Trigonify

Trigonify reference

Trigonometry identities & formulas

All 45 identities and formulas across 16 topics, each with its KaTeX-rendered notation and its own reference page. Every page here is free to read — no account needed.

Basics · 4

SOHCAHTOA

sin⁡θ=opphypcos⁡θ=adjhyptan⁡θ=oppadj\sin\theta = \dfrac{\text{opp}}{\text{hyp}}\quad \cos\theta = \dfrac{\text{adj}}{\text{hyp}}\quad \tan\theta = \dfrac{\text{opp}}{\text{adj}}

Mnemonic for the three core ratios in a right triangle.

Degrees ↔ Radians

1∘=π180 rad1 rad=180π ⁣ ⁣∘1^{\circ} = \dfrac{\pi}{180}\text{ rad}\qquad 1\text{ rad} = \dfrac{180}{\pi}^{\!\!\circ}

Arc length

s=rθs = r\theta

θ must be in radians.

Sector area

A=12r2θA = \tfrac{1}{2} r^{2} \theta

θ in radians.

Pythagorean · 3

Secant form

1+tan⁡2θ=sec⁡2θ1 + \tan^{2}\theta = \sec^{2}\theta

Cosecant form

1+cot⁡2θ=csc⁡2θ1 + \cot^{2}\theta = \csc^{2}\theta

Reciprocal · 3

Secant

sec⁡θ=1cos⁡θ\sec\theta = \dfrac{1}{\cos\theta}

Cosecant

csc⁡θ=1sin⁡θ\csc\theta = \dfrac{1}{\sin\theta}

Cotangent

cot⁡θ=1tan⁡θ\cot\theta = \dfrac{1}{\tan\theta}

Quotient · 2

Tangent

tan⁡θ=sin⁡θcos⁡θ\tan\theta = \dfrac{\sin\theta}{\cos\theta}

Cotangent

cot⁡θ=cos⁡θsin⁡θ\cot\theta = \dfrac{\cos\theta}{\sin\theta}

Even/Odd · 3

Sine is odd

sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta

Cosine is even

cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta

Tangent is odd

tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta

Cofunction · 3

sin/cos

sin⁡(π2−θ)=cos⁡θ\sin(\tfrac{\pi}{2} - \theta) = \cos\theta

cos/sin

cos⁡(π2−θ)=sin⁡θ\cos(\tfrac{\pi}{2} - \theta) = \sin\theta

tan/cot

tan⁡(π2−θ)=cot⁡θ\tan(\tfrac{\pi}{2} - \theta) = \cot\theta

Law of Sines · 2

Law of Sines

asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}

Circumradius form

asin⁡A=2R\dfrac{a}{\sin A} = 2R

R = circumradius of the triangle.

Law of Cosines · 2

For side a

a2=b2+c2−2bccos⁡Aa^{2} = b^{2} + c^{2} - 2bc \cos A

For angle A

cos⁡A=b2+c2−a22bc\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}

Area · 2

SAS area

Area=12absin⁡C\text{Area} = \tfrac{1}{2} a b \sin C

Heron's formula

Area=s(s−a)(s−b)(s−c),  s=a+b+c2\text{Area} = \sqrt{s(s-a)(s-b)(s-c)},\ \ s = \tfrac{a+b+c}{2}

Inverse · 3

arcsin range

arcsin⁡:[−1,1]→[−π2,π2]\arcsin : [-1, 1] \to [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]

arccos range

arccos⁡:[−1,1]→[0,π]\arccos : [-1, 1] \to [0, \pi]

arctan range

arctan⁡:R→(−π2,π2)\arctan : \mathbb{R} \to (-\tfrac{\pi}{2}, \tfrac{\pi}{2})

Sum & Difference · 3

Sine

sin⁡(α±β)=sin⁡αcos⁡β±cos⁡αsin⁡β\sin(\alpha \pm \beta) = \sin\alpha \cos\beta \pm \cos\alpha \sin\beta

Cosine

cos⁡(α±β)=cos⁡αcos⁡β∓sin⁡αsin⁡β\cos(\alpha \pm \beta) = \cos\alpha \cos\beta \mp \sin\alpha \sin\beta

Tangent

tan⁡(α±β)=tan⁡α±tan⁡β1∓tan⁡αtan⁡β\tan(\alpha \pm \beta) = \dfrac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha \tan\beta}

Double Angle · 3

sin 2θ

sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2 \sin\theta \cos\theta

cos 2θ

cos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos 2\theta = \cos^{2}\theta - \sin^{2}\theta = 2\cos^{2}\theta - 1 = 1 - 2\sin^{2}\theta

tan 2θ

tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^{2}\theta}

Half Angle · 3

sin θ/2

sin⁡θ2=±1−cos⁡θ2\sin\tfrac{\theta}{2} = \pm\sqrt{\dfrac{1 - \cos\theta}{2}}

cos θ/2

cos⁡θ2=±1+cos⁡θ2\cos\tfrac{\theta}{2} = \pm\sqrt{\dfrac{1 + \cos\theta}{2}}

tan θ/2

tan⁡θ2=1−cos⁡θsin⁡θ=sin⁡θ1+cos⁡θ\tan\tfrac{\theta}{2} = \dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1 + \cos\theta}

Power Reduction · 2

sin² θ

sin⁡2θ=1−cos⁡2θ2\sin^{2}\theta = \dfrac{1 - \cos 2\theta}{2}

cos² θ

cos⁡2θ=1+cos⁡2θ2\cos^{2}\theta = \dfrac{1 + \cos 2\theta}{2}

Product to Sum · 3

sin · cos

sin⁡αcos⁡β=12[sin⁡(α+β)+sin⁡(α−β)]\sin\alpha \cos\beta = \tfrac{1}{2}[\sin(\alpha+\beta) + \sin(\alpha-\beta)]

cos · cos

cos⁡αcos⁡β=12[cos⁡(α−β)+cos⁡(α+β)]\cos\alpha \cos\beta = \tfrac{1}{2}[\cos(\alpha-\beta) + \cos(\alpha+\beta)]

sin · sin

sin⁡αsin⁡β=12[cos⁡(α−β)−cos⁡(α+β)]\sin\alpha \sin\beta = \tfrac{1}{2}[\cos(\alpha-\beta) - \cos(\alpha+\beta)]

Sum to Product · 4

sin A + sin B

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2 \sin\tfrac{A+B}{2} \cos\tfrac{A-B}{2}

sin A - sin B

sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2 \cos\tfrac{A+B}{2} \sin\tfrac{A-B}{2}

cos A + cos B

cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2 \cos\tfrac{A+B}{2} \cos\tfrac{A-B}{2}

cos A - cos B

cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A - \cos B = -2 \sin\tfrac{A+B}{2} \sin\tfrac{A-B}{2}