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### Course: Trigonometry>Unit 4

Lesson 5: Using trigonometric identities

# Trig identity reference

Look up AND understand all your favorite trig identities

## Reciprocal and quotient identities

$\mathrm{sec}\left(\theta \right)=\frac{1}{\mathrm{cos}\left(\theta \right)}$

$\mathrm{csc}\left(\theta \right)=\frac{1}{\mathrm{sin}\left(\theta \right)}$

$\mathrm{cot}\left(\theta \right)=\frac{1}{\mathrm{tan}\left(\theta \right)}$

$\mathrm{tan}\left(\theta \right)=\frac{\mathrm{sin}\left(\theta \right)}{\mathrm{cos}\left(\theta \right)}$

$\mathrm{cot}\left(\theta \right)=\frac{\mathrm{cos}\left(\theta \right)}{\mathrm{sin}\left(\theta \right)}$

## Pythagorean identities

${\mathrm{sin}}^{2}\left(\theta \right)+{\mathrm{cos}}^{2}\left(\theta \right)={1}^{2}$
${\mathrm{tan}}^{2}\left(\theta \right)+{1}^{2}={\mathrm{sec}}^{2}\left(\theta \right)$
${\mathrm{cot}}^{2}\left(\theta \right)+{1}^{2}={\mathrm{csc}}^{2}\left(\theta \right)$

## 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
$\begin{array}{rl}\mathrm{sin}\left(\theta +\varphi \right)& =\mathrm{sin}\theta \mathrm{cos}\varphi +\mathrm{cos}\theta \mathrm{sin}\varphi \\ \\ \mathrm{sin}\left(\theta -\varphi \right)& =\mathrm{sin}\theta \mathrm{cos}\varphi -\mathrm{cos}\theta \mathrm{sin}\varphi \\ \\ \mathrm{cos}\left(\theta +\varphi \right)& =\mathrm{cos}\theta \mathrm{cos}\varphi -\mathrm{sin}\theta \mathrm{sin}\varphi \\ \\ \mathrm{cos}\left(\theta -\varphi \right)& =\mathrm{cos}\theta \mathrm{cos}\varphi +\mathrm{sin}\theta \mathrm{sin}\varphi \end{array}$
$\begin{array}{rl}\mathrm{tan}\left(\theta +\varphi \right)& =\frac{\mathrm{tan}\theta +\mathrm{tan}\varphi }{1-\mathrm{tan}\theta \mathrm{tan}\varphi }\\ \\ \mathrm{tan}\left(\theta -\varphi \right)& =\frac{\mathrm{tan}\theta -\mathrm{tan}\varphi }{1+\mathrm{tan}\theta \mathrm{tan}\varphi }\end{array}$
Double angle identities
$\mathrm{sin}\left(2\theta \right)=2\mathrm{sin}\theta \mathrm{cos}\theta$
$\mathrm{cos}\left(2\theta \right)=2{\mathrm{cos}}^{2}\theta -1$
$\mathrm{tan}\left(2\theta \right)=\frac{2\mathrm{tan}\theta }{1-{\mathrm{tan}}^{2}\theta }$
Half angle identities
$\begin{array}{rl}\mathrm{sin}\frac{\theta }{2}& =±\sqrt{\frac{1-\mathrm{cos}\theta }{2}}\\ \\ \mathrm{cos}\frac{\theta }{2}& =±\sqrt{\frac{1+\mathrm{cos}\theta }{2}}\\ \\ \mathrm{tan}\frac{\theta }{2}& =±\sqrt{\frac{1-\mathrm{cos}\theta }{1+\mathrm{cos}\theta }}\\ \\ & =\frac{1-\mathrm{cos}\theta }{\mathrm{sin}\theta }\\ \\ & =\frac{\mathrm{sin}\theta }{1+\mathrm{cos}\theta }\end{array}$

## Symmetry and periodicity identities

$\mathrm{sin}\left(-\theta \right)=-\mathrm{sin}\left(\theta \right)$
$\mathrm{cos}\left(-\theta \right)=+\mathrm{cos}\left(\theta \right)$
$\mathrm{tan}\left(-\theta \right)=-\mathrm{tan}\left(\theta \right)$
$\begin{array}{rl}\mathrm{sin}\left(\theta +2\pi \right)& =\mathrm{sin}\left(\theta \right)\\ \\ \mathrm{cos}\left(\theta +2\pi \right)& =\mathrm{cos}\left(\theta \right)\\ \\ \mathrm{tan}\left(\theta +\pi \right)& =\mathrm{tan}\left(\theta \right)\end{array}$

## Cofunction identities

$\begin{array}{rl}\mathrm{sin}\theta & =\mathrm{cos}\left(\frac{\pi }{2}-\theta \right)\\ \\ \mathrm{cos}\theta & =\mathrm{sin}\left(\frac{\pi }{2}-\theta \right)\\ \\ \mathrm{tan}\theta & =\mathrm{cot}\left(\frac{\pi }{2}-\theta \right)\\ \\ \mathrm{cot}\theta & =\mathrm{tan}\left(\frac{\pi }{2}-\theta \right)\\ \\ \mathrm{sec}\theta & =\mathrm{csc}\left(\frac{\pi }{2}-\theta \right)\\ \\ \mathrm{csc}\theta & =\mathrm{sec}\left(\frac{\pi }{2}-\theta \right)\end{array}$

## 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
• People don't invent mathematics, they discover it.
• 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?
• 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.
• 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!
• While they tend place the articles last, I've learned it's always best to read them first! (Or at least before the exercises.)
• How do you prove the half-angle identities?
• 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.
• 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?
• 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.
• In the Angle sum explanation diagram, why is the bottom triangle's adjacent side to angle theta (cos theta)(cos phi)?
• 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.
• I still don't know how to get the half angle identities.Who can help me?
• 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.
• 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?
• 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)
• Do you need to remember all these identities or at least know how to derive them using the unit circle.
• 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.
• 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.
• (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)