We want to find the exact values of cosine 5 Pi twelfths, so we need to figure out what angles add up to give me 5 Pi twelfths that we already know. I frequently will write down some of my known angles like π sixes 2π twelfths π fours would be 3 Pi twelfths. Well if I added together Pi 6 and π fours, I get the five Pi twelfths. So this is going to equal cosine of Pi 6 + π force. Now we know the cosine formula which says cosine of Pi 6, cosine π force minus sine of Pi 6, sine of π force. Well, what is cosine of Pi 6? It's root 3 / 2. What's cosine of π force root 2 / 2? Sine of Pi 6 is 1/2 sine of π force root 2 / 2. When we multiply and combine, we're going to get sqrt 6 -, sqrt 2 all over 4. How about a tangent 17 Pi twelves? Well, 17 Pi twelfths is a little bigger. I need to start with bigger numbers, maybe three π fourths, which in terms of twelfths would be 9 Pi twelfths. What do I need to add to 9 Pi twelfths to get 17 Pi twelfths? Well, I'd need to add 8 if I reduce my eight Pi twelfths. Is that something that I know? Well? That's 2π thirds. So 2π thirds is 8 Pi twelfths and three π fourths was nine Pi twelfths and nine Pi twelfths and eight Pi twelfths gave me the 17 Pi twelfths. So this is going to equal tangent of three π fourths plus 2π thirds. And our formula for tangents says tangent three π fourths plus tangent 2π thirds divided by 1 minus tangent three Pi force tangent 2π thirds. Now there are lots of different possible values, these just happen to be two that I found quickly. Tangent of three Pi force is -1 tangent of 2π thirds is negative root 3, so 1 - -1 times negative root 3. So that's going to give us a -1 minus root 3 / 1 minus root three. We don't ever leave a square root in the bottom, so we're going to rationalize this by multiplying by its conjugate one plus root 3 / 1 plus root 3. So we're going to get -1 minus root 3 minus root 3 -, 3 on the top and in the bottom we're going to get one plus root 3 minus root 3 - 3. So we get -4 -, 2 square roots of three all over -2 when we combine things up. So our final answer is going to be +2 plus root three -2 went into -4 a +2 times and -2 went into -2 root. 3A positive root three time sine of 195. Well, we need to think about angles. We know. We know 135° well, 135 plus what gives me 100 and 95135 + 60 So I can rewrite this a sine of 135° + 60 and our sine formula says sine of 135°, cosine 60 plus cosine of 135° sine 60. Well, sine of 135° is root 2 / 2, cosine of 60 is 1/2, cosine of 135 is negative root 2 / 2, and sine 60 is root 3 / 2. So this is going to simplify to root 2 minus root 6 all over 4. Looking at more examples, what if we had it as sine 20, cosine 80 minus cosine 20, sine 80? Well, this is one of our formulas. This is a sine of an angle, cosine of an angle, minus cosine of the angle, sine of the angle. So this is really a sine formula where we're going to have 20 -, 80. Well, 20 -, 80 is -60 and we know that sine of a negative is really the same thing as negative the sine because it's an odd identity. Sine of 60 is root 3 / 2. So we're going to get negative root 3 / 2 is our final answer. This next one, it's in the tangent formula. So we're going to have tangent of 40° -, 10°. That's really tangent of 30°. And we know that the tangent of 30° is root 3 / 3. This next type, they're going to give us two different trig functions with two different angles. And they want us to find sine of alpha plus beta, cosine alpha plus beta, tangent alpha plus beta, etcetera. So one of the easy ways to do this is to think about giving ourselves a triangle. If alpha is in between 0 and π halves, we're in the first quadrant. Here's alpha opposite over hypotenuse. I can always find this third side. So 5 ^2 -, 3 ^2 ^2 of 16, which would be 4 for this other angle where negative π halves less than beta, less than 0. So we're going to be over here somewhere and we're going to have cosine 2 root 5 over 5. We can find our third side, our 5 ^2 -, 2 square roots of 5 ^2. So we get 25 -, 2 square roots of 5 * 2 square roots of five is going to be 20. So this other side would be root 5. So if we had asked for cosine alpha plus beta, we'd know that that's cosine alpha, cosine beta minus sine alpha, sine beta. When we look at this triangle over here, we know our cosine alpha is 4/5. When we look at the other triangle, our cosine beta is 2 root 5 / 5. Sine alpha is going to be 3/5, sine beta is going to be negative root 5 / 5. When we simplify this all out, we get 8 root 5 + 3 root 5 / 25, or 11 root 5 / 25. If we had wanted this exact same thing, but we wanted cosine alpha minus beta would have been the exact same thing, but the sign would have changed in the middle. So when we get this off of our triangles, we'd have 4/5 two root 5 / 5 + 3/5 negative root 5 / 5. Simplifying it up again, we get 8 root 5 -, 3 root 5. This time over 25 8 - 3 is five root 5 / 25 which would reduce to root 5 / 5. Say we had been asked for sine of alpha plus beta. We look at our first triangle and we realize the formula. First of all, sine alpha cosine beta plus cosine alpha sine beta. So sine alpha was 3/5, cosine beta was 2, root 5 / 5, cosine alpha was 4/5, and sine beta was negative root 5 / 5. When we simplify this up, we would get 6 root 5 -, 4 root 5 over 25, or two square roots of 5 / 25. We can do the same thing for sine of alpha minus beta and tangents. This next one we're given F of X equals sine XG of X equal cosine X&H of X equal tangent X and we want to find F of a plus alpha plus beta. So we want sine of alpha plus beta. That's the formula that says sine alpha, cosine beta plus cosine alpha sine beta. Given our triangle, we know that X, one is our point. So if Y is one, can we find our Y or our X? Sorry, X ^2 is going to equal 3, so X is sqrt 3. So we came over sqrt 3. We went up one and this is a circle with the radius being 2 here because X ^2 + y ^2 equal 4. This other for beta we're given X ^2 + y ^2 equal 1 and we know that X is 1/3. If X is 1/3. Can we find our Y? So if X is over one third, because it's a unit circle in this case, we know that the hypotenuse is 1, so we're going to get y ^2 equaling 9 ninths -1 ninth or eight ninths Or Y is going to equal -2 square roots of 2 / 3. When I rationalize it and simplify. So looking at our two triangles, our sine alpha is just the opposite. Over the hypotenuse, our cosine beta is adjacent. Over the hypotenuse, our cosine alpha is adjacent. Over hypotenuse. Our sine beta is opposite over hypotenuse, so when we simplify this up, we're going to get 1/6 minus root 6 / 3 because the two on the bottom and the two on the top will cancel square roots. We multiply the numbers inside so minus root 6 / 3. Or if you prefer, you could have it as 1 -, 2 root sixes all over six by just getting a common denominator. What if we had H of alpha minus beta? Well, that would mean tangent of alpha minus beta. So tangent alpha minus tangent beta over 1 minus tangent alpha. Tangent beta. Tangent alpha, looking at this triangle, is going to be root 3 / 3 because it was 1 divided by root 3, and we rationalize. Tangent beta is going to be two root 2 / 3 / 1/3, and it is a negative because it's in the 4th quadrant. 1 minus root 3 / 3 * -2 root 2. If we simplify this up, we'd get root 3 / 3 + 2 square roots of 2 / 1 + 2, root 6 / 3. Now we don't ever leave a complex fraction, so we're going to multiply by 3 / 3. We get root 3 + 6 square roots of 2 / 3 + 2 square roots of 6. Can't leave a radical in the denominator, so we're going to multiply by the conjugate 3 - 2 square roots of 6 / 3 - 2 square roots of six. We're going to foil this out, so we're going to get 3 square roots of 3 - 2 square roots of 18 + 18 square roots of 2 - 12 square roots of 12, all over 9 - 6 root 6 + 6 root 6 - 24. So continuing on, we get 3 root 3 root 18. Simplifies into 3 sqrt 2 so -6 sqrt 218 sqrt 2 sqrt 12 is 2 square roots of three. So we end up with -24 square roots of three, and the six square roots of six in the bottom are going to cancel. So 9 - 24 is going to give us -15. Combining our like terms 3 and -24 gives us -21 root threes a -6 and a +18 is going to give us +12 root twos over -15. All of those coefficients are divisible by a -3, so we're going to get 7 root 3 -, 4 root 2 all over 5. As a final answer for that one, we have some identities. Next, if I have sine of three Pi halves plus Theta equal negative cosine Theta, If we use our formula, we get sine of three Pi halves, cosine Theta plus cosine 3 Pi halves sine Theta. Well, sine of three Pi halves is -1. Cosine of three Pi halves is 00 times anything is 0, so we end up with negative cosine Theta equaling negative cosine Theta. This next one secant alpha plus beta. We could think of that as one over cosine alpha plus beta. So that would be one over cosine alpha, cosine beta minus sine alpha, sine beta. Now we want to get cotangents everywhere or cotangents in the denominator. Before when we wanted to get tangents, we divided by cosine. So now if I want cotangents, we're going to divide by sine. So I'm going to take the top 1 divided by sine alpha, sine beta over cosine alpha, cosine beta divided by sine alpha, sine beta minus sine alpha, sine beta divided by sine alpha, sine beta. Well, one over sine alpha is cosecant, A1 over sine beta is cosecant beta. Cosine alpha over sine alpha is cotangent. So we have cotangent alpha and cotangent beta and the sine and the sine cancel. And that is indeed what we were trying to get. So cosecant alpha, cosecant beta over cotangent alpha, cotangent beta -1. OK now sine of sine inverse of 1/2 plus cosine inverse of zero. Well sine inverse of 1/2 is really really an angle. What angle gives us out sine inverse of 1/2? And that angle would be PI6. What angle gives us out cosine inverse of 0? Cosine inverse of 0 would be at Pi halves. Now we could possibly add those together, but we actually really want to practice this section. So we're going to say sine of Pi 6, cosine of π halves plus cosine of Pi 6, sine of π halves. Well, sine of Pi 6 is 1/2, cosine of π halves is 0, cosine of Pi 6 is root 3 / 2, sine of π halves is 1. So this is going to end up being root 3 / 2. Now if we had gone ahead and added, we would have gotten sine of four Pi 6, getting a common denominator, and that's really 2π thirds. And we know that sine of 2π thirds is root 3 / 2. So more than one way to do math, they really want us to be practicing the formulas. This next one tangent inverse of Four Thirds isn't as an angle off of our unit circle, so let's call that alpha cosine inverse of 12 thirteenths. Let's call it beta. If I make myself a triangle and have this angle being alpha, tangent is opposite over adjacent. That makes this a 345 right triangle. We could have done sqrt 3 ^2 + 4 ^2. If we do beta, we know that beta cosine inverse of 12 thirteenths would be adjacent over hypotenuse. So this is going to be a 5/12/13 triangle. Now the cosine of this, so we want it cosine alpha, cosine beta minus sine, alpha, sine beta. Once we have our triangle drawn, we just look what is our cosine A3 fifths, what's our cosine beta 12 thirteenths, what's our sine A4 fifths and our same sine beta 5 thirteenths. So we're going to get 36 -, 20 over 65 or 1660 fifths. What if we don't have numbers? We're going to do it the same way. We're going to let this one be alpha and this one be beta. So if we draw our triangle, we know that our alpha is going to have a being the opposite and one being the hypotenuse. We're going to have our beta being B being the adjacent and one being the hypotenuse. We should be able to find that third side by Pythagorean theorem. So in this case sqrt 1 -, b ^2 over here, sqrt 1 -, a ^2. So if I have tangent of alpha minus beta, that's just tangent alpha minus tangent beta over one plus tangent alpha tangent beta. So tangent alpha a over square root 1 -, a ^2 minus square root 1 -, b ^2 / b all over 1 + a sqrt 1 - a ^2 * sqrt 1 -, b ^2 / b. To simplify that up, it probably is going to be easiest to multiply the top and the bottom by a common denominator. The common denominator here would be B square roots of 1 -, a ^2. If I do the bottom, I'm going to do the top. So when I multiply sqrt 1 -, a ^2. So cancel in the first term, leaving me AB. The B's cancel in the second term, leaving me sqrt 1 - b ^2 * sqrt 1 -, a ^2. In the bottom we're going to get b sqrt 1 - a ^2 and then plus a times sqrt 1 - b ^2. Then we would rationalize and I'll let you guys do that one. The last one we're going to have solving between 0 and 2π, so 0 less than or equal to Theta less than 2π. If I think about taking one of the terms to one side and getting one of the trig terms by itself, then we're going to square each side. So we get sine squared Theta equal 1 -, + 2 sqrt 3 cosine Theta plus three cosine squared Theta by foiling it out on the right hand side. Sine squared Theta is really one minus cosine squared Theta. So then we're going to take everything to one side. We're going to get 4 cosine squared Theta +2 root 3 cosine Theta equaling 0. These each have a two cosine Theta in common. So if I pull out what they have in common, I get 2 cosine Theta plus root 3 left. So cosine Theta is going to equal 0 from the term we pulled out and cosine Theta is going to equal negative root 3 / 2 from that second parenthesis. Well, cosine Theta is zero at π halves and also three Pi halves. Cosine Theta is negative root 3 / 2 when Theta is five Pi 6 and also seven Pi 6.