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Trig identity reference

Look up AND understand all your favorite trig identities

Reciprocal and quotient identities

sec(θ)=1cos(θ)

csc(θ)=1sin(θ)

cot(θ)=1tan(θ)

tan(θ)=sin(θ)cos(θ)

cot(θ)=cos(θ)sin(θ)

Pythagorean identities

sin2(θ)+cos2(θ)=12
tan2(θ)+12=sec2(θ)
cot2(θ)+12=csc2(θ)

Identities that come from sums, differences, multiples, and fractions of angles

These are all closely related, but let's go over each kind.
Angle sum and difference identities
sin(θ+ϕ)=sinθcosϕ+cosθsinϕsin(θϕ)=sinθcosϕcosθsinϕcos(θ+ϕ)=cosθcosϕsinθsinϕcos(θϕ)=cosθcosϕ+sinθsinϕ
tan(θ+ϕ)=tanθ+tanϕ1tanθtanϕtan(θϕ)=tanθtanϕ1+tanθtanϕ
Double angle identities
sin(2θ)=2sinθcosθ
cos(2θ)=2cos2θ1
tan(2θ)=2tanθ1tan2θ
Half angle identities
sinθ2=±1cosθ2cosθ2=±1+cosθ2tanθ2=±1cosθ1+cosθ=1cosθsinθ=sinθ1+cosθ

Symmetry and periodicity identities

sin(θ)=sin(θ)
cos(θ)=+cos(θ)
tan(θ)=tan(θ)
sin(θ+2π)=sin(θ)cos(θ+2π)=cos(θ)tan(θ+π)=tan(θ)

Cofunction identities

sinθ=cos(π2θ)cosθ=sin(π2θ)tanθ=cot(π2θ)cotθ=tan(π2θ)secθ=csc(π2θ)cscθ=sec(π2θ)

Appendix: All trig ratios in the unit circle

Use the movable point to see how the lengths of the ratios change according to the angle.

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  • leaf green style avatar for user Jesse Chien
    why do ppl invent triangles and calculus T.T
    (131 votes)
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  • blobby green style avatar for user pwu25
    okay this article is great... but i really wish i Had seen it before some of the exercises that came before it. i had to puzzle a lot of those out and it took me longer than it would have had i seen this article. it seems (at least to me) that its a little out of place. vote if you agree!
    (44 votes)
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  • mr pink red style avatar for user Vivian Hir
    I am kind of struggling on solving sinusoidal equations (advanced) since I don't do all the identities. I don't check all of the solutions.
    Here is some that I know:
    sin(θ)=(θ+360)
    sin(θ)=pi-θ
    sin(θ)=θ+2pi
    cos(θ)=2pi-θ
    cos(θ)=θ+2pi
    is there any others missing? am I doing anything wrong?
    (24 votes)
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    • blobby green style avatar for user Ross Henderson
      First of all, we should probably make the notation a bit more rigorous, because the way you've phrased it isn't quite correct. Instead, write:
      sin(θ)=sin(θ+360)=sin(θ+2pi)
      sin(θ)=sin(pi-θ)
      sin(θ)=sin(θ+2pi) see above
      cos(θ)=cos(2pi-θ)
      cos(θ)=cos(θ+2pi)
      ... and yes, there are lots of others - technically, an infinite number of others since sin and cos are periodic and repeat every 2pi, positive or negative. So, for example, sin(θ)=sin(θ+2npi), where n is any integer.

      I'd probably add to the list:
      sin(-θ)=-sin(θ)
      cos(-θ)=cos(θ),
      if we're sticking to sin(a)=sin(b) and cos(a)=cos(b) style identities.
      (35 votes)
  • blobby green style avatar for user liam obaid
    How do you prove the half-angle identities?
    (25 votes)
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    • leaf blue style avatar for user Matthew Daly
      The easiest way is to see that cos 2φ = cos²φ - sin²φ = 2 cos²φ - 1 or 1 - 2sin²φ by the cosine double angle formula and the Pythagorean identity. Now substitute 2φ = θ into those last two equations and solve for sin θ/2 and cos θ/2. Then the tangent identity just follow from those two and the quotient identity for tangent.
      (20 votes)
  • starky ultimate style avatar for user Jesse
    I have a table of trig identities in my Calculus textbook that has the double cosine identity as:
    cos 2x = cos^2 x - sin^2 x
    Makes sense, because that's the way you would get it if you applied the rule of adding 2 different angles. How do you get from there to what they have here:
    cos 2x = 2cos^2 x - 1?
    (3 votes)
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    • mr pink red style avatar for user andrewp18
      The Pythagorean identity states:
      sin²𝑥 + cos²𝑥 = 1
      This means:
      sin²𝑥 = 1 - cos²𝑥
      The standard double cosine identity is:
      cos 2𝑥 = cos²𝑥 - sin²𝑥
      Substituting for sin²𝑥:
      cos 2𝑥 = cos²𝑥 - (1 - cos²𝑥)
      cos 2𝑥 = 2cos²𝑥 - 1
      Comment if you have questions.
      (43 votes)
  • mr pants teal style avatar for user Rodrigo De los Santos
    In the Angle sum explanation diagram, why is the bottom triangle's adjacent side to angle theta (cos theta)(cos phi)?
    (11 votes)
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    • leaf blue style avatar for user Matthew Daly
      The bottom triangle is a right triangle with hypotenuse length h = cos phi. So if x were your unknown side, doing normal trig on it gives cos theta = x/h = x / (cos phi), or in other words x = (cos theta)(cos phi). All of the sides in that diagram are defined in the same way, relative to the one side that was defined to be of length 1.
      (10 votes)
  • aqualine ultimate style avatar for user John He
    I still don't know how to get the half angle identities.Who can help me?
    (6 votes)
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    • leaf yellow style avatar for user Howard Bradley
      I won't do all three, but you can get the idea from the cosine case.
      I assume you're comfortable with the double angle formula: cos(2θ) = cos²θ - sin²θ

      Make the substitution φ = 2θ or θ = φ/2

      cos(φ) = cos²(φ/2) - sin²(φ/2)
      Using the Pythagorean identity we get
      cos(φ) = cos²(φ/2) - (1 - cos²(φ/2))
      = 2cos²(φ/2) -1

      Solving for cos(φ/2) gives
      cos²(φ/2) = (cos(φ) + 1)/2
      or
      cos(φ/2) = ±√((cos(φ) + 1)/2)
      Which is the result we wanted.

      Now once you have that, you can get the sine case by substituting for sin(φ/2) in terms of cosines

      ie √(1 - sin²(φ/2)) = √((cos(φ) + 1)/2)

      or (1 - sin²(φ/2)) = (cos(φ) + 1)/2
      or sin²(φ/2) = 1 - (cos(φ) + 1)/2

      or sin²(φ/2) = 2/2 - (cos(φ) + 1)/2
      = (1 - cos(φ))/2

      Hence sin(φ/2) = ±√((1 - cos(φ))/2)
      Note that the half angle formula for sine gives a result in terms of cosine.

      I'll leave the case of tan(φ/2) to you.
      (20 votes)
  • blobby green style avatar for user redmaroon16
    My teacher, as well as textbook, say that the cosine symmetry identity is cos(−θ)=-cos(θ)
    but this says that it's cos(−θ)=+cos(θ) which one is correct?
    (3 votes)
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  • duskpin tree style avatar for user Moe
    Do you need to remember all these identities or at least know how to derive them using the unit circle.
    (4 votes)
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    • mr pink red style avatar for user David Lee
      It's okay just know how to derive them using the unit circle but if you remember all of them, it'll be faster when you solve questions. And I recommend you to remember it because when you are taking a test, you don't have time to derive using the unit circle.
      + It's not that hard to remember though. There is a pattern. And once you proved why it is then it'll be way easier to memorize it.
      (11 votes)
  • primosaur seedling style avatar for user Thaao Hanshew
    This isn't exactly related to this, but I don't know where else to ask. I was doing problems related to this on KA, and needed to find the tangent of 15 degrees. I used tan(45-30) in order to find it, which gave me (3-sqrt3)/(3+sqrt3). In the KA "hints," they used tan(60-45) and got (sqrt(3)-1)/(sqrt(3)+1) ... These seem to be two ways of expressing the same value, as putting both into a calculator returns the same result. But for the life of me, I cannot seem to algebraically manipulate my answer to get KA's answer. If I start with tan(60-45), I get that form easily, but how can I prove (3-sqrt3)/(3+sqrt3) equals (sqrt(3)-1)/(sqrt(3)+1) ? I want to be able to more easily choose the right answer in the future, without having to evaluate all of the multiple choices with a calculator and compare them to the evaluation of my own expression.
    (6 votes)
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    • leaf green style avatar for user kubleeka
      (3-√3)/(3+√3)
      Multiply numerator and denominator by 3-√3 and multiply out to get:
      (12-6√3)/6
      Cancel the 6s to get:
      (2-√3)/1

      Multiply and divide by √3 +1 to get:
      (2-√3)*(√3+1)/(√3+1) =(2√3+2-3-√3)/(√3+1) =(√3-1)/(√3+1)
      (7 votes)