Mathematics

Trigonometry

Angle measurement, trigonometric ratios, fundamental identities, special angles, reference angles, and solving triangles using the Law of Sines and Law of Cosines. Includes sum, difference, double angle, half angle, and sum-to-product formulas.

2π  (rad)=360∘2\pi \; (rad) = 360^\circ
Radians to degreesDegrees to radians
α(∘)=180πα(rad)\displaystyle \alpha(^\circ) = \frac{180}{\pi} \alpha(rad)α(rad)=π180α(∘)\displaystyle \alpha(rad) = \frac{\pi}{180} \alpha(^\circ)

Arc length

L=R⋅α(rad)L=R⋅π⋅α180(degrees)L = R \cdot \alpha(rad) \qquad L = \frac{R \cdot \pi \cdot \alpha}{180}(degrees)
Right triangle with acute angle alpha
Functions
Sinesin⁡α=oppositehypotenuse=ac\sin \alpha = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{a}{c}
Tangenttan⁡α=oppositeadjacent=ab\tan \alpha = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{a}{b}
Secantsec⁡α=hypotenuseadjacent=cb\sec \alpha = \dfrac{\text{hypotenuse}}{\text{adjacent}} = \dfrac{c}{b}
Reciprocal Functions
Cosinecos⁡α=adjacenthypotenuse=bc\cos \alpha = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{b}{c}
Cotangentcot⁡α=adjacentopposite=ba\cot \alpha = \dfrac{\text{adjacent}}{\text{opposite}} = \dfrac{b}{a}
Cosecantcsc⁡α=hypotenuseopposite=ca\csc \alpha = \dfrac{\text{hypotenuse}}{\text{opposite}} = \dfrac{c}{a}
ReciprocalPythagorean & Quotient
sec⁡α=1cos⁡α\displaystyle \sec \alpha = \frac{1}{\cos \alpha}
csc⁡α=1sin⁡α\displaystyle \csc \alpha = \frac{1}{\sin \alpha}
cot⁡α=1tan⁡α\displaystyle \cot \alpha = \frac{1}{\tan \alpha}

tan⁡α=sin⁡αcos⁡α\displaystyle \tan \alpha = \frac{\sin \alpha}{\cos \alpha}

sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1

tan⁡2α+1=sec⁡2α\tan^2 \alpha + 1 = \sec^2 \alpha

1+cot⁡2α=csc⁡2α1 + \cot^2 \alpha = \csc^2 \alpha

0º30º45º60º90º
sen α\text{sen } \alpha012\displaystyle \frac{1}{2}22\displaystyle \frac{\sqrt{2}}{2}32\displaystyle \frac{\sqrt{3}}{2}1
cos⁡α\cos \alpha132\displaystyle \frac{\sqrt{3}}{2}22\displaystyle \frac{\sqrt{2}}{2}12\displaystyle \frac{1}{2}0
tan α\text{tan } \alpha033\displaystyle \frac{\sqrt{3}}{3}13\sqrt{3}∄\nexists
Complementary Angles (α+β=90∘\alpha+\beta=90^\circ)

sin⁡α=cos⁡(90∘−α)\sin \alpha = \cos(90^\circ - \alpha)
cos⁡α=sin⁡(90∘−α)\cos \alpha = \sin(90^\circ - \alpha)
tan⁡α=cot⁡(90∘−α)\tan \alpha = \cot(90^\circ - \alpha)
sec⁡α=csc⁡(90∘−α)\sec \alpha = \csc(90^\circ - \alpha)
Complementary angles unit circle
Supplementary Angles (α+β=180∘\alpha+\beta=180^\circ)

sin⁡(180∘−α)=sin⁡α\sin(180^\circ - \alpha) = \sin \alpha
cos⁡(180∘−α)=−cos⁡α\cos(180^\circ - \alpha) = -\cos \alpha
Supplementary angles unit circle
Opposite Angles (Even/Odd Identities)

sin⁡(−α)=−sin⁡α\sin(-\alpha) = -\sin \alpha
cos⁡(−α)=cos⁡α\cos(-\alpha) = \cos \alpha
Opposite angles unit circle
Angles differing by 180∘180^\circ

sin⁡(180∘+α)=−sin⁡α\sin(180^\circ + \alpha) = -\sin \alpha
cos⁡(180∘+α)=−cos⁡α\cos(180^\circ + \alpha) = -\cos \alpha
Angles differing by 180 degrees
Angles differing by 90∘90^\circ

sin⁡(90∘+α)=cos⁡α\sin(90^\circ + \alpha) = \cos \alpha
cos⁡(90∘+α)=−sin⁡α\cos(90^\circ + \alpha) = -\sin \alpha
Angles differing by 90 degrees
Any triangle
Law of Sines
asin⁡A=bsin⁡B=csin⁡C\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Law of Cosines
a2=b2+c2−2bc⋅cos⁡Ab2=a2+c2−2ac⋅cos⁡Bc2=a2+b2−2ab⋅cos⁡C\begin{aligned} a^2 &= b^2 + c^2 - 2bc \cdot \cos A \\ b^2 &= a^2 + c^2 - 2ac \cdot \cos B \\ c^2 &= a^2 + b^2 - 2ab \cdot \cos C \end{aligned}

It is a circle of radius 1 centered at the origin. The coordinates (x, y) of any point on the circle correspond to the cosine and sine, respectively, of the angle formed by the radius.
Full unit circle with angles in radians and degrees
Geometric interpretation of trigonometric ratios
Sign of sine and cosine in the 4 quadrants
sin⁡(α+β)=sin⁡α⋅cos⁡β+sin⁡β⋅cos⁡αcos⁡(α+β)=cos⁡α⋅cos⁡β−sin⁡α⋅sin⁡βtan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β\begin{aligned} \sin(\alpha + \beta) &= \sin \alpha \cdot \cos \beta + \sin \beta \cdot \cos \alpha \\ \cos(\alpha + \beta) &= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta \\ \tan(\alpha + \beta) &= \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta} \end{aligned}
sin⁡(α−β)=sin⁡α⋅cos⁡β−sin⁡β⋅cos⁡αcos⁡(α−β)=cos⁡α⋅cos⁡β+sin⁡α⋅sin⁡βtan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡α⋅tan⁡β\begin{aligned} \sin(\alpha - \beta) &= \sin \alpha \cdot \cos \beta - \sin \beta \cdot \cos \alpha \\ \cos(\alpha - \beta) &= \cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta \\ \tan(\alpha - \beta) &= \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \cdot \tan \beta} \end{aligned}
sin⁡(2α)=2sin⁡α⋅cos⁡αcos⁡(2α)=cos⁡2α−sin⁡2αtan⁡(2α)=2tan⁡α1−tan⁡2α\begin{aligned} \sin(2\alpha) &= 2\sin \alpha \cdot \cos \alpha \\ \cos(2\alpha) &= \cos^2 \alpha - \sin^2 \alpha \\ \tan(2\alpha) &= \frac{2\tan \alpha}{1 - \tan^2 \alpha} \end{aligned}
sin⁡(α2)=±1−cos⁡α2cos⁡(α2)=±1+cos⁡α2tan⁡(α2)=±1−cos⁡α1+cos⁡α\begin{aligned} \sin\left(\frac{\alpha}{2}\right) &= \pm\sqrt{\frac{1-\cos \alpha}{2}} \\ \cos\left(\frac{\alpha}{2}\right) &= \pm\sqrt{\frac{1+\cos \alpha}{2}} \\ \tan\left(\frac{\alpha}{2}\right) &= \pm\sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}} \end{aligned}
sin⁡A+sin⁡B=2sin⁡(A+B2)⋅cos⁡(A−B2)sin⁡A−sin⁡B=2cos⁡(A+B2)⋅sin⁡(A−B2)cos⁡A+cos⁡B=2cos⁡(A+B2)⋅cos⁡(A−B2)cos⁡A−cos⁡B=−2sin⁡(A+B2)⋅sin⁡(A−B2)\begin{aligned} \sin A + \sin B &= 2\sin\left(\frac{A+B}{2}\right) \cdot \cos\left(\frac{A-B}{2}\right) \\ \sin A - \sin B &= 2\cos\left(\frac{A+B}{2}\right) \cdot \sin\left(\frac{A-B}{2}\right) \\ \cos A + \cos B &= 2\cos\left(\frac{A+B}{2}\right) \cdot \cos\left(\frac{A-B}{2}\right) \\ \cos A - \cos B &= -2\sin\left(\frac{A+B}{2}\right) \cdot \sin\left(\frac{A-B}{2}\right) \end{aligned}