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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.

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  • leaf yellow style avatar for user Steven *
    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))
    (29 votes)
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    • mr pants teal style avatar for user ObiWanPEZ
      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).
      (31 votes)
  • blobby green style avatar for user qudrat mommandi
    In which video does he talk about why cos2x = cos^2 x = sin^2 x?
    (16 votes)
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  • hopper cool style avatar for user 🤔 ᴄᴏᴅᴇᴅ ɢᴇɴɪᴜȿ 😎
    How are you supposed to do: cos²(θ)? Is it the same as cos(θ)²?
    (4 votes)
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  • leafers ultimate style avatar for user Abhishek
    Please tell me the video cos(2*theta)=cos^2(theta)-sin^2(theta)
    (4 votes)
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    • leaf blue style avatar for user Aperneto
      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)
      (5 votes)
  • blobby green style avatar for user maltesedog246
    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! :)
    (3 votes)
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    • leaf green style avatar for user bhvima
      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.
      (4 votes)
  • old spice man blue style avatar for user Timmy
    is cos^2(θ) the same as cos(θ)^2? If not then shouldn't the identity be cos(θ)^2?
    (2 votes)
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  • leaf grey style avatar for user J005311
    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!
    (3 votes)
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  • piceratops ultimate style avatar for user Travis Petersen
    Are there any particular applications for sin(2a)=2sin(a)cos(a)? Is this less significant than cos(2a)=cos^2(a)-sin^2(a)?
    (1 vote)
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    • purple pi purple style avatar for user doctorfoxphd
      There are many applications from engineering, architecture and projects that involve design-to-specifications. The struts under a highway need to retrofitted to be a certain length, and currently the angle between the struts is such and such. To meet the requirement for stability under a certain earthquake, the angle cannot be greater than such and such, but to reach the other side of the road bed, the supports need to be lengthened by how much? The angle between two cross members is some measure, and in order to meet specs, it has to be increased, so how much longer do the rods need to be in order to create that angle? Even design of a chicken coop might be simplified if you can use calculations like these.
      (3 votes)
  • blobby green style avatar for user REiDtardedNo.13
    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)
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  • aqualine tree style avatar for user Wardiyah
    So I'm having problems with a question I got on a worksheet about Multiple-Angle Identities. I need to solve each equation for 0 ≤ theta ≤ 2pi, but I'm not sure about this one: -sin2θ = (√2)sinθ - 2sin2θ
    Not sure if I'm going about this the right way but I substituted a double angle identity for sin2θ to get:
    -(2sinθcosθ)= (√2)sinθ - 2(2sinθcosθ)
    then I moved everything on one side to get:
    0 = (√2)sinθ -(2sinθcosθ)
    to get rid of the radical, I squared the right side but I'm not sure if it should look like this:
    2sin^2θ - 4sin^2θcos^2θ
    or:
    2sinθ^2 - 4sinθ^2cosθ^2
    or even:
    2sin^2θ^2 - 4sin^2θ^2cos^2θ^2
    Which is the right one? and am I doing this right so far?
    Thanks.
    (1 vote)
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    • leaf green style avatar for user kubleeka
      None of your suggestions are correct, and moreover, squaring both sides like this will not get rid of the radical.

      Remember that if we square a quantity a+b, the result is (a+b)²
      =(a+b)(a+b)
      =a²+ab+ba+b²
      =a²+2ab+b².

      So if we square √2(sinθ)-2sinθcosθ, we get
      2sin²θ-4√2sin²θcosθ+4sin²θcos²θ, which still has a √2 in it.

      Instead of squaring, try factoring the expression on the right-hand side. You have a common factor of sinθ. So you can rewrite your equation as
      0=(sinθ)(√2-2cosθ)

      Since you have two things whose product is 0, you'll get solutions when either sinθ=0 or when √2-2cosθ=0. What does θ have to be for that to happen?
      (3 votes)

Video transcript

We have triangle ABC here, which looks like a right triangle. And we know it's a right triangle because 3 squared plus 4 squared is equal to 5 squared. And they want us to figure out what cosine of 2 times angle ABC is. So that's this angle-- ABC. Well, we can't immediately evaluate that, but we do know what the cosine of angle ABC is. We know that the cosine of angle ABC-- well, cosine is just adjacent over hypotenuse. It's going to be equal to 3/5. And similarly, we know what the sine of angle ABC is. That's opposite over hypotenuse. That is 4/5. So if we could break this down into just cosines of ABC and sines of ABC, then we'll be able to evaluate it. And lucky for us, we have a trig identity at our disposal that does exactly that. We know that the cosine of 2 times an angle is equal to cosine of that angle squared minus sine of that angle squared. And we've proved this in other videos, but this becomes very helpful for us here. Because now we know that the cosine-- Let me do this in a different color. Now, we know that the cosine of angle ABC is going to be equal to-- oh, sorry. It's the cosine of 2 times the angle ABC. That's what we care about. 2 times the angle ABC is going to be equal to the cosine of angle ABC squared minus sine of the angle ABC squared. And we know what these things are. This thing right over here is just going to be equal to 3/5 squared. Cosine of angle a ABC is 3/5. So we're going to square it. And this right over here is just 4/5 squared. So it's minus 4/5 squared. And so this simplifies to 9/25 minus 16/25, which is equal to 7/25. Sorry. It's negative. Got to be careful there. 16 is larger than 9. Negative 7/25. Now, one thing that might jump at you is, why did I get a negative value here when I doubled the angle here? Because the cosine was clearly a positive number. And there you just have to think of the unit circle-- which we already know the unit circle definition of trig functions is an extension of the Sohcahtoa definition. X-axis. Y-axis. Let me draw a unit circle here. My best attempt. So that's our unit circle. So this angle right over here looks like something like this. And so you see its x-coordinate-- which is the cosine of that angle-- looks positive. But then, if you were to double this angle, it would take you out someplace like this. And then, you see-- by the unit circle definition-- the x-coordinate, we are now sitting in the second quadrant. And the x-coordinate can be negative. And that's essentially what happened in this problem.