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# Trig limit using double angle identity

AP.CALC:
LIM‑1 (EU)
,
LIM‑1.E (LO)
,
LIM‑1.E.1 (EK)

## Video transcript

alright let's see we can find the limit of 1 over square root of 2 sine of theta over cosine of 2 theta as theta approaches negative PI over 4 and like always try to give it a shot before we go through it together well one take on it is well let's just let's just say that this is going to be the same thing as the limit as theta approaches negative PI over 4 of 1 plus square root of 2 sine theta over the limit as theta approaches negative PI over 4 make sure we can see that negative there of cosine of 2 theta and both of these expressions are if these were function definitions or if we were to graph y equals 1 plus square root of sine square root of 2 times sine theta or y equals cosine of 2 theta we would get continuous functions especially at theta is equal to negative PI over 4 so we could just substitute in and say well this is going to be equal to this expression evaluated it negative PI over 4 so 1 plus square root of 2 times sine of negative PI over 4 over cosine of 2 times negative PI over 4 now negative PI over 4 sine of negative PI over 4 is going to be negative square root of 2 over 2 so this is negative square root of 2 over 2 we're assuming this is in radians if we're thinking degrees this would be a negative 45 degree angle so this is one of the one of the the trig values that it's good to know and so if you have if you had one so see well X to me just rewrite it so this is going to be equal to one plus square root of two times that is going to be negative two over two so this is going to be minus one that's the numerator over here all of this stuff simplifies to negative one over this is going to be cosine of negative PI over two all right this is negative PI over two sine of negative PI over 2 if you thought it agrees that's going to be negative 90 degrees well cosine of that is just going to be is zero so what we end up with is equal to zero over zero and as we've talked about before if we had something non zero divided by zero we'd say okay that's undefined we might as well give up but we have this indeterminate form it does not mean that the limit does not exist it's usually a clue that we should use some tools in our toolkit one of which is to do some manipulation here to get an expression that maybe is defined at theta is equal to or it is does not it is not an indeterminate form at theta is equal to negative PI over 4 and we'll see other tools in our toolkit in the future so let me algebraically manipulate this a little bit so if I have one plus the square root of two sine theta over cosine two theta as you can imagine the things that might be useful here are our trig identities and in particular cosine of two theta seems interesting let me write some trig identities involving cosine of two theta I'll write it over here so we know that cosine of two theta is equal to cosine squared of theta minus sine squared of theta which is equal to one minus two sine squared of theta which is equal to two cosine squared theta minus one and you can go from this one to this one to this one just using the Pythagorean identity and we proved that in earlier videos and intr Gotama tree on Khan Academy now do any of these look useful well all of these three are going to be differences of squares so we can factor them in interesting ways and remember our goal in the end of the day is maybe cancel things out that are making us get this zero over zero and if I could factor this into something that involves a 1 plus square root of 2 sine theta that I'm going to be in business and it looks like it looks like this right over here that can be factored as 1 plus square root of 2 sine theta times 1 minus square root of 2 sine theta so let me use this cosine of 2 theta is the same thing cosine of 2 theta is the same thing as 1 minus 2 sine squared theta which is just a difference of squares we can rewrite that as this is a squared minus B squared this is a plus B times a minus B so I can just replace this with 1 plus the square root of 2 sine theta times 1 minus square root of 2 sine theta and now we have some nice canceling or potential canceling that can occur so we could say that cancels with that and we could say that that is going to be equal let me do this in a new color this is going to be equal to the numerator we just have 1 in the denominator we just start left with 1 minus square root of 2 sine theta and if we want these expressions to truly be equal we would have to have them to have the same if you view them as function definitions is having the same domain so this one right over here this one we already saw is not defined at theta is equal to negative PI over 4 and so this one in order for these to be equivalent we have to say that this one is also not and actually other places but let's just let's just say theta does not does not equal negative negative PI over 4 and we could think about all of this happening in some type of an open interval around negative PI over 4 if we wanted to get very precise but if we for for this particular case well let's just say all everything we're doing is in the open interval so in an open interval in open interval between theta or say negative 1 and 1 and I think that covers it because if we have PI if we have PI over 4 that is not going to get us the 0 over 0 form and PI over 4 would make this denominator equal to 0 but it also makes let's see over for also we'll make this denominator equal to zero because we would get one minus one so I think I think we're good if we're just assuming or restricting to this open interval and that's okay because we're taking the limit as it approaches something within this open interval and I'm being extra precise because I'm trying to explain it to you and it's important to be precise but obviously if you're working this out on a test or notebook you wouldn't be taking putting at or taking as much trouble to put you putting all of these caveats in so what we've now realized is that okay this expression actually let's think about this let's think about the limit the limit as theta approaches negative PI over 4 of this thing without the restriction of 1 over 1 minus square root of 2 sine of theta if we're dealing with this over you know in this open interval we achieve in your disregarding that theta this data or this expression is continuous at it is defined and it is continuous at theta is equal to negative PI over 4 so this is just going to be equal to 1 over 1 minus the square root of 2 times sine of negative PI over 4 sine of negative PI over 4 sine of negative PI over 4 we've already seen is negative square root of 2 over 2 and so this is going to be equal to 1 over 1 minus square root of 2 times the negative square root of 2 over 2 so negative negative you get a positive square root of 2 times square root of 2 is 2 over 2 is going to be 1 so this is going to be equal to 1/2 and so I want to be very clear this expression is not the same thing as this expression they are the same thing at all values of theta is especially if we're dealing in this open interval except at theta equals negative PI over 4 this one is not defined and this one is defined but as we've seen multiple times before if we find a function that is equal to our original or an expression that's equal equal to our original expression and all values of theta except except whether the original one was not defined at a certain point but this new one is defined and is continuous there well then these two limits are going to be equal so if this limit is 1/2 then this limit is going to be 1/2 and I've said this in previous videos it might be very tempting so yeah well I'm just going to algebraically simplify this in some way to get this and I'm not going to worry about too much about these constraints and then I'm just going to substitute negative PI over 4 and you will get this answer which is the correct answer but it's really important to recognize that this expression and this expression are not the same thing and what allows you to do this is is the truth that if you have two functions if you have F and G two functions equal let me write it this way equal equal for all X except for all alright let me just write this way for all X except for a then the limit then and let me write it this way equal for all for all except for all X except a and F continuous continuous at a then then the limit of f of X as X approaches a is going to be equal to the limit of G of X as X approaches a and I said this in multiple videos and that's what we are doing right here but just so you can make sure you you got it right the answer here is 1/2
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