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### Course: Precalculus > Unit 2

Lesson 10: Using trigonometric identities- Finding trig values using angle addition identities
- Using the tangent angle addition identity
- Find trig values using angle addition identities
- Using trig angle addition identities: finding side lengths
- Using trig angle addition identities: manipulating expressions
- Using trigonometric identities
- Trig identity reference
- Trigonometry: FAQ

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# Trig identity reference

Look up AND understand all your favorite trig identities

## Reciprocal and quotient identities

## Pythagorean identities

## 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**

**Double angle identities**

**Half angle identities**

## Symmetry and periodicity identities

## Cofunction identities

## 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.

## Want to join the conversation?

- why do ppl invent triangles and calculus T.T(129 votes)
- People don't invent mathematics, they discover it.(79 votes)

- 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?(25 votes)- 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.(36 votes)

- How do you prove the half-angle identities?(26 votes)
- 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.(21 votes)

- 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!(29 votes)
- While they tend place the articles last, I've learned it's always best to read them first! (Or at least before the exercises.)(15 votes)

- 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)- 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.(41 votes)

- In the Angle sum explanation diagram, why is the bottom triangle's adjacent side to angle theta (cos theta)(cos phi)?(11 votes)
- 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)

- I still don't know how to get the half angle identities.Who can help me?(6 votes)
- 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.(19 votes)

- 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)- The correct identity is:

cos(−θ)=+cos(θ)

Your teacher and the book probably mixed up sine and cosine. The sine identity is:

sin(−θ)=-sin(θ)(16 votes)

- Do you need to remember all these identities or at least know how to derive them using the unit circle.(4 votes)
- 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.(12 votes)

- 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)
- (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)