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Addition and half-angle identities compute the sine, cosine, and tangent of $\pi/12$

Example

Using $\pi/12=\pi/3-\pi/4$ , $\sin(\pi/12)=\frac{\sqrt6-\sqrt2}{4},\quad \cos(\pi/12)=\frac{\sqrt6+\sqrt2}{4},\quad \tan(\pi/12)=2-\sqrt3.$

Facts & Assumptions

Given: The positive number $\pi$ and the angles $\pi/4,\pi/3\in(0,\pi)$ .

[L1]

Quarter-turn values and shifts by pi/2 and pi gives $\sin(\pi/2)=1$ and $\cos(\pi/2)=0$ .

[L2]

Half-angle identities with the sign determined by the quadrant gives $\cos(x/2)=\varepsilon_c\sqrt{(1+\cos x)/2}$ and $\sin(x/2)=\varepsilon_s\sqrt{(1-\cos x)/2}$ , with the signs of the corresponding half-angle functions.

[L3]

Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions gives $\cos(\pi/2-x)=\sin x$ and $\cos(\pi-x)=-\cos x$ .

[L4]

Double-angle and quadratic power-reduction identities gives $\cos(2x)=2\cos^2x-1$ .

[L5]

Pi is the first positive zero of sine gives $\sin x>0$ whenever $0<x<\pi$ .

[L6]

The subtraction formulas for sine and cosine gives $\sin(u-v)=\sin u\cos v-\cos u\sin v$ and $\cos(u-v)=\cos u\cos v+\sin u\sin v$ .

[L7]

Tangent, cotangent, secant, and cosecant on their exact natural domains defines $\tan x=\sin x/\cos x$ when $\cos x\ne0$ .

Verification

1.1

By [L5], $\sin(\pi/4)>0$ , and [L3] gives $\cos(\pi/4)=\sin(\pi/4)>0$ . Applying [L2] to $x=\pi/2$ and using [L1] therefore gives $\sin(\pi/4)=\cos(\pi/4)=\sqrt2/2$ .

L1L2L3L5algebra

1.2

Put $c=\cos(\pi/3)$ . Then $c>0$ because [L3] writes it as $\sin(\pi/6)$ and [L5] makes that value positive. Also [L3] and [L4] give $-c=\cos(2\pi/3)=2c^2-1$ , hence $(2c-1)(c+1)=0$ ; positivity rules out $c=-1$ , so $\cos(\pi/3)=1/2$ .

L3L4L5algebra

2.1

From [L3] and step 1.2, $\cos(2\pi/3)=-1/2$ . Since $\sin(\pi/3)>0$ by [L5], [L2] applied to $x=2\pi/3$ gives $\sin(\pi/3)=\sqrt{(1+1/2)/2}=\sqrt3/2$ .

L2L3L5step 1.2algebra

3.1

Substitute steps 1.1, 1.2, and 2.1 into [L6] with $u=\pi/3$ and $v=\pi/4$ . This gives $\sin(\pi/12)=(\sqrt6-\sqrt2)/4$ and $\cos(\pi/12)=(\sqrt6+\sqrt2)/4>0$ .

L6step 1.1step 1.2step 2.1algebra

4.1

By [L7] and step 3.1, $\tan(\pi/12)=(\sqrt6-\sqrt2)/(\sqrt6+\sqrt2)$ ; rationalizing gives $(8-4\sqrt3)/4=2-\sqrt3$ .

L7step 3.1algebra∎

Depends on

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Sources

NIST Digital Library of Mathematical Functions, Chapter 4 (standard reference, not scraped)

Addition and half-angle identities compute the sine, cosine, and tangent of π/12\pi/12π/12