Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Trigonometric functions of any angle. When we're doing reference angles, we're always going to refer back to the X axis. So when we look at the various quadrants, we have Theta, we have the opposite side being the height, which is the Y, the adjacent side being the left and right distance or the X. And the radius can always be found by using the Pythagorean theorem. So the radius is going to equal sqrt X ^2 + y ^2. Now depending on which quadrant we're in depends on the signs, whether they're positive or negative of our six trig functions. But as you can see, on all four of these graphs, the sign is always the opposite side over the hypotenuse, so y / r The height over the radius cosine is X / r or the left and right over the radius tangent opposite over adjacent y / X. The cosecant is the reciprocal of sine, so r / y secant reciprocal of cosine r / X, cotangent the reciprocal of tangent X / y. Frequently we say things like all students take calculus to help us remember which signs are positive and which quadrants. So in the first quadrant they're all positive. Cosine, sine, and tangent are all positive. In the second quadrant, the sign is positive, and so was its reciprocal function of the cosecant. So we'd have a negative, positive, negative, negative because we've gone left, positive because we've gone up, and a negative because a positive divided by a negative is a negative. In the 3rd quadrant, tangent and its reciprocal function are positive. So tangent and cotangent we've gone left. We've gone down South, negative and negative, and then a negative divided by a negative is the positive, which is why the tangent is positive. Cosine, we've gone right, IE positive, and then we've gone down negative, and a negative divided by a positive is a negative. So all students take calculus as a nice way to remember that they're all positive in the first sine and it's reciprocal in the second tangent, and it's reciprocal in the third cosine and it's reciprocal in the fourth. Reference angles always refer back to the X axis. So if we have our Theta, we're going to look. If we're in the second quadrant, we're going to always look at doing 180° minus Theta to get our reference angle, or π minus Theta to get our reference angle if we're in the second quadrant. If we're in the third quadrant, we're going to do our Theta -180° or our Theta minus π to get the angle that's between the X axis and the third quadrant. In the fourth quadrant, it's going to be 2π minus Theta or 360° minus Theta. We're always trying to find an acute angle that'll go back to the X axis. So if we're given some point negative 25 S -2 five, we're going to find the six trig functions. So there's a -2 because we're going left. So we're in the second quadrant. So we want to find cosine, sine, tangent, and then the reciprocals of that. The first thing we want to do is we want to figure out this hypotenuse. By Pythagorean theorem, we know that a ^2 + b ^2 = C ^2, so 5 ^2 is 25. Negative 2 ^2 is 4, so that hypotenuse is sqrt 29. So our cosine is going to be -2 / sqrt 29, and then we always rationalize it so -2 sqrt 29 / 29. Sine is the opposite side 5 over the hypotenuse sqrt 29, so 5 sqrt 29 / 29 tangent opposite over adjacent, so 5 / -2 are -5 halves. We don't leave a negative in the bottom secant is just the reciprocal, so let's have it be negative sqrt 29 / 2. If we look at this one here to take the reciprocal, we don't have to rationalize it. So cosecant is going to be squared of 29 / 5 and cotangent is going to be -2 / 5. If we were trying to figure out what quadrant we're in, cotangent being positive, well, we know the cotangent is positive and the first because they're all positive here students take calculus. So in the third quadrant also tangent and cotangent are positive. But now I want sign negative, while all of them are positive in the first quadrant. So it can't be there. So it must be the third quadrant. And if we think about it, the sine is negative in the third quadrant. So our answer there is quadrant 3. This next example, we want sine to be negative, so we know that we're going to be all students take calculus, so we know we're going to be negative in the 3rd and the 4th, but we want our tangent to be positive. So that has to be in our third quadrant. The tangent being 8 seventeenths. So we know the opposite side is going to be 8, the hypotenuse is going to be 17. So we can find this other side by the square root of 17 ^2 - -8 ^2. It's minus here because we knew the hypotenuse and we knew one side. If we go into our calculator and we go 17 ^2 - -8 ^2 and then the square root of that, I should have put that in first. Well, it's going to be 15, the square root 225, 15. So going back to our problem, this is just going to be 15. Now it wants us to find all of our remaining trig functions. So cosine Theta, Cosine in the third quadrant is negative adjacent over hypotenuse so -15 seventeens tangent Theta tangent in our third quadrant is positive. So we're going to have opposite over adjacent or 8 / 15 Cosecant. Theta is just the reciprocal of sine so -17 eights. Secant is just the reciprocal of cosine -17 fifteenths. Cotangent is just the reciprocal a tangent 15 eighths. If we want to find reference angles 12π sevenths, well, we know that one Pi would be 7 Pi sevenths and 2π would be 14 Pi sevenths, so 12π sevenths. If we thought about making this into fourteenths, we'd get 14 Pi fourteenths and 28 Pi fourteenths. Halfway between there is going to be 21 Pi fourteenths. So 12π sevens would be 24 Pi fourteenths. So it's got to be and this 4th quadrant right here, 24 Pi fourteenths. So I want to find this angle in here, my reference angle. The way to do it is going to be 2π minus my 12π sevenths. So 14 Pi sevenths -12π sevenths is going to give me a reference angle of 2π sevenths. For this next example, 3 Pi FIS. We know that straight over here is five Pi FIS. SO3 Pi FIS is here. In order to find this reference angle, we're going to take π and subtract 3 Pi FIS. So my reference angle is 2π FIS 570. Well, we know one full circle is 360. We know that an straight line would make it another 180, so that'd be 540. And then we want 30° more. So if we started with here, one full circle is 360. 360 + 180 is 540. To get to 570, I need to go 30° more, so the reference angle here is going to be 30°. So in this case, we're actually going to take the 570 and subtract a full circle, which would give us 210, and then we're going to take the 210 and subtract 180, that straight line to get our 30°. We want to use our reference angles to find exact values on these. So sine of -13 Pi force, we know that right here is going to be 8 Pi force and here is going to be 12π force. So -1 full circles 8 Pi force, another line or another π worth is -12π force. So we're going to be here in the second quadrant. Sine in the second quadrant is going to be positive and our reference angle is going to be π force. So this is going to be the same thing as sine of π force and sine of π force is root 2 / 2 cosine of 31 Pi six. Well, if we go one full circle, we're at 12 Pi 6. If we go 2 full circles, we're at 24 Pi 6. So we want to go 31 Pi 6 if we go 1/2 a circle. If we were at 24 and we go 1/2 a circle, now we're at 30 Pi 6, so we're going to go a PI6 more. So my reference angle is going to be PI6. Cosine in the third quadrant is going to be negative. So I want the negative of cosine PI6 and cosine Pi 6 is root 3 / 2. So cosine of 31 Pi 6 is going to be negative root 3 / 2 tangent of -13 Pi thirds. One full circle would be 6 Pi thirds. Another full circle would be 12π thirds, so another third would be thirteen Pi thirds. Our reference angle then is going to be π thirds. Tangent in the 4th quadrant is negative, so that's the same thing as negative tangent π thirds. Tangent of π thirds is going to be the same thing as sine of π thirds over cosine of π thirds. And our sine of π thirds is root 3 / 2 and our cosine of π thirds is 1/2. So we get negative root 3. This one cosine of Pi -1 tangent of eight Pi thirds. We need to find our reference angle one full circle 6 Pi thirds, 7 Pi thirds 8 Pi thirds. So we're in our second quadrant with a reference angle of π thirds. And our second quadrant we know that tangents negative, so negative tangent π thirds minus the sine. So we're going to look at -5 Pi six -5 Pi 6 is going to put us into the third quadrant with a reference angle of Pi 6 and the third quadrant we know sine is negative, so we're going to have minus a negative sine of Pi 6. So tangent of π thirds is root 3 -. A negative makes it a plus. Sine of Pi 6 is 1/2. So our answer here is going to be root 3 + 1/2 when we do compositions. If we wanted F of four Pi thirds, we're just going to say what is sine of four Pi thirds? Well, four Pi thirds is π plus another third, so we're in the third quadrant, and sine in the third quadrant's negative. So negative sine of π thirds, sine of π thirds is root 3 / 2. So sine of four Pi thirds negative root 3 / 2. If we wanted G of Pi 6. That's just saying what's cosine of Pi 6, cosine of Pi 6? Our reference angle is π six, and the cosine of Pi 6 is root 3 / 2 because in the first quadrant cosine's positive. If we wanted H of G of 13 Pi force, we're going to take H of the G equation, which is cosine of 13 Pi force. So we're going to take 2 cosine 13 Pi force and evaluate this. If we're talking about 13 Pi force, a full circle is 8 Pi force. Another half a circle would make it 12π force. So we're going to be in the third quadrant. Cosine in the third quadrant is negative and we would have a reference angle of π force. So we get 2 times negative root 2 / 2 for a final answer of negative root 2. Thank you and have a wonderful day.