Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. We have sine Theta equal 12 thirteenths and we're in between π halves and Pi. We would know that we're in our second quadrant, and if we know two of the sides of a right triangle, the opposite in the hypotenuse, we can find the third side. It's sqrt 13 ^2 - 12 ^2, which is going to give us a 5. So if we wanted to, if we were asked to find sine of two Theta, we would use our identity that says sine of two Theta is really just two sine Theta cosine Theta. So sine of two Theta is 2 times the sine Theta, which would be 12 thirteenths times the cosine Theta, which would be -5 thirteenths. So we'd get -120 over 169. If we wanted cosine 2 Theta, we would have the cosine squared Theta -5 thirteenths minus the sine squared Theta 12 thirteenths. So 25160 ninths -144 a 160 ninths, negative 100 and 19160 ninths for tangent to Theta, two are tangents -12 fifths over 1 - -12 fifths squared. So we'd have negative twenty 4/5 / 1 -, 100 and 4420 fifths. So negative twenty 4/5 / -1 nineteen 20 fifths. We would invert and multiply, so -24 fifths times 25 / -119 five would go into the top and the bottom to reduce, so we'd get 120 / 119 if we started with. Oh, let's go back for a second here. If we were trying to find two Theta, two Theta here, we could multiply by two all the way through. So we would know that two Theta would have to be in between π and 2π if the sine is negative and the cosine's negative and the tangent's positive. We know that for this problem, two Theta is in the 3rd quadrant because the third quadrant is where sine and cosine are negative and tangents positive. If we look at our next example of sine, Theta equal -3 fifths where we're three Pi halves less than Theta less than 2π, so we're going to be in the 4th quadrant. This time our sine is opposite over hypotenuse, so we know our adjacent side is going to be a four. We could find that by the Pythagorean theorem if we needed to. So when we look at sine of Theta over two, we know that this formula was plus or minus. So the easiest way to do that is to look at the restriction on the original Theta. If I want Theta over two, I'm going to divide everything through by A2SO3, Pi, fours, and π. So we know that Theta over two is going to be in quadrant 2. So Theta over two is in quadrant 2. If it's in quadrant 2, the sine of quadrant 2 is going to be positive, the cosine in quadrant 2 is going to be negative, and the tangents going to be a negative. If we look at our sine, we'd have one minus our cosine of our our given angle, which would be 1 -, 4 fifths all over two. That one is really 5 fifths, so 5 fifths -4 fifths would give us 1/5 / 2, or the square root of 110th. We would rationalize that by multiplying by sqrt 10 / sqrt 10 to get sqrt 10 / 10. For the cosine. It's going to be negative because it was in the 2nd quadrant 1 plus cosine of 4/5 / 2. That's going to give me 9 fifths divided by two or 9/10. Now sqrt 9 is 3, and we have to rationalize that root 10, so root 10 over root 10, so -3 root 10 all over 10. If we look at tangent Theta over 2, the easiest formula, there were three of them, but the easiest one is 1 minus cosine Theta over sine Theta, because that way we don't have to worry about whether it's positive or negative. You'll see in a moment that our answer will come out as a negative and sine Theta is a single term, so if we have to rationalize it, it's much easier than if it was a binomial or two terms. So 1 minus cosine, cosine being 4/5 over sine being -3 fifths. So we don't want a complex fraction. We get 1/5 / -3 fifths. The fifths would cancel. The fives in the denominator cancel, so we'd get -1 third. We expected the tangent in the second quadrant to be negative, and it shows up that it is. If we wanted cosine of 105, we don't know 105, but we do know cosine of 210 / 2 because cosine of 210 divided by two is really just a half angle formula. So we're going to think of this as one plus cosine 210 / 2. Now 105 is going to be in our second quadrant. So cosine in our second quadrant we're going to expect to be negative. What is the cosine of 210? The cosine of 210 is going to be negative root 3 / 2 all over 2. So here the easiest way to do this is to rationalize by multiplying by sqrt 2 / sqrt 2. So we're going to get negative square root 2 -, sqrt 3 all over sqrt 4, which is really just two. So our final answer for the cosine of 105° is going to be negative square root 2 minus root 3 all over two tangent of five Pi eights. We don't know tangent of five Pi eights, but we could figure out tangent of 10 Pi eights divided by two. The tangent Formula One minus cosine 10 Pi eights which is really five Pi force over sine 10 Pi eights once again is five Pi force. So cosine of five Pi force is going to be negative root 2 / 2. Sine is negative root 2 / 2. The easiest way to rationalize this is to multiply by 2 / 2 to get rid of the complex fraction. So we're going to have two plus root 2 all over negative root 2. Now we're going to multiply the top and the bottom by root 2. So we're going to get 2 root 2 + 2 all over -2 which will simplify to negative root 2 -. 1 sine of three Pi eights. We don't know that, but we do know sine of six Pi eights divided by two. Well, sine of six Pi eights divided by two. We're going to have to think about is it going to be positive or negative? Well, three Pi eights is in the first quadrant, so it's going to be positive, and we're going to have one minus cosine. 6 Pi eights is really three π fourths all over two cosine of three Pi force is just going to be negative root 2 / 2 all over two. So if we multiply by A2 on top and bottom, we'd get 2 plus root 2 all over sqrt 4 and sqrt 4 is just two. Given two equations of circles and points on each circle along with which quadrant the angle will fall in, we can find double angles and half angles. So if I know that X ^2 + y ^2 equal 10, and I'm given the point -3 A and we're in the third quadrant, I can stick that -3 and for X the A and for Y and solve for my A. My a ^2 is 1, so a is positive or -1. Because we're in the third quadrant, I know my A is going to be -1. So if I thought about drawing in my triangle, I went -3 to the left and one down and a hypotenuse of 10 on this other one. If I have X ^2 + y ^2 equal 1 and I'm given that my Y is 1/4, I can find my B by plugging it in. So b ^2 plus 116th equal 1B squared is going to equal 15 sixteenths. B is positive or negative squared of 15 / 4. Well, because we're in the second quadrant, we're going to have it be -15 square roots of 15 / 4. And we know that we went up 1/4. And in this case, our hypen hypotenuse is one over here. Our hypotenuse would have been sqrt 10. Sorry. So if we wanted F of two Theta, that's really asking for sine of two Theta because it's given that F of X is sine X. So if I want sine of two Theta, that's two times sine Theta, cosine Theta or alpha. Sorry. So 2 the sine alpha is going to be -1 / sqrt 10. The cosine alpha is -3 / sqrt 10, so -1 square roots of ten -3 square roots of 10. All times 2 is going to come up with 6/10, which would reduce to 3/5 if I wanted F of beta over 2. That's really sine of my beta over two, and my sine formula says it's positive or negative. Well, if beta originally was in between π halves and π, then when we think of beta over two, we're in between π force and Pi halves, taking this whole thing and dividing each piece by two, so π force, beta over 2, and π halves. So this is really going to be in the first quadrant. Our sine in our first quadrant is going to be positive. So we're going to use the positive square root 1 minus cosine alpha over two. Our cosine alpha though, is oh, beta, cosine beta because we were using betas. Our cosine beta here is negative sqrt 15 force over 1 / 2, so the easiest way to rationalize that is to think of multiplying by sqrt 4 / sqrt 4. So we'd get 4 plus square roots of 15 all over eight and eight. We know is 2 square roots of two, so to rationalize that, we have to multiply by root 2 over root 2. So we're going to get sqrt 8 + 2 square roots of 15 all over 2 times sqrt 4. Sqrt 4 is 2, sqrt 8 + 2 square roots of 15 all over 4. If we have 8 sine 4th X, the directions are going to be. We don't want to have any powers for the sine 4X, so we're going to use the power reducing formula. And we know that sine to the 4th X we could think of as sine squared X times sine squared X. And the power reducing formula says if I have sine squared X, that's really one minus cosine 2X over two. And that once again is 1 minus cosine of 2X over two. 2 * 2 in the bottom is the four, which would reduce with the 8 up here, 8 / 4 is going to be two, SO 2 * 1 minus cosine of 2X the whole quantity squared. If we continued this, we'd realize that that's really foiled out to two 1 -, 2 cosine 2X plus cosine squared 2X. But we don't want any powers, so we need to get rid of that. Cosine squared 2X. Well, that's still a formula. We're going to go ahead and distribute the two out. That's a formula that says cosine squared 2X is going to be one plus cosine two times the angle of 2X all over two. SO 2 -, 4 cosine 2X. The twos here will cancel, so plus one plus cosine 4X. So a final answer of 3 -, 4 cosine 2X plus cosine 4X. If we had this next example, we want to get rid of all of the powers. So sine 2 or sine squared X is 1 minus, cosine of 2X all over two, and cosine squared X is one plus cosine 2X all over two. If we foil this out, the two in the times two in the denominator will cancel with the four. So we end up with 1 minus plus cosine 2X, minus cosine 2X minus cosine squared 2X. But once again, we don't want a square, so we're going to replace that cosine squared with the fact that that's really just one plus cosine of two times the angle. In this case, the angle was 2X all over 2, so we get 1 -, 1 plus cosine 4X all over two. If we distribute out that negative, we get 1 -, 1/2 minus cosine 4X over two. When we combine 1 -, 1/2, we get a half minus cosine 4X over 2. That's our final answer. We have some identities here. If we have cosine 4 Theta minus sine 4 Theta to the 4th power of Theta equal cosine 2 Theta, we could think of this as difference of squares cosine squared Theta minus sine squared Theta times cosine squared Theta plus sine squared Theta, cosine squared Theta minus sine squared Theta is one of our identities. It's cosine 2 Theta, cosine squared Theta plus sine squared Theta is also an identity. It's the Pythagorean identity of one. So anything times one is itself and that's what we were trying to prove. This next one, if I have tangent alpha over 2 and I want to get it to be tangent alpha over secant alpha plus one, I'm going to use my tangent alpha over 2 formula that has two terms in the bottom one plus cosine alpha because over here I wanted two terms in the bottom. Well I want to get my sine to turn into tangent. To get sine to turn into tangent, I would need to divide by cosine. But if I do one term divided by cosine, I've got to do them all. And if we look at this, sine over cosine is tangent, one over cosine is secant, and cosine over cosine is 1. And that's what we were trying to do with this identity. Here we have a solving equation, so the angles aren't the same. We're going to use our double angle formula to say that sine 2 Theta is really two sine Theta cosine Theta. I'm going to take the other sine Theta to the same side so that it can all equal 0. At this point I'm going to factor out a sine Theta and get 2 cosine Theta -1 left. Each term that has a Theta in it is then going to get equal to 0, and we're going to solve when is sine 0, sine is 0 when Theta is 0, or π when is cosine Theta 1/2. That's true when Theta is π thirds and also five. Thank you and have a wonderful day.