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## AP®︎ Calculus BC (2017 edition)

### Course: AP®︎ Calculus BC (2017 edition) > Unit 1

Lesson 9: Trigonometric limits & squeeze theorem- Limits of trigonometric functions
- Limits of trigonometric functions
- Trig limit using Pythagorean identity
- Trig limit using double angle identity
- Limits using trig identities
- Squeeze theorem intro
- Squeeze theorem example
- Squeeze theorem
- Limit of sin(x)/x as x approaches 0

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# Trig limit using double angle identity

In this video, we dive into finding the limit at θ=-π/4 of (1+√2sinθ)/(cos2θ) by employing trigonometric identities. We use the cosine double angle identity to rewrite the expression, allowing us to simplify and cancel terms. This approach helps us overcome the indeterminate form and find the limit, showcasing the power of trig identities in solving limit problems.

## Want to join the conversation?

- Wow, my Trig course never taught me fully the formula at3:40, at least I only have part of it in my notes. I've never seen the =1-2sinx^2=2cosx^2-1 before.

Can anyone direct me to which video he discusses that? And is there a similar formula for Sin2x? I am wondering if there is a portion of that I never knew as well. Thanks :)(66 votes)- Below are the derivation for the "angle addition formula for sin and cosine" respectively. https://www.youtube.com/watch?v=R0EQg9vgbQw

https://www.youtube.com/watch?v=0VBQnR2h8XM&t=38s

Hope this might help(11 votes)

- At6:11how did Sal come up with the open interval at (-1,1)? What is the logic behind the open interval?(27 votes)
- It is in a unit circle which has a radius of 1. As a result, the ranch of sin, cos and tan is limited between these two values. This is just a rough approximation but in this case it is the precisest value we get.(6 votes)

- At1:48, how did Sal know off the top of his head what sin((-pi/4)) was? Is there a series of videos I could watch to learn these radian values?(10 votes)
- -π/4 is an angle on the unit circle. Most of us have it memorized. You may want to memorize them as well for future benefits if you continue to do math. You can google "unit circle" and get a lot of pic/charts showing those special angles and their cosine's and sine's values.(22 votes)

- At8:20in the video, Sal begins to explain that it is important to understand that the factored equation is not the same as the original unfactored equation. In his explanation he speaks of two functions f(x) and g(x). I’m assuming that g(x) is the original unfactored equation and that f(x) is the resulting factored equation. In the video, Sal states that f(x) and g(x) are equivalent except at ‘a’, and that f(x) is continuous at ‘a’. I'm assuming this means that g(x) is defined for all real numbers except 'a', and therefore is not continuous at 'a'. And that f(x) is defined and is continuous at ‘a’. Sal goes on to say that the limits of f(x) and g(x) are equivalent at 'a'.

Is there a mathematical theorem or proof that states the limit of a factored equation is equivalent to the limit of the respective unfactored equation as long as the factored equation is continuous at the point for which the limit is taken? Is this distinction covered by the Epsilon-Delta proof of limits. For I learned from the video on Epsilon-Delta proofs that a limit can still exists at 'a' despite a function being undefined at 'a'.

I want to better understand this distinction as Sal has identified this situation several times in multiple videos, i.e. where a factored equation has removed a discontinuity which was present in the original unfactored equation, and that the limits are considered equivalent. I believe mathematicians refer to such a discontinuity as a ‘removable discontinuity’.

I am comfortable performing such manipulations, as it makes inherent sense to me that the limits are equivalent; but I wouldn’t know how to mathematically state or prove that such manipulations are correct. If a function can be factored, and the factoring removes a discontinuity, it seems more logical to me to state that the original function had a false or phantom discontinuity. In other words, the discontinuity doesn’t really exist, it only seems to exist due to the equation not being in its simplest, i.e. factored, form.(17 votes)- I have already given a same answer to a similar question, but still I'll hope this helps you out.

As Sal has mentioned in his introductory videos that "The beauty of limits is:They don't depend on the actual value of the function at that limit. They describes how the function behaves when it gets close to the limit."

So here in this case, Sal created an analogy by creating the refactored function f(x) which is continuous at x=a and then as g(x) is same as f(x) except it is discontionuous at a, so from the definition stated above, limit of g(x) has to be equal to f(x) at that point.

And the way we can evaluate the limit of f(x) at x=a is by various methods like epsilon delta, graphs, tables etc. which are logically as well as mathematically correct.

So, long story short, Sal evaluated the limit of the original function the other way around(which is correct :-)).

I'm sorry if i was unclear. Just a small effort to help. :-)(10 votes)

- how does 1-2sin^2(theta) get factored into squrt(2)sin(theta)? that doesn't make sense to me(9 votes)
- we are seeing 1-2sin^2(theta) in the form of a^2-b^2 which is (a+b)(a-b) and we can even write 2 as ( sqrt(2) )^2 and 1 as 1^2 , since square and square root will cancel each other and 1^2 is 1.

now we can write 1-2sin^2(theta) = 1^2-( sqrt(2)sin(theta) )^2= (1+sqrt(2)sin(theta))(1-sqrt(2)sin(theta).

and after that we can cancel out the common factor from numerator and denominator.(10 votes)

- at3:30how is cos2(theta)= cos^2(theta)-sin^2(theta) what is the difference between cos2 and cos^2 ? thank you(7 votes)
- It's not cos2(θ), it's cos(2θ). Hence the term "double angle identity."

See this page, about a quarter of the way down: https://www.khanacademy.org/math/precalculus/trig-equations-and-identities-precalc/using-trig-identities-precalc/a/trig-identity-reference(7 votes)

- at2:33you have conver 2 cos pi/4 as the same as cos pi/2...

my question is why do must you convert in this way as the cos pi/4 is sq root 2 over 2... x 2 is the sq rt of 2.. and the cos of the sq rt of 2 is not 0(4 votes)- It is cos(2θ) which is not the same as 2cos(θ). Sal should had use parenthesis like I did to avoid confusion for students. cos(2•π/4) , you simplify what inside the parenthesis first then compute cosine function.(10 votes)

- what is sal referring to
**in open interval**at5:56.(5 votes)- The interval (a,b) is just the set of real numbers between a and b. If we mean to include the endpoints a and b in the set, we write it as [a,b] and call it a closed interval. If we mean to exclude them, we write (a,b) and call it an open interval.(7 votes)

- How do we know when a function is continuous?(4 votes)
- Graphically, a function is continuous if you can trace it with a pencil without having to lift it up. There are no holes, gaps, asymptotes, or jumps.

If you don’t have a graph, use limits. A function is continuous at a point if the “limit as x approaches c” = f(c). If this applies to all points on an interval, then the function is continuous over that interval.

Some functions you should be able to tell are continuous right away. All polynomials, linear, sine and cosine, radical, and exponential functions are continuous at every point in their domain.

Hope this helps!(5 votes)

- At6:13, Sal talks about something called
*open interval.*Can someone explain it to me?

By the way, is the initial function also 0/0 at 7π/4 and 15π/4 etc.? If so, then shouldn't the constraint be θ ≠ -π/4 + 2πk, where k is an integer? are are we not being this specific here?(3 votes)- An open interval doesn't include the endpoints.

Example:

The open interval 𝑥 ∈ (−2, 3) ⇔ −2 < 𝑥 < 3

The closed interval 𝑥 ∈ [−2, 3] ⇔ −2 ≤ 𝑥 ≤ 3

I don't know why Sal keeps reminding us that we're looking at this function of ours over an open interval, because it could just as well be a closed interval.

– – –

The function 𝑓(𝜃) = (1 + √2 sin 𝜃)∕cos 2𝜃 is defined for all 𝜃 such that

cos 2𝜃 ≠ 0 ⇔ 2𝜃 − 𝜋𝑘 ≠ arccos 0 = 𝜋∕2 ⇔

⇔ 𝜃 ≠ (2𝑘 + 1)𝜋∕4, 𝑘 ∈ ℤ

However, by restricting the domain to 𝜃 ∈ (−1, 1) Sal tried to make all discontinuities redundant apart from 𝜃 = −𝜋∕4, and had he instead chosen something like 𝜃 ∈ (−1, 0) he would have succeeded, because there's no need to list 𝜃-values for which 𝑓(𝜃) is undefined if they're not in the domain.(6 votes)

## Video transcript

- [Voiceover] All right, let's
see if we can find the limit of 1 over the 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. We'll see well this is
going to be equal to this expression evaluated
at 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 in
degrees, this would be a negative 45-degree angle,
so this is one of the, one of the trig values
that it's good to know and so if you have, if you have 1, so let's see, actually, let me just rewrite it, so this is going to be equal to 1 plus square root of 2 times that is going to be negative 2 over 2, so this is going to be minus 1, that's the numerator over here. All of this stuff simplifies to negative 1 over, this is going to be cosine of negative pi over 2, right? This is negative pi over 2, cosine of negative pi over
2, if you thought in degrees, that's going to be negative 90 degrees. Well, cosine of that is
just going to be 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 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 does not, is not an indeterminate form, that theta is equal to 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 1 plus the square root of 2, sine theta, over cosine 2 theta, as you can imagine, the things that might be useful
here are our trig identities and in particular, cosine of
2 theta seems interesting. Let me write some trig
identities involving cosine of 2 theta. I'll write it over here. So we know that cosine of 2 theta is equal to cosine squared of theta minus sine squared of theta which is equal to 1 minus 2 sine squared of theta which is equal to 2 cosine squared theta minus 1, and you can go from this
one to this one to this one just using the Pythagorean identity. We proved that in earlier
videos in trigonometry 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
at 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 involved
a 1 plus square root of 2 sine theta, then 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 square root of 2 sine theta times 1 minus square root of 2 sine theta, and now, we have some nice cancelling, or potential cancelling that can occur, so we could say that cancels with that and we could say that
that is going to be equal, and let me do this in a new color, this is going to be equal to, in the numerator we just have 1, in the denominator we just are 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, as 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 wanted to, for this particular case, well, let's just say, everything we're doing
is in the open interval, so in, in 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 zero over zero form, and pi over 4 would make this denominator equal to zero but it also makes, let's see, pi over 4 also
will make this denominator equal to zero, 'cause we would get 1 minus 1, so I think, I think we're good if we're just assuming, if we're restricted 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, or taking as much trouble to be 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 the square root of 2 sine of theta. If we're dealing with this over, you know, with this open interval, wait, actually, even disregarding that, theta, this theta, 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, 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, is equal to our original expression, and all values of theta except, except where 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 to say, 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, wait, let me just write this this way, for all x except for a, then the limit, then, and let me write it this way, equal 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 got it right, the answer here is 1/2.