- [Voiceover] All right, let's see if we can find the limit

of 1 over the square root of 2 sine of theta

And like always, try to give it a shot before we go through

Well, one take on it is well, let's just,

let's just say that this is going to be the same thing

Make sure we can see that negative there,

especially at theta is equal to negative pi over 4,

We'll see well this is going to be equal to

this expression evaluated at negative pi over 4,

sine of negative pi over 4 is going to be

so this is negative square root of 2 over 2,

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

All of this stuff simplifies to negative 1

cosine of negative pi over 2, if you thought in degrees,

Well, cosine of that is just going to be zero,

if we had something non-zero divided by zero,

but we have this indeterminate form, it does not mean

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

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,

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

and you can go from this one to this one to this one

We proved that in earlier videos in trigonometry

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

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1.64 使用雙角恆等式的三角極限（Trig limit using double angle identity | Limits and continuity | AP Calculus AB | Khan Academy）