Half-Angle Identities

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.

Sample Problem

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.

Sample Problem

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.)