Module 6: Lesson 4 Using Trigonometric Identities Part 1-­‐Video Transcript Hey everyone this video is on module 6 lesson 4, using trig identities. These examples are like the example that you seen in part 1 notes. So lets look at the first one, it says given that cos x is .36 and angle x is in quadrant 1, find sine of x, tangent of x, secant of x, cosecant of x, and cotangent of x. Round each answer to three decimal places. So if I looking at this particular problem, I know that the cosine of some angle equals .36 and that angle is in quadrant 1. I know that I am looking in this quadrant and finding these things (pointing to sinx, tanx, secx, cscx, and cotx) for angles in quadrant 1. So if I was asked to find cosine of x, or find these 5 things after given this, I would find the secant of x first, because I know the relationship between the secant of x and cosine is related, secant of x is equal to 1 over the cosine of x. So to find the cosecant of x I could just do 1 over .36. When I do that, I do 1 divided by .36 and get 2.778 since I am rounding to three decimal places, so 2.778. Now think about what sign this needs to be for secant, if I am in quadrant 1, secant is positive as well. This is an approximate answer. So now the next one that I would find would be sine. I can relate the cosine and sine together by the Pythagorean identity. So I could have, sine squared of x plus cosine squared of x equals 1. So if I know that cosine of x is .36, I could put .36 in for cosine of x, and then all I need to do is get sine of x by itself. So I am going to do this in 2 steps in one. So the sine of x would equals 1 minus .36 squared and I would take the square root of that, so its going to be plus or minus (positive or negative) in front. So lets go ahead and put that in, so if you do square root of 1 minus .36 squared, you get .933, .933 yeah. So that is approximately .933, and I have a plus or a minus here. Well which one do I choose, well I’m doing sine, I know that the angle has to be in quadrant 1, so if I am in quadrant 1 and finding sine it’s always going to be positive. So the only one that I take from that is positive .933. Now I find the cosecant just like I did the secant. Cosecant is related to sine as being 1 over the sine of x. So if I know that sine is positive .933, put that in, do the math there, hit enter, and I get 1.072. All there answers are approximate, I know that cosecant is positive too because we are in quadrant 1. And then to find the tangent, tangent is related to sine and cosine by this, tangent of x is the sine of x over the cosine of x. Let me write that a little bit better. So I am going to put that in, sine of x is .933, and cosine if .36. So if I do .933 divided by .36 I get 2.59. Oh three decimal places, sorry about that, 2.592. Now tangent if I’m in quadrant 1 again is going to be positive. It’s a positive divided by a positive here, so I have approximately 2.593 and to get the cotangent of that, cotangent is going to be 1 over the tangent which would be 1 over 2.592, which will be approximately, .389. And once again cotangent would be positive as well because we are in quadrant 1.