Who are you? Who am I? Where did we come from, and where are we going? Could I eat an entire pizza by myself? These are all questions about our identity. Some of them are easier to answer than others. The easy answers may lead to intestinal distress, though.
The identities we're really interested in today are the trig identities. We want more tricks for working with trig functions, and the double-angle and half-angle identities fit the bill. The addition formulas continue to be our MVFs (Most Valuable Formulas).
Small questions can lead to big things happening. Asking your friend, "How was your day today?" can lead to a 20-minute rant about the evils of capitalism, and how their manager, in particular, is the most twisted despot this side of Sauron.
Here's another question that seems simple enough: what happens if we try to add an angle to itself? That seems simple, but it leads us straight to the double-angle identities.
sin (α + α) = sin α cos α + cos α sin α = 2sin α cos α
Or, to have an actual double angle involved in this:
sin (2α) = 2sin α cos α
That's it, seriously. Easiest proof ever, right? Cosine is the same way:
cos (2α) = cos (α + α) = cos α cos α – sin α sin α
= cos2 α – sin2 α
Oh hey, squared trig functions. Remember that trig2 α is the exact same as (trig α)2. But they aren't the same as trig (α2). Got it?
An easy string of discoveries in a row can't last forever, can it? If anyone is going to ruin this, it's going to be tangent. Tangent is like a professional party ruiner.
Tangent must be on break or something, though, because the double-angle identity for tangent is easy too.
We're not complaining.
Find the exact value of sin (120°).
We know all about sine and cosine of 60°, so let's use those and our shiny and new double-angle formula for sine to get sin (120°):
sin (120) = sin (2(60)) = 2sin (60) cos (60)
Notice that it this actually the same value as sin 60°. What could be behind such a coincidence? As always, our first thought is aliens. Our second thought is that the reference angle for 120° is 60°, and that sine is positive in Quadrant II.
Maybe that does explain it. If aliens were involved, they must have been super sneaky, and super useless. They didn't even do anything. How lazy.
We may think we already know the Pythagorean identity. Pythagoras was that old Greek dude who obsessed over triangles. A lot. But that's not the identity we're talking about. It's this one:
sin2 θ + cos2 θ = 1
One is the loneliest number, and so are sin2 and cos2, together. We're not sure how the math on that works out.
We can find out, though. Let's go back to our original definitions for sine and cosine. We'll save going back to the future for another time.
If we square them both and add them together, we get the start of a great party (because it's hip to be square).
And we've got it. That stuff up top? That's a2 + b2. Down bottom is c2. Old man Pythagoras has something to say about those: they're equal. Because they are equal, the whole thing equals 1.
At least, that proves it for angles in right triangles. Pythagoras didn't have nearly as much to say about all the other triangles and angles out there. At this point, we're just going to wave our hands, tell you that it's actually true all the time, and ask you to trust us. We've earned that much trust, haven't we? Forget about the shaving cream incident; that's ancient history by now.
Now we've done it. We were playing baseball in the house, again, and we broke our mother's favorite angle in half. She is going to kill us when she gets home, unless we can fix this mess. We don't know anything about working with half of an angle, though.
Quick, let's learn the half-angle identities so that we can save our bacon. Shmoopy math powers, don't fail us now.
To start off this journey of discovery, let's put the cosine double-angle identity side-by-side with the Pythagorean identity.
sin2 α + cos2 α = 1
cos (2α) = cos2 α – sin2 α
Hmm. HMM. We're starting to get some ideas here. What if we stick one of them inside of the other, turducken-style? Let's substitute in (1 - sin2 α) from the first equation for (cos2 α) in the second.
cos (2α) = 1 – sin2 α – sin2 α
cos (2α) = 1 – 2sin2 α
With a little switcheroo, we can start solving for sine. Stick with us here, we'll get to the point soon. Ish.
2sin2 α = 1 – cos (2α)
We have sin α related to cos (2α). That's great and all, but the name of the game is "half-angle identities," not "here's another way to find sine." We can divide our angles by 2, though, and get:
Ring the bell, we have the half-angle identity for sine.
We're not done yet, though, because we can find the cosine half-angle identity too.
Let's back things up and, instead of replacing cos2 α with (1 – sin2 α), let's replace sin2 α with (1 – cos2 α):
cos (2α) = cos2 α – (sin2 α)
cos (2α) = cos2 α – (1 – cos2 α)
cos (2α) = 2cos2 α – 1
Now let's shuffle-board stuff around to solve for cos α:
We don't really need this whole angle. How about we share it 50-50 with you?
There we go. Half the angle for double the fun. Now we can find even smaller angles, like this one.
Find the exact value of cos (22.5°).
Ack! We've never had to deal with half a degree before. It's just awful. But now we have the half-angle identities, so we can do this. As long as we realize that α = 45°, and we do realize it, we are good to go.
That plus/minus dealie at the beginning isn't necessary now that we know we're dealing with a degree of 22.5. We're in Quadrant I, so our answer must be positive. We also know that cos 45 is definitely . That's all we need to deliver a good solving-down to this equation.
Yeesh, that's not good. Time to spruce it up a bit:
That's better, but having a radical inside of a radical just feels wrong. Like wearing two pairs of socks. We can't do anything about it, though, so this is as good an answer as we'll get.
Speaking of "we can't do anything about it," we also need to find the tangent half-angle identity. We kid, we kid. You're cool, tangent. We'll once again use the sine over cosine trick to get things done.
And that's the last identity to find. Good thing, too. We're almost all identitied out.
Find the exact value of tan (15°).
If we were feeling particularly saucy, we would go ahead and solve this using the double-angle identity we already know. That would just be mean, though, so we'll try out our new identity. Sorry, we can't tell you what our new identity is, it's a secret.
(It's Batman. We're totally Batman.)
15° is half of 30°, so we will use that as our angle / half-angle combo. We'll need cos 30°, but we know that's . Special triangles FTW. And, we're in the first quadrant, so the result will be positive.
Make like a Marine and "Simpli Fi."
We still have way too many radicals in there, so let's multiply the top and bottom inside the big radical by :
And Batman solves another case. Uh, wait, Shmoop solves the case. Totally not Batman.
(We're totally Batman.)