If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Using the cosine double-angle identity

The cosine double angle formula tells us that cos(2θ) is always equal to cos²θ-sin²θ. For example, cos(60) is equal to cos²(30)-sin²(30). We can use this identity to rewrite expressions or solve problems. See some examples in this video. Created by Sal Khan.

Want to join the conversation?

• Instead of doing Sin(A+B) and following these long formulas, couldn't we simply deduce any angles directly via Arcsin(y/h) and then add them together?

Eg: Sin(60 + Angle ABC) is simply Sin(60 + Arcsin(BC/AB))
• The only problem I can find with this is that arc-functions are typically not introduced until after this point. Realistically, though, they could be introduced almost immediately after sines & cosines. I can't find a good, pedagogical reason why this happens.

Edit: Two years on, I can now see that I was talking about a special case. Your example of sin(60 + arcsin(BC/AB)) is actually making a simpl-ish case harder. I think what you were trying to end up doing is split the sine function over its respective inputs, i.e. sin(60) + sin(arcsin(ABC)), which does not work. The ONLY way to resolve trigonometric functions containing addition or subtraction is to use the formulas above (https://www.youtube.com/watch?v=ulQkjvHjWEc).
• In which video does he talk about why cos2x = cos^2 x = sin^2 x?
• How are you supposed to do: cos²(θ)? Is it the same as cos(θ)²?
• cos²(θ) is different with cos(θ)²
For example: θ = 60 degrees, then cos(θ) = 1/2
then cos²(θ) = (1/2)^2 = 1/4
and cos(θ)² = cos (60)(60) = cos 3600 degree = cos 0 degree = 1
3600 degree equals 10 full circles
• Please tell me the video cos(2*theta)=cos^2(theta)-sin^2(theta)
• I don't think there is one yet. But you arrive at that trig identity by applying the sum formula. Watch:

1. First we apply the sum formula, cos(a+b) = cos(a) * cos(b) - sin(a) * sin(b):
cos(2*phi) = cos(phi + phi) = cos(phi) * cos(phi) - sin(phi) * sin(phi)
2. Now you can see that you are multiplying cos(phi) by itself and sin(phi) by itself. So,
cos(phi) * cos(phi) - sin(phi) * sin(phi) = cos^2(phi) - sin^2(phi)

What's interesting about this trig identity is that you can use it to calculate cos(phi) in terms of cos or sin, by applying the Pythagorean identity. So, you have two other trig identities derived from this one that are very useful. Sal could have also used any of them to solve the problem:

A You only know the cosine of the angle:
cos(2 * phi) = 2 * cos^2(phi) - 1
B. You only know the sine of the angle (you can actually calculate cos(2*phi) by just knowing the sine:
cos(2 * phi) = 1 - 2 * sin^2(phi)

In contrast, the sin of a product is not nearly as exciting:
sin(2*phi) = 2 * sin(phi) * cos(phi)
• There are three different double angle formulas for cosine:
cos2x = cos^2 x - sin^2 x
= 2cos^2 x - 1
= 1 - 2 sin^2 x
Do we have to memorize all formulas or if not, which one can be used for all scenarios?
Thanks in advance! :)
• You don't have to memorize all formulas but it helps to do so.
If you remember,
``1 = cos^2 x + sin^2 x``

So we have, `cos^2 x = 1 - sin^2 x` and `sin^2 x = 1 - cos^2 x`

If we replace `cos^2 x` in the first double angle formula `cos2x = cos^2 x - sin^2 x` with `1 - sin^2 x` we get:

`cos2x = 1 - 2 sin^2 x`

Similarly, if we replace `sin^2 x` in the first double angle formula `cos2x = cos^2 x - sin^2 x` with `1 - cos^2 x` we get:

`cos2x = 2 cos^2 x - 1`

Hope this helps.
• is cos^2(θ) the same as cos(θ)^2? If not then shouldn't the identity be cos(θ)^2?
• cos²(θ) is another notation for [cos(θ)]².
• What does cos and sin mean? Did I miss a video?
• Wait, this gives another explanation for cos(0) = 1. cos(a-b) = cos(a)cos(b) + sin(a)sin(b); cos(a-a) = cos(a)cos(a)+sin(a)sin(a) = cos²(a) + sin²(a) or [cos(a)]² + [sin(a)]²; cos(a-a) = cos²(a) + sin²(a), cos(0) = cos²(a) + sin²(a). And cos²(θ) + sin²(θ) = 1. Therefore, cos(0) = 1!
• I understand how the double angles work but how do I do triple angles? One of our homework questions is, cos(3x)cos(2x)-sin(3x)sin(2x) and it wants us to express it as a single trigonometric ratio. How do I do this?
(1 vote)
• You can just use the sum and difference identity here. You know that
cos(α+β)=cos(α)cos(β)-sin(α)sin(β) which means it also works the other way
cos(α)cos(β)-sin(α)sin(β)=cos(α+β)
so... let α=3x and β=2x
cos(3x)cos(2x)-sin(3x)sin(2x)=cos(3x+2x)=cos(5x)