The arc length is based off the circumference of a circle, which is 2π R If we think about wanting an arc length, we only want a portion of the full circle, so we want a proportion of it. Let's say we want Theta as the angle. So if we want the arc length, the letter that's usually used as S equal, we could think of that Theta out of a whole circle. And if we're doing this in radians, the full circle in radians would be 2π. So we want a ratio, we want the Theta out of 2π, and then that is going to be multiplied by the circumference of the full circle because we're wanting a ratio or a portion of the full. What happens next is the 2π's cancel. So S equals Theta R if we're in radiance, that's the easiest equation. Now if in a unit circle, we know that in a unit circle our R equal 1, so thus S equals Theta in a unit circle. So the angle measure equals the arc length in a unit circle. Terminal rays and initial rays. The initial ray is usually thought of as the X axis. The terminal ray is where the array actually ends or the angle ends. Positive angles go counterclockwise, negative angles go clockwise. Our six trigonometric functions we have sine being opposite over hypotenuse, cosine being adjacent over hypotenuse, tangent being opposite over adjacent cosecant, hypotenuse over opposite secant, hypotenuse over adjacent, cotangent adjacent over opposite. So sine Theta is just one over cosecant and cosecant is one over sine, the reciprocal functions of each other. Cosine Theta is one over secant and secant Theta is one over cosine those once again reciprocal functions. Tangent Theta equal 1 over cotangent which is also sine over cosine. Cotangent is one over tangent or cosine over sine. Once again, reciprocal functions. We're going to look at some standard values and we get our standard values based off of a 4545 right triangle and also a 3060 right triangle. If we start with our 4545 right triangle, we know that the two sides opposite the equal angles have to be equal, so we could call those two sides one. Now by Pythagorean theorem we can find the hypotenuse because a ^2 + b ^2 = C ^2 and a right triangle and we know that a is 1 ^2 and B is 1 and we're going to square it and define the C we're going to square root. So this hypotenuse is going to be sqrt 2. Now if we do our trig functions, we can do sine of 45, which would be one over root 2. To rationalize, we're going to multiply the top and the bottom each by root 2. So we get sqrt 2 / sqrt 4 or sqrt 2 / 2. Cosine 45 is going to be the same thing one over root 2, which simplifies to root 2 / 2, tangent 45 opposite over adjacent. Or one secant 45 is going to be root 2 / 1 or just root 2, cosecant 45, root 2 / 1 again and cotangent 45 one over one. Now we also know that 45° is really the same thing as Pi 4th radiance and our 3060 right triangle. If we thought about starting with an equilateral triangle and letting each side be a length of two, then when we dropped this altitude or the height we split the 60° into 30 and 30. So we had to split the base into one and one. If we do that, we can now find the height based off of the Pythagorean theorem. So the square root. The hypotenuse here was the two, SO 2 ^2 -, 1 ^2 or sqrt 3. So if we want to do sine of 60°, that's the opposite over hypotenuse. Cosine of 60 would be adjacent over hypotenuse, tangent of 60, opposite over adjacent. So root 3 / 1 which is just root 3. Cosecant of 60 is going to be two root threes over three. If we rationalize, secant of 60 is 2 and cotangent of 60 is root 3 / 3. If we do, the 30° sine of the 30° sine is 1/2, cosine of 30° is root 3 / 2, tangent of 30° is root 3 / 3, cosecant of 30° is 2, secant of 30° to root 3 / 3, and cotangent of 30° root 3. We know that 60° is π thirds and we know that 30° is π six. If we forgot how to convert degrees to radians, think about taking whatever your degrees is 60° and multiplying it by the fact that 180° is the same thing as π radians, so 60 and 180 reduce to 1/3 times the Pi. We have also to consider the quadrants that things are positive and negative in. In a quadrant and a Cartesian plane, all of the trig functions are positive in the first quadrant. So positive, positive, positive. The cosine is usually thought of as our left and right. The sine are up and down, and then the tangent is just the ratio of the sine divided by cosine. So if we look here in the second quadrant, we've gone left, so the cosine is negative. We've gone up, so the sine is positive and a positive divided by a negative is a negative. In the third quadrant, we've gone left, so negative we've gone down negative, and a negative divided by negative is a positive. In this last quadrant, cosine, we've gone right, so positive. We've gone down South negative. Negative divided by positive is a negative. So one way to remember this is all students take calculus. All of them are positive in the first sine, is positive in the second tangent, positive in the third cosine, positive in the 4th. And that also includes the reciprocal functions. Periodic functions are functions that repeat. So if we have sine of Theta plus 2K Pi, that's going to really equal sine Theta because we know that the sine values repeat every 2π, where K is just some integer. Cosine repeats every 2π also. So cosine of Theta plus 2K Pi is really cosine Theta. Tangent repeats every π. So tangent of Theta plus K π is really tangent Theta. When we look at a unit circle, we know that the r = 1. But if we looked at any circle and put in a right triangle, we could identify the Theta, the X, the Y and the R on our right triangle. X being the distance left and right, Y being up and down, R being the radius. So cosine Theta is adjacent over hypotenuse or X / r. So X = r cosine Theta sine Theta equal y / r or R sine Theta is Y. Now in a right triangle, we know that the two sides squared, the sum of them equals the hypotenuse squared. So X ^2 + y ^2 equal R-squared. But in this case, we showed that X was really R cosine Theta, and we're going to square that. Y was R sine Theta, and we're going to square that and that's going to equal R-squared. So we get cosine squared Theta plus sine squared Theta equaling one. If we divided everything through by an R-squared, that's a Pythagorean identity. We have two more Pythagorean identities that come relatively easy from starting with this one and dividing everything through by cosine squared for each of those 3 terms. And then we're going to do the same thing by dividing everything through by sine squared. Cosine squared over cosine squared is 1. Sine over cosine is tangent, so one plus tangent squared Theta equal 1 over cosine squared Theta is secant squared Theta. So if we did the same thing but divided every term by sine squared Theta, we'd get our third Pythagorean identity. And our third Pythagorean identity says cotangent squared Theta plus one equal cosecant squared Theta. So those are our three Pythagorean identities. Cosine squared Theta plus sine squared Theta equal 1, one plus tangent squared Theta equals secant squared Theta and cotangent squared Theta plus one equal cosecant squared Theta. Addition and subtraction formulas. Cosine alpha plus or minus beta is cosine alpha, cosine beta minus plus sine alpha, sine beta. When we read this, we need to think about the top sine goes with the top sine. As we read across the bottom sine goes with the bottom sine. Sine alpha plus or minus beta is sine alpha, cosine beta plus minus cosine alpha sine beta, tangent alpha plus or minus beta is tangent alpha plus or minus tangent beta over 1, minus plus tangent alpha, tangent beta. The double angle formulas cosine 2 Theta, cosine squared Theta minus sine squared Theta, which is also equivalent to two cosine squared thetas -1. We can get that one by using the Pythagorean identity we just developed and replacing sine squared Theta as one minus cosine squared Theta. Those are both also equivalent to 1 - 2 sine squared Theta. Sine 2 Theta is 2 sine Theta, cosine Theta and tangent 2 Theta is 2 tangent Theta divided by 1 minus tangent squared Theta half angle formulas. Cosine squared Theta is one plus cosine 2 Theta over 2 Sine squared Theta is 1 minus cosine 2 Theta over 2 tangent squared Theta 1 minus cosine 2 Theta over one plus cosine 2 Theta. The law of cosines C ^2 equal a ^2 + b ^2 - 2 AB cosine C Law sines sine a / a equals sine angle b / b equals sine angle C over side C. Transformation of a graph. If we have Y equal a times the function of the quantity B times the quantity X + C + d, the A in front tells us a vertical stretch of compression, the B tells us a horizontal stretcher compression, the C tells us the horizontal shift, and the D tells us the vertical shift. The C is actually needing to be taken into account with the B because the horizontal stretch and compression is going to affect the horizontal shift also. OK. We're going to develop a couple special inequalities that we'll use. So we're going to start with the fact that we know that s = r Theta and then a unit circle R equal 1. So we know that the arc length is really the same thing as the angle measure. That's going to be important because now we're going to look at the fact that PQ is really just sine Theta and OQ is really just cosine Theta. So if I wanted, QAQA would be the whole distance of 1 minus the cosine Theta. Now we see that it's a right triangle. So we know that PQ squared plus QA squared has got to equal line segment Pennsylvania squared. Well, PQ squared is really just sine squared, Theta and QA squared is just going to be 1 minus cosine Theta squared and PA squared. We're going to leave as PA the line segment Pennsylvania squared. Now we know that the line segment PA has got to be less than the arc length PA and the arc length PA is equal to Theta. So we know that the line segment PA has to be less than Theta. So we know that sine squared, Theta, or a piece of this addition problem has to be less than Theta squared. We also know that one minus cosine Theta squared would have to be less than Theta squared because the two had to add up together to give us something that was smaller still than the Theta. To get rid of the sine squared, we're going to take the square root. So we're going to have negative the absolute value of Theta less than sine, Theta less than the absolute value of Theta. Doing the same process for the other equation, we're going to get negative absolute value of Theta less than 1 minus cosine, Theta less than the absolute value of Theta.