Examples of product to sum formulas and sum to product formulas. If I have sine times cosine, this is really just a formula and the formula says this is equal to 1/2 times the quantity of sine of the two angles added together. If we add those two angles together, we're going to get 270° plus sine of the two angles subtracted. If we subtract these two angles, we're going to get 120°. Now we actually want to evaluate that sine of 270 is -1 and sine of 120 is root 3 / 2. So our final answer should be negative 1/2 plus root 3 / 4. Or if you wanted a common denominator, we'd have -2 plus root 3 all over 4 for the sine times sine. This once again is just a formula. It's 1/2 cosine, the difference of the two angles minus cosine of the sum of the two angles. In this case, cosine of 120 is negative 1/2, cosine of 270 is 0, so we get negative 1/4 this next one cosine times cosine. It's also a formula 1/2 cosine of the two angles subtracted plus cosine of the two angles added. This case 1/2 cosine of two 120 is negative and a half cosine of 270 is 0. So this is actually going to be the same as the previous. If we look at the next one, this is a formula of sine minus sine. So this just equals 2 sine of the two angles subtracted divided by two. So if we subtract those angles divide by two, we get 120 cosine of the two angles added together to 70 / 235. So now the sine of 120 is root 3 / 2 and the cosine of 135 is negative root 2 / 2. So a final answer of negative root 6 all over 2. If we look at this next one, cosine minus cosine, this is just -2 sine of the two angles added together, so 270 / 2 135 times sine of the two angle subtracted. So 240 divided by two 120. So -2 sine of 135 is going to be root 2 / 2. Sine of 120 is root 3 / 2. So this one's going to give us the final answer of negative root 6 / 2. This last one cosine 255 plus cosine 15 to cosine of the two angles added together divided by two, cosine of the two angles subtracted divided by two. So cosine of 135, negative root 2 / 2, cosine of 120, negative 1/2. So this one simplifies to negative root 2 / 2. Nope, positive root 2 / 2. If we have just generic with thetas instead cosine 2 Theta times sine 3 Theta, this is just going to be 1/2 sine of those two angles added together. So 5 Theta plus sine of the two angles subtracted SO3 Theta -2 Theta or 1/2 sine 5 Theta plus sine Theta. This next one assign plus assign it's two sine of the two angles added together divided by two SO2 Theta +4 Theta divided by two cosine of the two angles subtracted divided by two. So 2 Theta +4 Theta over 2 is just three Theta cosine of two Theta -4 Theta over 2 is going to be negative Theta. But by our even identity, that ends up being 2 sine Theta, cosine Theta, because cosine is an even function. This last one they want us to actually solve for Theta when we're in between 0 and 2π. So if I have sine 2 Theta minus sine 4 Theta, that's really the same thing as sine 2 sine 2 Theta -4 Theta over 2 cosine 2 Theta +4 Theta over 2 equaling 0. So we'd get 2 sine. This is going to give us a -2 Theta over 2 or negative Theta cosine 6 Theta over 2, which is 3 Theta. By our even and odd identity, we know that this is really going to turn into two sine Theta, cosine -2, sine Theta, cosine 3 Theta because our sine function is an odd function. So if it's negative Theta it turns into negative sine Theta. So now we want sine Theta equaling zero and we want cosine 3 Theta equaling 0. And sine is 0 at positive negative K π and cosine is zero at π halves plus K Pi. When we divide by three, we get Pi 6 + K Pi thirds. So now if we want everything between 0 and 2π, we're going to think about if K = 0, we'd end up with π, and we'd end up with PI6 if K equal 1. If K = 0, we ended up with 0 * π or 0 and PI6. If K equal 1, we end up with π and π thirds plus Pi 6, so π thirds plus PI6. This is really 2 Pi 6 + π six or three Pi 6, so that'd be a Pi halves. If K equal 2, we'd end up with 2π. That one's going to be too big. But over here, we'd end up with Pi 6 + 2π thirds, Pi 6 + 2π thirds. That's going to be 5 Pi 6. If K equal 3, we'd end up with Pi 6 + π, which would be 7 Pi 6. If K equal 4, we'd end up with Pi 6 + 4 Pi thirds, or 8 Pi 6 + 1 is 9 Pi 6, which would reduce to three Pi halves. If K equal 5, we'd have Pi 6 + 5 Pi thirds 1011 Pi 6. So if K = 6, we're going to be too much because we're going to have Pi 6 + 2π. So our answer here are going to be 0 PI6 Pi halves, 5 Pi 6PI7PI63 Pi halves and 11 Pi 6.