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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 1

Lesson 7: Determining limits using algebraic manipulation

# Trig limit using double angle identity

Sal finds the limit at θ=-π/4 of (1+√2sinθ)/(cos2θ) by rewriting the expression using the cosine double angle identity.

## Want to join the conversation?

• Wow, my Trig course never taught me fully the formula at , 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 :) • At how did Sal come up with the open interval at (-1,1)? What is the logic behind the open interval? • At , 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? • -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.
• At in 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. • 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. :-)
• how does 1-2sin^2(theta) get factored into squrt(2)sin(theta)? that doesn't make sense to me • 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.
• at how is cos2(theta)= cos^2(theta)-sin^2(theta) what is the difference between cos2 and cos^2 ? thank you • at you 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 • what is sal referring to in open interval at . • At , 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? • 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. 