Look up AND understand all your favorite trig identities.

Reciprocal and quotient identities

sec(θ)=1cos(θ)\sec(\theta)= \dfrac{1}{\cos(\theta)}
Play with the point on the unit circle to see how cosine and secant change together. Notice how when cosine is small, secant is big, and vice versa. It turns out they always multiply to exactly 11.
We can also see this identity using similar triangles. Slide the dot below the graph to see one triangle transform into the other. Watch carefully to see which segments correspond to each other.

csc(θ)=1sin(θ)\csc(\theta)= \dfrac{1}{\sin(\theta)}
Play with the point on the unit circle to see how sine and cosecant change together. Notice how when sine is small, cosecant is big, and vice versa. It turns out they always multiply to exactly 11.
We can also see this identity using similar triangles. Slide the dot below the graph to see one triangle transform into the other. Watch carefully to see which segments correspond to each other.

cot(θ)=1tan(θ)\cot(\theta)= \dfrac{1}{\tan(\theta)}
Play with the point on the unit circle to see how tangent and cotangent change together. Notice how when tangent is small, cotangent is big, and vice versa. It turns out they always multiply to exactly 11.
We can also see this identity using similar triangles. Slide the dot below the graph to see one triangle transform into the other. Watch carefully to see which segments correspond to each other.

tan(θ)=sin(θ)cos(θ)\tan(\theta)= \dfrac{\sin(\theta)}{\cos(\theta)}
We can see this identity using similar triangles. Slide the dot below the graph to see one triangle transform into the other. Watch carefully to see which segments correspond to each other.

cot(θ)=cos(θ)sin(θ)\cot(\theta)= \dfrac{\cos(\theta)}{\sin(\theta)}
We can see this identity using similar triangles. Slide the dot below the graph to see one triangle transform into the other. Watch carefully to see which segments correspond to each other.

Pythagorean identities

sin2(θ)+cos2(θ)=12\sin^2(\theta) + \cos^2(\theta)=1^2
This identity comes from writing down the Pythagorean theorem for the triangle below.
tan2(θ)+12=sec2(θ)\tan^2(\theta) + 1^2=\sec^2(\theta)
This identity comes from writing down the Pythagorean theorem for the triangle below.
cot2(θ)+12=csc2(θ)\cot^2(\theta) + 1^2=\csc^2(\theta)
This identity comes from writing down the Pythagorean theorem for the triangle below.

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+cosθsinϕ\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi
sin(θϕ)=sinθcosϕcosθsinϕ\sin(\theta-\phi)=\sin\theta\cos\phi-\cos\theta\sin\phi
cos(θ+ϕ)=cosθcosϕsinθsinϕ\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi
cos(θϕ)=cosθcosϕ+sinθsinϕ\cos(\theta-\phi)=\cos\theta\cos\phi+\sin\theta\sin\phi
The figure below shows a way of proving the angle sum identities. The left and right sides of the rectangle are equal, giving us:
sin(θ+ϕ)=sinθcosϕ+cosθsinϕ\purple{\sin(\theta+\phi)}=\blue{\sin\theta}\red{\cos\phi}+\blue{\cos\theta}\red{\sin\phi}
The top and bottom sides are also equal, giving us:
cos(θ+ϕ)=cosθcosϕsinθsinϕ\purple{\cos(\theta+\phi)}=\blue{\cos\theta}\red{\cos\phi}-\blue{\sin\theta}\red{\sin\phi}
It's easiest to make sense of this diagram if you start with the right triangle in the middle of the diagram and build outward using the right triangle definitions of the trig functions (SOH CAH TOA).
A similar diagram could show the angle difference identities. Can you draw one?
(Technically, this isn't a proof for all possible values of θ\theta and ϕ\phi, but the identities do apply for all values.)
tan(θ+ϕ)=tanθ+tanϕ1tanθtanϕ\tan(\theta+\phi)=\dfrac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}
tan(θϕ)=tanθtanϕ1+tanθtanϕ\tan(\theta-\phi)=\dfrac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi}
The figure below shows a way of proving the angle sum identity for tangent.
This one is a little bit tricky to make sense of. It's all about building up to the triangle in the upper-left part of the diagram. When you apply the right triangle definition of tangent to that triangle, you get:
It's easiest to make sense of this diagram if you start with the segment at the bottom of the diagram and build upward using the right triangle definitions of the trig functions (SOH CAH TOA).
A similar diagram could show the angle difference identity. Can you draw one?
(Technically, this isn't a proof for all possible values of θ\theta and ϕ\phi, but the identities do apply for all values.)
Double angle identities
sin(2θ)=2sinθcosθ\sin(2\theta)=2\sin\theta\cos\theta
We can get this identity if we take the angle sum identity for sine, but make both angles the same.
sin(θ+ϕ)=sinθcosϕ+cosθsinϕ\sin(\blue\theta+\red\phi)=\sin\blue\theta\cos\red\phi+\cos\blue\theta\sin\red\phi
sin(θ+θ)=sinθcosθ+cosθsinθ\sin(\blue\theta+\blue\theta)=\sin\blue\theta\cos\blue\theta+\cos\blue\theta\sin\blue\theta
sin(2θ)=2sinθcosθ\sin(2\blue\theta)=2\sin\blue\theta\cos\blue\theta
cos(2θ)=2cos2θ1\cos(2\theta)=2\cos^2\theta-1
We can also get this one from the angle sum identity, but we need to do a little extra manipulation.
First, let's make both angles the same.
cos(θ+ϕ)=cosθcosϕsinθsinϕ\cos(\blue\theta+\red\phi)=\cos\blue\theta\cos\red\phi-\sin\blue\theta\sin\red\phi
cos(θ+θ)=cosθcosθsinθsinθ\cos(\blue\theta+\blue\theta)=\cos\blue\theta\cos\blue\theta-\sin\blue\theta\sin\blue\theta
cos(2θ)=cos2θsin2θ\cos(2\blue\theta)=\cos^2\blue\theta-\sin^2\blue\theta
Now, we can use the identity sin2θ+cos2θ=1\sin^2\theta +\cos^2\theta= 1 to put the right-hand side in terms of just cosines.
cos(2θ)=cos2θ(1cos2θ)\cos(2\blue\theta)=\cos^2\blue\theta-(1-\cos^2\blue\theta)
cos(2θ)=cos2θ1+cos2θ\cos(2\blue\theta)=\cos^2\blue\theta-1+\cos^2\blue\theta
cos(2θ)=2cos2θ1\cos(2\blue\theta)=2\cos^2\blue\theta-1
tan(2θ)=2tanθ1tan2θ\tan(2\theta)=\dfrac{2\tan\theta}{1-\tan^2\theta}
This one also comes from the angle sum identity.
tan(θ+ϕ)=tanθ+tanϕ1tanθtanϕ\tan(\blue\theta+\red\phi)=\dfrac{\tan\blue\theta+\tan\red\phi}{1-\tan\blue\theta\tan\red\phi}
tan(θ+θ)=tanθ+tanθ1tanθtanθ\tan(\blue\theta+\blue\theta)=\dfrac{\tan\blue\theta+\tan\blue\theta}{1-\tan\blue\theta\tan\blue\theta}
tan(2θ)=2tanθ1tan2θ\tan(2\blue\theta)=\dfrac{2\tan\blue\theta}{1-\tan^2\blue\theta}
Half angle identities
sinθ2=±1cosθ2\sin\dfrac\theta2=\pm\sqrt{\dfrac{1-\cos\theta}{2}}
cosθ2=±1+cosθ2\cos\dfrac\theta2=\pm\sqrt{\dfrac{1+\cos\theta}{2}}
tanθ2=±1cosθ1+cosθ=       1cosθsinθ=       sinθ1+cosθ\begin{aligned} \tan\dfrac{\theta}{2}&=\pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}\\ \\ &=~~~~~~~\dfrac{1-\cos\theta}{\sin\theta}\\ \\ &=~~~~~~~\dfrac{\sin\theta}{1+\cos\theta}\end{aligned}
We can get all of these by taking the double angle formulas and making the substitution:
θθ2\blue\theta\rightarrow\red{\dfrac{\theta}{2}}
Then there's some rearranging to do.
For example:
sin(2θ)=2sinθcosθ\sin(2\blue\theta)=2\sin\blue\theta\cos\blue\theta
turns into...
sin(2θ2)=2sinθ2cosθ2\sin(\cancel2\red{\dfrac{\theta}{\cancel2}})=2\sin\red{\dfrac{\theta}{2}}\cos\red{\dfrac{\theta}{2}}
And then we need to solve for sinθ2\sin\red{\dfrac{\theta}{2}}. Give it a shot, and see if you can figure out how to show the other identities.

Symmetry and periodicity identities

sin(θ)=sin(θ)\sin(-\theta)=-\sin(\theta)
We can see this by looking at a unit circle diagram.
cos(θ)=+cos(θ)\cos(-\theta)=+\cos(\theta)
We can see this by looking at a unit circle diagram.
tan(θ)=tan(θ)\tan(-\theta)=-\tan(\theta)
This one is harder to see on a unit circle diagram, but we can get it by writing tangent in terms of sine and cosine, then applying the sine and cosine identities for negative angles.
tan(θ)=sin(θ)cos(θ)\tan(\theta)= \dfrac{\sin(\theta)}{\cos(\theta)}
So,
tan(θ)=sin(θ)cos(θ)=sin(θ)cos(θ)=tan(θ)\tan(-\theta)= \dfrac{\sin(-\theta)}{\cos(-\theta)}= \dfrac{-\sin(\theta)}{\cos(\theta)}=-\tan(\theta)

sin(θ+2π)=sin(θ)\sin(\theta+2\pi)=\sin(\theta)
cos(θ+2π)=cos(θ)\cos(\theta+2\pi)=\cos(\theta)
tan(θ+π)=tan(θ)\tan(\theta+\pi)=\tan(\theta)

Cofunction identities

sinθ=cos(π2θ)\sin\theta= \cos(\dfrac{\pi}{2}-\theta)
cosθ=sin(π2θ)\cos\theta= \sin(\dfrac{\pi}{2}-\theta)
tanθ=cot(π2θ)\tan\theta= \cot(\dfrac{\pi}{2}-\theta)
cotθ=tan(π2θ)cot\theta= \tan(\dfrac{\pi}{2}-\theta)
secθ=csc(π2θ)\sec\theta= \csc(\dfrac{\pi}{2}-\theta)
cscθ=sec(π2θ)\csc\theta= \sec(\dfrac{\pi}{2}-\theta)
All these identities look the same!
The key to understanding these identities is realizing that going from θ\theta to π2θ\dfrac \pi 2 -\theta is equivalent to reflecting your angle over the line y=xy=x.
The interactive diagram below attempts to show how the angles relate to each other and that sinθ=cos(π2θ)\red{\sin\theta}= \blue{\cos(\dfrac{\pi}{2}-\theta)}. The other co-function pairs work the same way, leading to the other 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.
Loading