points like how can we use these angles to evaluate our trigonometric functions well just remember sine of theta represents the Y chord or sine of theta equals the y coordinate of any coordinate point on the other Circle cosine of theta represents the x coordinate and tangent of theta equals the um tangent Theta is y over X then we have the reciprocal functions so we have the cosecant of theta which is 1 over y secant of theta which is 1 over X and then we have the cotangent of theta which is an X over y okay so how much room do we got we got a lot of room okay I guess I got a decent amount of room here that we'll get into okay so let's go ahead and evaluate all six of these so I think the main thing that we want to do for each of these points before we go into value is figure out the corner points right so hopefully you recognize seven pi over 6 that is going to be in the third quadrant all right so everything's going to be negative and has a reference angle of pi over 6. so if I go into the unit circle I can see that this coordinate point is going to be negative square root of 3 over 2 comma a negative one half okay so again I'm doing things rather quickly hopefully you can kind of follow along so if I want to find the um the sign of this that's going to represent the y coordinate so sine I'll just do sine of T because we already know T right there so sine of T in this case is going to be a negative one-half cosine of T is going to equal to negative square root of 3 over 2 and the tangent of t now this one's going to be a little bit more work here right that's going to be the Y over the X so that's a negative one-half divided by negative square root of 3 over 2. the negatives go to positive right and now I'm going to left with a 1 the 2's divide out so on left let's say 1 over negative square root of 3 over 2. so 1 over a squared oops let's put an equal sign here so therefore this is going to be a 1 over a square root of 3 which you can rationalize the denominator by multiplying the square root of 3 on the top and bottom and it's going to give you square root of 3 over 3. again that's a positive now let's go through the reciprocal functions so the cosecant of t um so remember the cosecantity is just going to be reciprocal right so that's going to be a negative 2. the secant of t secant T is going to be reciprocal over here so that one's going to be a 2 over negative square root of 3 which again rationalize the denominator so multiply by a square root of 3 on the top and bottom that's going to be a 2 so it's still a negative negative square root of 3 over 2 over a 3. and then we have the cotangent of T So cotan T is basically this exact same one just flipped right so now the two so again will divide out and that's just going to be left with a square root of 3. now I don't want to write out all those again so I am going to copy and paste this hopefully I can do this perfectly I think I did and let's do it all over again haha oh I did miss up one that's okay uh we can go and add a nice little line here there you go all right so put a line there all right so now let's go and do this point at this point I'll do a different label here so negative 5 pi over 4. so negative 5 pi over 4 here is that's going to be in the second quadrant I think that's maybe where is that um so that's going to be in second quadrant the reference angle is going to be pi over 4 so that angle is square root of two over two comma square root of 2 over 2 but since it's in the second quadrant my X is negative so that's a negative square root of three I'm sorry negative square root of two over 2 comma a square root of 2 over 2. okay so the sine represents the y coordinate square root of 2 over 2 cosine x coordinates negative square root of 2 over 2. tangent which is y over X well guys these are the exact same thing right so anything divided by itself it's just going to equal one in this case one of them is negative so it's negative one um cosecant is going to be 1 over y uh so let's go and do that so that's a 1 over square root of 2 over 2. again multiply by your reciprocals here or multiply multiply over here you're basically going to have a 2 over square root of 2 you rationalize the denominator on the top and on the bottom and you should be left with a let's see let's actually rewrite this out actually I forgot this one so if you multiply so you're going to have a 2 over a square root of 2 rationalize the denominator right and you're just gonna be left with the square root of two yeah okay so rationalize the denominator and that's going to leave you with the square root of 2. now the cool thing about pi over four is whatever you get for cosecant you're going to get for secant right they're basically the exact same answers except notice obviously cosecant as negative right so that's a negative square root of two and then cotangent is going to be X over Y which is again same thing over the same thing so that's going to be a negative one and then let's go ahead and play this game one more time and evaluate now for this angle at 3 pi over 2. now this angle is interesting because we need to this angle is not a get this is not a coordinate point on um within the unit circle this is actually on an axis and this coordinate point here is three power two that's going to be at zero comma negative one okay so kind of a little interested in here so my sign in this example is going to be my y coordinate which is negative 1. cosine x coordinate which is 0. tangent which is y over X so that's going to be negative 1 over 0 which ladies and gentlemen you can't divide by zero so that is undefined cosecant though is zero over negative one so that's going to be a zero secant is one over negative one which is just negative one but cotangent guys I'm sorry secant did that wrong secant sorry let me slow down here tangent was undefined good cosecant is zero over negative one which is one oh no I got those wrong cosecant or sorry cosecant is one over y right so that's one over negative one which is negative one secant is one over X which is one over zero which is also undefined and cotangent is X over Y which is zero over negative one which is equal to a zero okay so that was going through all of them