 In this video we want to solve the trigonometric equation sine theta minus cosine theta equals one. We want to do this on the domain zero to two pi. So clearly we want to solve the problem for radians here. Now how are we going to go about doing this? The we're going to have to use some identities but the identity we're going to use isn't not obvious at the beginning of the problem here because you'll notice that the angles are just theta in play here. So since it's just theta no angle sum, angle difference, half angle, double angle, those identities don't seem to make any sense here. You might try to do like a could I do like a sum to product or difference to product type identity but those really only work when you have a sine and a sine and a cosine and a cosine since we have a mismatch here. What do you do here? Well the key is we want to use the Pythagorean identity but it's not obvious how we use the Pythagorean identity yet. So that's our going to be our first approach on this problem. How do we solve this using the Pythagorean identity? So the first thing we're going to do is we want to isolate one of the trigonometric functions because what I want you to think of here is that sine and cosine, sine and cosine these things kind of act like square roots, right? They kind of act like the square root of something because after all by the Pythagorean relationship if you solve for sine you end up with the square root of one minus cosine squared. If you solve for cosine you end up with the square root of one minus sine squared. So they kind of behave like square roots and this we see this a lot like sine theta can equal root two over two it can equal root three over two just as some examples right for specific angles. So because of this reason we we can kind of pretend like trigonometric functions sine and cosine are like square roots. So if you have the equation the square root of x minus the square root of x plus one and this is equal to one the strategy you would employ would to be isolate one of the radicals like so and then you square both sides foil in and out you get an x on the left hand side notice how the square root disappeared on the right hand side you do have to foil it out you end up with one plus two times the square root of x plus one and then you end up with an x plus one you can combine some like terms notice how the x's cancel from both sides and so you end up with something like negative two is equal to is equal to two times the square root of x plus one and you can keep on going from there that's the idea here we want to separate the trig function square things foil things out of necessary and then go from there you'll notice that this equation actually has no solution because it's negative equal to positive but that's beside the points the technique that matters here we want to think about this in terms of sine and cosine if we pretend those are square roots then we would isolate one of the radicals on the left and then you're going to have one plus cosine on the other side so the way to solve these radical equations in algebra is the same way you do it in government right if you have a bunch of radicals the idea is to isolate the radicals and take them out secretly but when they're by themselves uh so we we get signed by itself we're going to square the left hand side we have to do the same thing to the right hand side what's good for the goose is good for the gander now on the left hand side of course we just have a sine squared theta on the right hand side we do have to foil things out so we end up with one plus we're going to get two cosine theta plus cosine squared theta like so and so now we want to try to combine like terms in some regard like how in the world are we going to do that well this is the point of squaring things and this is where the pythagorean equation comes into play notice we have a cosine theta right here and we have a cosine squared but we have the sine squared here if we could combine sine squared and cosine together sine squared and cosine squared sine squared plus cosine squared would be one if we could do that that'd be great but there's no opposite sides of the equation if we try to move one to the other side we're going to get a minus and that's that's that's the double angle identity it's not the pythagorean identity so instead what we're going to do is just starting with our pythagorean identity cosine squared plus sine squared equals one notice if you solve for sine squared you end up with one minus cosine squared theta for which you're going to substitute that in for sine squared in which case it's the sine square is now gone and it's just a cosine so we end up with one minus cosine squared theta is equal to one plus two cosine two cosine plus cosine squared theta and so now we're going to combine like terms set the left hand side equal to zero minus one minus one so those are actually going to cancel we're going to add cosine squared theta to both sides add cosine squared theta and so we end up with on the left hand side i'm going to swap to the side so we're going to have two cosine squared theta we're going to have two cosine theta this equals zero the constant term disappears that makes factoring a lot easier on the left hand side let's recognize the common the greatest common divisor between them we can take out the factor of two we can also take out a cosine and so we take out two cosine theta that leaves behind a cosine theta plus one equal to zero so we set each of these factors equal to zero and go from there what happens if two cosine theta equals zero well if two cosine theta equals zero divide by two you get cosine theta cosine theta theta equals zero well the theme of the unit circle one is the x coordinate zero we're willing on the y-axis that'll happen at the top of the unit circle pi halves and happens at the bottom three pi halves we're solving this one in radians of course what about the other one then cosine theta plus one well if cosine theta plus one equals zero we get that cosine theta is equal to negative one so when is the x coordinate on the unit circle equal to negative one when you're on the far left of the unit circle that happens at the angle pi so theta equals pi in that situation so putting our solutions together we get that theta could equal pi halves comma pi comma three pi halves like so and so we were to solve this equation remember we were trying to solve the equation sine theta minus cosine theta equals one on one rotation of this unit circle from zero to two pi and how did we do this well the idea is we treated the trigonometric function sine and cosine like they were radicals we squared things now with regular radicals square roots the square would make them go away for the trigonometric functions we have to utilize the Pythagorean identity to get rid of them and so we have to utilize the Pythagorean identity to simplify this and make it into a quadratic equation but in order to do that we had to square things to create it but this this causes a potential problem for us here going back to the idea of solving a radical equation if you have an equation like the following say x equals one if you square both sides you get x squared equals one squared which equals well x squared equals one right but there's actually two solutions to this equation right this this equation right here has the solution x equals one and negative one but the original equation just had just had an x equals one there so when you square both sides of an equation the thing is you could introduce solutions that are actually not authentic solutions i like to think of these as party crashers uh party crashers meaning that someone was invited to the party and some was not who has an invitation right we have to investigate this this is honestly good practice to do on any equation check your solution but when it comes to equations involving squaring both sides you have these potential party crashers it's essential that we check these so let's try these solutions pi halves pi three pi halves to see who's the right one or who's not right so if we were to try pi halves try that one for a second uh theta equals pi halves so we're going to take sine of pi halves minus cosine of pi halves like so sine of pi halves that's equal to one cosine of pi halves is zero that equals one which is the right hand side so pi halves checks out pi halves has an invitation um what about pi theta equals pi well if they equals pi you're going to take sine of pi you're going to get minus cosine of pi sine of pi is zero cosine of pi is negative one so you get a double negative which is one that checked out too just because one works out doesn't mean one doesn't or anything like that so you we do need to check all of these ones what about three pi halves here well if you take sine of three pi halves minus cosine of three pi halves in that situation we'll sine of three pi halves that's going to be a negative one cosine of three pi halves is zero so you get a negative one that does not equal one so aha three pi halves is a party crasher three pi halves has to go to party jail um for for intrusion right now so we have to remove three pi halves this is a very common mistake that students often see is that they did everything right to get pi halves pi and three pi halves but if you don't check your solution you won't see why three pi halves didn't work right you have to check your solution whenever you square both sides and that's what makes this problem a lot more advanced is that when you square both sides it introduces these potentially incorrect solutions um so we have to check our answers when we're done again that's good practice for every equation you try to solve but it's critical for this one right here so it turns out the actual solutions are theta equals pi halves and pi three pi halves needs to be removed