Alphabeta Math
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17 results · all verified · 0 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 17 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Fundamental Trigonometric Identities

1 · Prerequisites

2 · Summary

The sine--cosine addition formulas and the quotient-function definitions supply the basic algebra on this page. The development keeps every tangent, cotangent, secant, and cosecant denominator visible, and uses the real polynomial and finite-counting prerequisites for the Chebyshev and binomial arguments.

The page derives subtraction, double- and triple-angle, signed half-angle, product-to-sum, and rational half-angle identities before connecting the unit circle with amplitude--phase form. It then develops real polynomial roots and Chebyshev recurrences, multiple-angle identities, and the minimax normalization of TnT_n.

3 · Logical flowchart

4 · Definitions, theorems and proofs

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Pythagorean and parity identities for all six trigonometric functions on their natural domains

Statement

For every real xx, sin2x+cos2x=1\sin ^2x+\cos ^2x=1, sin(x)=sinx\sin(-x)=-\sin x, and cos(x)=cosx\cos(-x)=\cos x. Wherever the displayed quotients are defined, tan(x)=tanx\tan(-x)=-\tan x, cot(x)=cotx\cot(-x)=-\cot x, sec(x)=secx\sec(-x)=\sec x, csc(x)=cscx\csc(-x)=-\csc x, 1+tan2x=sec2x1+\tan^2x=\sec^2x, and 1+cot2x=csc2x1+\cot^2x=\csc^2x. The conventions and prerequisite facts used below are recorded in Parity and the Pythagorean identity for sine and cosine, Tangent, cotangent, secant, and cosecant on their exact natural domains.

Facts & Assumptions

Given: A real xx and the quotient definitions on their stated nonzero-denominator domains.

Proof

technique · direct
1.1

The sine--cosine Pythagorean and parity identities give the first three equalities.

given
1.2

On the domain of tangent, divide sin2x+cos2x=1\sin ^2x+\cos ^2x=1 by cos2x\cos ^2x; on the domain of cotangent divide it by sin2x\sin ^2x.

algebra
2.1

Applying the quotient definitions to the parity equalities gives the four quotient parity laws without introducing a zero denominator.

given
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions

Statement

For every real xx, sin(π/2x)=cosx,cos(π/2x)=sinx,sin(πx)=sinx,cos(πx)=cosx,\sin(\pi/2-x)=\cos x,\quad \cos(\pi/2-x)=\sin x,\quad \sin(\pi-x)=\sin x,\quad \cos(\pi-x)=-\cos x, and the quarter-turn and reflection formulas are sin(x+π/2)=cosx,cos(x+π/2)=sinx,sin(x)=sinx,cos(x)=cosx.\sin(x+\pi/2)=\cos x,\quad \cos(x+\pi/2)=-\sin x,\quad \sin(-x)=-\sin x,\quad \cos(-x)=\cos x. On the common natural domains of the two sides, tan(π/2x)=cotx,cot(π/2x)=tanx,sec(π/2x)=cscx,csc(π/2x)=secx,\tan(\pi/2-x)=\cot x,\quad \cot(\pi/2-x)=\tan x,\quad \sec(\pi/2-x)=\csc x,\quad \csc(\pi/2-x)=\sec x, and tan(πx)=tanx,cot(πx)=cotx,sec(πx)=secx,csc(πx)=cscx.\tan(\pi-x)=-\tan x,\quad \cot(\pi-x)=-\cot x,\quad \sec(\pi-x)=-\sec x,\quad \csc(\pi-x)=\csc x. The conventions and prerequisite facts used below are recorded in Pythagorean and parity identities for all six trigonometric functions on their natural domains, Quarter-turn values and shifts by pi/2 and pi, The addition formulas for sine and cosine, The zero sets of sine and cosine and the least positive common period 2 pi, Tangent, cotangent, secant, and cosecant on their exact natural domains.

Facts & Assumptions

Given: A real xx.

[L1]

Quarter-turn values and shifts by pi/2 and pi states that, for every real xx, sin(x+π/2)=cosx\sin(x+\pi/2)=\cos x and cos(x+π/2)=sinx\cos(x+\pi/2)=-\sin x.

[L2]

The addition formulas for sine and cosine states that, for all real a,ba,b, sin(a+b)=sinacosb+cosasinb\sin(a+b)=\sin a\cos b+\cos a\sin b and cos(a+b)=cosacosbsinasinb\cos(a+b)=\cos a\cos b-\sin a\sin b.

[L3]

Pythagorean and parity identities for all six trigonometric functions on their natural domains gives sin(x)=sinx\sin(-x)=-\sin x and cos(x)=cosx\cos(-x)=\cos x.

[L4]

Tangent, cotangent, secant, and cosecant on their exact natural domains defines tant=sint/cost\tan t=\sin t/\cos t, cott=cost/sint\cot t=\cos t/\sin t, sect=1/cost\sec t=1/\cos t, and csct=1/sint\csc t=1/\sin t on their natural domains.

Proof

technique · direct
1.1

By [L1] with xx replaced by x-x, and then [L3], sin(π/2x)=cosx\sin(\pi/2-x)=\cos x and cos(π/2x)=sinx\cos(\pi/2-x)=\sin x; [L1] also gives the displayed quarter-turn formulas.

L1L3
1.2

Apply [L2] to π+(x)\pi+(-x) and use the values at π\pi supplied by [L1] (put x=π/2x=\pi/2) together with [L3]. This gives sin(πx)=sinx\sin(\pi-x)=\sin x and cos(πx)=cosx\cos(\pi-x)=-\cos x.

L1L2L3
2.1

Substitute the cofunction and supplementary sine--cosine equalities into [L4]. The stated nonvanishing conditions are exactly those that make both quotient or reciprocal expressions defined, so this yields all eight displayed identities.

L4step 1.1step 1.2
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The subtraction formulas for sine and cosine

Statement

For all real u,vu,v, sin(uv)=sinucosvcosusinv,cos(uv)=cosucosv+sinusinv.\sin(u-v)=\sin u\cos v-\cos u\sin v,\qquad \cos(u-v)=\cos u\cos v+\sin u\sin v. The conventions and prerequisite facts used below are recorded in The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine.

Facts & Assumptions

Given: Reals u,vu,v.

Proof

technique · direct
1.1

Apply the addition formulas to u+(v)u+(-v).

given
2.1

Replace sin(v)\sin(-v) and cos(v)\cos(-v) by their parity values and simplify.

algebra
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Addition and subtraction formulas for tangent, cotangent, secant, and cosecant on their exact domains

Statement

If cosucosvcos(u+v)0\cos u\cos v\cos(u+v)\ne0, then tan(u+v)=tanu+tanv1tanutanv,sec(u+v)=secusecv1tanutanv;\tan(u+v)=\frac{\tan u+\tan v}{1-\tan u\tan v},\qquad \sec(u+v)=\frac{\sec u\sec v}{1-\tan u\tan v}; if cosucosvcos(uv)0\cos u\cos v\cos(u-v)\ne0, then tan(uv)=tanutanv1+tanutanv,sec(uv)=secusecv1+tanutanv.\tan(u-v)=\frac{\tan u-\tan v}{1+\tan u\tan v},\qquad \sec(u-v)=\frac{\sec u\sec v}{1+\tan u\tan v}. If sinusinvsin(u+v)0\sin u\sin v\sin(u+v)\ne0, then cot(u+v)=cotucotv1cotu+cotv,csc(u+v)=cscucscvcotu+cotv;\cot(u+v)=\frac{\cot u\cot v-1}{\cot u+\cot v},\qquad \csc(u+v)=\frac{\csc u\csc v}{\cot u+\cot v}; if sinusinvsin(uv)0\sin u\sin v\sin(u-v)\ne0, then cot(uv)=cotucotv+1cotvcotu,csc(uv)=cscucscvcotvcotu.\cot(u-v)=\frac{\cot u\cot v+1}{\cot v-\cot u},\qquad \csc(u-v)=\frac{\csc u\csc v}{\cot v-\cot u}. The conventions and prerequisite facts used below are recorded in Tangent, cotangent, secant, and cosecant on their exact natural domains, The addition formulas for sine and cosine, The subtraction formulas for sine and cosine.

Facts & Assumptions

Given: Reals u,vu,v satisfying the relevant displayed nonvanishing hypothesis.

[L1]

Tangent, cotangent, secant, and cosecant on their exact natural domains defines tant=sint/cost\tan t=\sin t/\cos t, cott=cost/sint\cot t=\cos t/\sin t, sect=1/cost\sec t=1/\cos t, and csct=1/sint\csc t=1/\sin t on their natural domains.

[L2]

The addition formulas for sine and cosine gives the sine and cosine formulas for u+vu+v.

[L3]

The subtraction formulas for sine and cosine gives sin(uv)=sinucosvcosusinv\sin(u-v)=\sin u\cos v-\cos u\sin v and cos(uv)=cosucosv+sinusinv\cos(u-v)=\cos u\cos v+\sin u\sin v.

Proof

technique · direct
1.1

Under the cosine nonvanishing hypothesis, divide the two formulas in [L2] by cosucosv\cos u\cos v. Their quotient gives the tangent addition formula, and their cosine formula gives cos(u+v)=cosucosv(1tanutanv)\cos(u+v)=\cos u\cos v(1-\tan u\tan v). Taking reciprocals by [L1] gives the secant addition formula.

L1L2
1.2

Under the sine nonvanishing hypothesis, divide the formulas in [L2] by sinusinv\sin u\sin v. Their quotient gives the cotangent addition formula, and their sine formula gives sin(u+v)=sinusinv(cotu+cotv)\sin(u+v)=\sin u\sin v(\cot u+\cot v). Taking reciprocals by [L1] gives the cosecant addition formula.

L1L2
2.1

The two formulas in [L3], divided by the same nonzero products as in steps 1.1 and 1.2, give respectively the tangent--secant and cotangent--cosecant subtraction formulas. Every denominator used is nonzero by the hypothesis displayed next to that formula.

L1L3
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Double-angle and quadratic power-reduction identities

Statement

For every real xx, sin(2x)=2sinxcosx,cos(2x)=cos2xsin2x=2cos2x1=12sin2x,\sin(2x)=2\sin x\cos x,\quad \cos(2x)=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x, and sin2x=(1cos2x)/2,cos2x=(1+cos2x)/2.\sin^2x=(1-\cos2x)/2,\qquad \cos^2x=(1+\cos2x)/2. When defined, tan(2x)=2tanx/(1tan2x)\tan(2x)=2\tan x/(1-\tan^2x). The conventions and prerequisite facts used below are recorded in The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Tangent, cotangent, secant, and cosecant on their exact natural domains.

Facts & Assumptions

Given: A real xx.

Proof

technique · direct
1.1

Put u=v=xu=v=x in the addition formulas.

given
1.2

Use sin2x+cos2x=1\sin^2x+\cos^2x=1 to rewrite the cosine identity and solve both resulting equalities for the squares.

algebra
2.1

Divide the double-angle sine identity by the double-angle cosine identity only when both quotient expressions are defined.

algebra
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Triple-angle identities for sine, cosine, and tangent

Statement

For every real xx, sin3x=3sinx4sin3x,cos3x=4cos3x3cosx.\sin3x=3\sin x-4\sin^3x,\qquad \cos3x=4\cos^3x-3\cos x. If tanx\tan x and tan3x\tan3x are defined and 13tan2x01-3\tan^2x\ne0, then tan3x=(3tanxtan3x)/(13tan2x)\tan3x=(3\tan x-\tan^3x)/(1-3\tan^2x). The conventions and prerequisite facts used below are recorded in Double-angle and quadratic power-reduction identities, The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Tangent, cotangent, secant, and cosecant on their exact natural domains.

Facts & Assumptions

Given: A real xx satisfying the displayed tangent hypotheses where required.

Proof

technique · direct
1.1

Expand sin(2x+x)\sin(2x+x) and cos(2x+x)\cos(2x+x) by addition, then substitute the double-angle identities.

given
1.2

Use sin2x=1cos2x\sin^2x=1-\cos^2x and cos2x=1sin2x\cos^2x=1-\sin^2x to collect the stated cubic forms.

algebra
2.1

Divide the two cubic identities under the stated nonzero conditions and cancel the common cosine power.

algebra
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Half-angle identities with the sign determined by the quadrant

Statement

For every real xx, cos(x/2)=εc(1+cosx)/2,sin(x/2)=εs(1cosx)/2,\cos(x/2)=\varepsilon_c\sqrt{(1+\cos x)/2},\qquad \sin(x/2)=\varepsilon_s\sqrt{(1-\cos x)/2}, where εc,εs{1,0,1}\varepsilon_c,\varepsilon_s\in\{-1,0,1\} are respectively the signs of cos(x/2)\cos(x/2) and sin(x/2)\sin(x/2) (so sgn(0)=0\operatorname{sgn}(0)=0). Thus the positive square root is valid only where the relevant half-angle function is nonnegative. The conventions and prerequisite facts used below are recorded in Double-angle and quadratic power-reduction identities, Square roots exist: a unique a0\sqrt{a} \ge 0 with (a)2=a(\sqrt{a})^2 = a; the positives are {x2:x0}\{x^2 : x \neq 0\}, Signs, monotonicity intervals, and ranges of sine and cosine, Tangent, cotangent, secant, and cosecant on their exact natural domains.

Facts & Assumptions

Given: A real xx.

[L1]

Double-angle and quadratic power-reduction identities gives cos2t=(1+cos2t)/2\cos^2t=(1+\cos2t)/2 and sin2t=(1cos2t)/2\sin^2t=(1-\cos2t)/2 for every real tt.

Proof

technique · direct
1.1

Apply [L1] with t=x/2t=x/2. Then cos2(x/2)=(1+cosx)/2\cos^2(x/2)=(1+\cos x)/2 and sin2(x/2)=(1cosx)/2\sin^2(x/2)=(1-\cos x)/2, so both radicands are nonnegative.

L1
2.1

By [L2], cos(x/2)=(1+cosx)/2|\cos(x/2)|=\sqrt{(1+\cos x)/2} and sin(x/2)=(1cosx)/2|\sin(x/2)|=\sqrt{(1-\cos x)/2}.

L2step 1.1
3.1

Multiplying each equality of step 2.1 by its sign, with sign 00 when the corresponding value is 00, yields the displayed identities. The sign ranges are therefore exactly {1,0,1}\{-1,0,1\}.

step 2.1
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Product-to-sum and sum-to-product identities

Statement

For all real u,vu,v, sinusinv=cos(uv)cos(u+v)2,cosucosv=cos(uv)+cos(u+v)2,\sin u\sin v=\frac{\cos(u-v)-\cos(u+v)}2,\quad \cos u\cos v=\frac{\cos(u-v)+\cos(u+v)}2, sinucosv=sin(u+v)+sin(uv)2,\sin u\cos v=\frac{\sin(u+v)+\sin(u-v)}2, and, with a=(u+v)/2a=(u+v)/2, b=(uv)/2b=(u-v)/2, the reverse sum-to-product identities follow by solving these formulas for the sums. The conventions and prerequisite facts used below are recorded in The addition formulas for sine and cosine, The subtraction formulas for sine and cosine.

Facts & Assumptions

Given: Reals u,vu,v.

Proof

technique · direct
1.1

Add and subtract the addition and subtraction formulas for cosine.

algebra
1.2

Add the addition and subtraction formulas for sine.

algebra
2.1

Substitute u=a+bu=a+b, v=abv=a-b to obtain every reverse identity.

algebra
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The tangent half-angle identities and rational parametrization of the unit circle away from (1,0)(-1,0)

Statement

If t=tan(θ/2)t=\tan(\theta/2) is defined, then cosθ=1t21+t2,sinθ=2t1+t2.\cos\theta=\frac{1-t^2}{1+t^2},\qquad \sin\theta=\frac{2t}{1+t^2}. Conversely every point (x,y)(x,y) on x2+y2=1x^2+y^2=1 with x1x\ne-1 has the unique parameter t=y/(1+x)t=y/(1+x) and equals ((1t2)/(1+t2),2t/(1+t2))((1-t^2)/(1+t^2),2t/(1+t^2)). The conventions and prerequisite facts used below are recorded in Double-angle and quadratic power-reduction identities, Tangent, cotangent, secant, and cosecant on their exact natural domains, Parity and the Pythagorean identity for sine and cosine.

Facts & Assumptions

Given: The indicated real parameter or unit-circle point.

Proof

technique · direct
1.1

Divide the double-angle identities by cos2(θ/2)\cos^2(\theta/2) to obtain the rational formulas; 1+t2>01+t^2>0.

algebra
1.2

For a point with x1x\ne-1, put t=y/(1+x)t=y/(1+x) and use x2+y2=1x^2+y^2=1 to simplify both rational expressions to xx and yy.

algebra
2.1

The same formula t=y/(1+x)t=y/(1+x) proves uniqueness.

algebra
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

t(cost,sint)t\mapsto(\cos t,\sin t) is a bijection from [0,2π)[0,2\pi) onto the real unit circle

Statement

The map t(cost,sint)t\mapsto(\cos t,\sin t) is a bijection from [0,2π)[0,2\pi) onto S1={(x,y)R2:x2+y2=1}S^1=\{(x,y)\in\mathbb R^2:x^2+y^2=1\}. The conventions and prerequisite facts used below are recorded in Parity and the Pythagorean identity for sine and cosine, Signs, monotonicity intervals, and ranges of sine and cosine, The zero sets of sine and cosine and the least positive common period 2 pi, Pi as twice the smallest positive zero of cosine.

Facts & Assumptions

Given: A point (x,y)(x,y) on S1S^1 or parameters s,t[0,2π)s,t\in[0,2\pi).

Proof

technique · direct
1.1

The Pythagorean identity puts every image point on S1S^1.

given
1.2

The range and sign theorem supplies an angle in the stated half-open interval with prescribed cosine and the compatible sine, proving surjectivity.

given
2.1

If two parameters have the same pair, the sine--cosine period theorem says their difference is a multiple of 2π2\pi; the half-open interval forces that multiple to be zero.

given
CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Every acosx+bsinxa\cos x+b\sin x has the amplitude-phase form Rcos(xϕ)R\cos(x-\phi)

Statement

For reals a,ba,b not both zero, put R=a2+b2>0R=\sqrt{a^2+b^2}>0. There is a unique ϕ[0,2π)\phi\in[0,2\pi) with cosϕ=a/R\cos\phi=a/R and sinϕ=b/R\sin\phi=b/R, and acosx+bsinx=Rcos(xϕ)a\cos x+b\sin x=R\cos(x-\phi) for every real xx. For a=b=0a=b=0 the left side is identically zero. The conventions and prerequisite facts used below are recorded in t(cost,sint)t\mapsto(\cos t,\sin t) is a bijection from [0,2π)[0,2\pi) onto the real unit circle, Square roots exist: a unique a0\sqrt{a} \ge 0 with (a)2=a(\sqrt{a})^2 = a; the positives are {x2:x0}\{x^2 : x \neq 0\}, The addition formulas for sine and cosine.

Facts & Assumptions

Given: Reals a,b,xa,b,x.

Proof

technique · direct
1.1

If a=b=0a=b=0, the claimed zero identity is immediate.

algebra
1.2

Otherwise R>0R>0 and (a/R)2+(b/R)2=1(a/R)^2+(b/R)^2=1, so the unit-circle parametrization gives the stated unique ϕ\phi.

given
2.1

Expand Rcos(xϕ)R\cos(x-\phi) and substitute the two defining coordinates of ϕ\phi.

algebra
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials

Definition

A formal real polynomial is either the zero polynomial 00, or a finite coefficient list (a0,,an)(a_0,\ldots,a_n) with an0a_n\ne0; we write the latter as p(X)=k=0nakXkp(X)=\sum_{k=0}^{n}a_kX^k. The list, rather than the function it induces, is the polynomial object. Its evaluation at xRx\in\mathbb R is the real number p(x):=k=0nakxkp(x):=\sum_{k=0}^{n}a_kx^k.

For nonzero p=(a0,,an)p=(a_0,\ldots,a_n), define degp:=n\deg p:=n and lc(p):=an\operatorname{lc}(p):=a_n. The zero polynomial has no degree and no leading coefficient. A nonzero polynomial is monic when lc(p)=1\operatorname{lc}(p)=1. Thus degree and leading coefficient are defined from the displayed finite list, without asserting that distinct formal polynomials define distinct functions. The conventions for finite sums and integer powers used in the evaluation are recorded in Finite sums and finite products, by recursion, Integer powers ama^m.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

A real polynomial vanishing at aa is divisible by xax-a

Statement

If p(x)=k<nakxkp(x)=\sum_{k<n}a_kx^k and p(a)=0p(a)=0, then there is a real polynomial qq with p(x)=(xa)q(x)p(x)=(x-a)q(x) for every real xx. The conventions and prerequisite facts used below are recorded in Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials, Factorisation of bnanb^n - a^n, and the resulting Lipschitz estimate, Laws of finite sums and finite products.

Facts & Assumptions

Given: A polynomial pp and a real root aa.

Proof

technique · constructive
1.1

For each k1k\ge1, the power-difference factorization gives xkak=(xa)j<kajxk1jx^k-a^k=(x-a)\sum_{j<k}a^jx^{k-1-j}.

given
1.2

Since p(a)=0p(a)=0, write p(x)=k<nak(xkak)p(x)=\sum_{k<n}a_k(x^k-a^k).

algebra
2.1

Substitute the factorization from step 1.1 and collect the finite coefficient sums into a polynomial qq.

constructdischarge-construct
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

A nonzero real polynomial of degree nn has no more than nn distinct real roots

Statement

A nonzero real polynomial of degree nn has at most nn distinct real roots. The conventions and prerequisite facts used below are recorded in Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials, A real polynomial vanishing at aa is divisible by xax-a, The principle of mathematical induction.

Facts & Assumptions

Given: A nonzero real polynomial of degree nn.

Proof

technique · induction
1.1

At degree 00 the polynomial is a nonzero constant and has no root.

base
1.2

Assume the claim at degree nn.

ih
2.1

If a degree-n+1n+1 polynomial has a root aa, the factor lemma writes it as (xa)q(x)(x-a)q(x) with qq of degree nn; every other root is a root of qq.

step 1.2given
3.1

The induction hypothesis gives at most nn other roots, hence at most n+1n+1 roots in all.

discharge-induction
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-02Open item page →

Chebyshev polynomials of the first and second kinds by their three-term recurrences

Definition

Let PR\mathcal P_{\mathbb R} be the set of formal real polynomials of Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials. Associate to every PPRP\in\mathcal P_{\mathbb R} an eventually zero coefficient sequence (pk)(p_k): use its displayed finite coefficient list and extend it by zeros when P0P\ne0, and put pk=0p_k=0 for every kk when P=0P=0. Do the same for QQ, with coefficients (qk)(q_k). Define P+QP+Q coefficientwise and define PQPQ by the convolution coefficients

(P+Q)k=pk+qk,(PQ)k=j=0kpjqkj.(P+Q)_k=p_k+q_k,\qquad (PQ)_k=\sum_{j=0}^{k}p_jq_{k-j}.

using the real finite sum of Finite sums and finite products, by recursion. In either construction, discard trailing zero coefficients; if every coefficient is zero, the result is the zero polynomial. Scalar multiplication and subtraction are the corresponding coefficientwise operations. Thus these formulas define operations on the formal coefficient-list objects, independently of evaluation.

Write 1=(1)1=(1) and X=(0,1)X=(0,1), and define

Φ(P,Q):=(Q,2XQP)(P,QPR).\Phi(P,Q):=(Q,2XQ-P)\qquad(P,Q\in\mathcal P_{\mathbb R}).

Apply The recursion theorem to the set PR×PR\mathcal P_{\mathbb R}\times\mathcal P_{\mathbb R}, first with initial value (1,X)(1,X) and then with initial value (1,2X)(1,2X), always using the function Φ\Phi. This gives unique pair sequences HT,HUH^T,H^U. Define TnT_n and UnU_n to be the first coordinates of HnTH^T_n and HnUH^U_n, respectively. Since the first coordinate of Φ(P,Q)\Phi(P,Q) is QQ, the second coordinate of HnTH^T_n is Tn+1T_{n+1} and the second coordinate of HnUH^U_n is Un+1U_{n+1}. Consequently

T0=1,T1=X,Tn+2=2XTn+1Tn,T_0=1,\quad T_1=X,\quad T_{n+2}=2XT_{n+1}-T_n, U0=1,U1=2X,Un+2=2XUn+1UnU_0=1,\quad U_1=2X,\quad U_{n+2}=2XU_{n+1}-U_n

for every nNn\in\mathbb N. These unique sequences are the Chebyshev polynomials of the first and second kinds.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Degrees and leading coefficients of the Chebyshev polynomials

Statement

For n1n\ge1, TnT_n has degree nn and leading coefficient 2n12^{n-1}, while UnU_n has degree nn and leading coefficient 2n2^n. The conventions and prerequisite facts used below are recorded in Chebyshev polynomials of the first and second kinds by their three-term recurrences, The principle of mathematical induction, Integer powers ama^m.

Facts & Assumptions

Given: A natural nn.

[L1]

Chebyshev polynomials of the first and second kinds by their three-term recurrences defines T0=1T_0=1, T1=xT_1=x, Tn+1=2xTnTn1T_{n+1}=2xT_n-T_{n-1} and U0=1U_0=1, U1=2xU_1=2x, Un+1=2xUnUn1U_{n+1}=2xU_n-U_{n-1}.

Proof

technique · induction
1.1

The initial values in [L1] give the asserted degrees and leading coefficients at n=1n=1 (and the recurrence needs the consecutive base indices 0,10,1).

L1base
1.2

Assume the degree and leading-coefficient assertions at consecutive indices.

ih
2.1

In each recurrence of [L1], 2x2x times the degree-nn term has degree n+1n+1, whereas the subtracted predecessor has degree n1n-1. Thus no leading-term cancellation is possible, and the next leading coefficients are 22n1=2n2\cdot2^{n-1}=2^n for Tn+1T_{n+1} and 22n=2n+12\cdot2^n=2^{n+1} for Un+1U_{n+1}.

L1step 1.2algebra
3.1

This proves the stated degree and leading-coefficient formulas at every index.

discharge-induction
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Tn(cosθ)=cos(nθ)T_n(\cos\theta)=\cos(n\theta) and Un(cosθ)sinθ=sin((n+1)θ)U_n(\cos\theta)\sin\theta=\sin((n+1)\theta) for every nNn\in\mathbb N

Statement

For every nNn\in\mathbb N and real θ\theta, Tn(cosθ)=cos(nθ),Un(cosθ)sinθ=sin((n+1)θ).T_n(\cos\theta)=\cos(n\theta),\qquad U_n(\cos\theta)\sin\theta=\sin((n+1)\theta). In particular, for n1n\ge1 and 0jn0\le j\le n, Tn(cos(jπ/n))=(1)jT_n(\cos(j\pi/n))=(-1)^j. The conventions and prerequisite facts used below are recorded in Chebyshev polynomials of the first and second kinds by their three-term recurrences, The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, The principle of mathematical induction.

Facts & Assumptions

Given: A natural nn and a real θ\theta.

Proof

technique · induction
1.1

The two identities follow directly from the initial polynomial values at n=0n=0 and n=1n=1.

base
1.2

Assume the identities at nn and n1n-1.

ih
2.1

The recurrences and the addition formulas give the usual second-order recurrences for cos((n+1)θ)\cos((n+1)\theta) and sin((n+2)θ)\sin((n+2)\theta).

step 1.2algebra
3.1

Hence the identities hold at n+1n+1, so induction proves them for every natural nn; substituting θ=jπ/n\theta=j\pi/n gives the stated alternating values.

discharge-induction
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Finite binomial formulas for cos(nθ)\cos(n\theta) and sin(nθ)\sin(n\theta)

Statement

For every nNn\in\mathbb N and real θ\theta, cos(nθ)=2jn(1)j(n2j)cosn2jθsin2jθ,\cos(n\theta)=\sum_{2j\le n}(-1)^j\binom n{2j}\cos^{n-2j}\theta\sin^{2j}\theta, sin(nθ)=2j+1n(1)j(n2j+1)cosn2j1θsin2j+1θ.\sin(n\theta)=\sum_{2j+1\le n}(-1)^j\binom n{2j+1}\cos^{n-2j-1}\theta\sin^{2j+1}\theta. The conventions and prerequisite facts used below are recorded in The addition formulas for sine and cosine, Pascal's rule (n+1k+1)=(nk)+(nk+1)\binom{n+1}{k+1} = \binom{n}{k} + \binom{n}{k+1}, and the hockey-stick identity in(ik)=(n+1k+1)\sum_{i \le n}\binom{i}{k} = \binom{n+1}{k+1}, The set [A]k[A]^{k} of kk-element subsets and the binomial coefficient (nk):=[n]k\binom{n}{k} := \lvert [n]^{k}\rvert, Finite sums and finite products, by recursion, Integer powers ama^m, The principle of mathematical induction.

Facts & Assumptions

Given: A natural nn and real θ\theta.

[L1]

The addition formulas for sine and cosine gives the formulas for sin(a+b)\sin(a+b) and cos(a+b)\cos(a+b) for all real a,ba,b.

Proof

technique · induction
1.1

At n=0n=0 the two displayed finite sums give 11 and 00.

base
1.2

Assume the two formulas at nn.

ih
2.1

Apply [L1] to nθ+θn\theta+\theta and insert the two induction sums. Collecting the coefficient of each monomial cosn+1rθsinrθ\cos^{n+1-r}\theta\sin^r\theta leaves the sum of the two adjacent binomial coefficients.

L1step 1.2algebra
3.1

By [L2], those adjacent sums are exactly (n+1r)\binom{n+1}r; even rr contribute to cosine with sign (1)r/2(-1)^{r/2} and odd rr contribute to sine with sign (1)(r1)/2(-1)^{(r-1)/2}. This proves both formulas at n+1n+1 without using complex numbers.

L2step 2.1discharge-induction
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

For n1n\ge1, 21nTn2^{1-n}T_n is the minimax monic polynomial of degree nn on [1,1][-1,1]

Statement

For n1n\ge1, the polynomial Pn=21nTnP_n=2^{1-n}T_n is monic of degree nn and for every monic real polynomial qq of degree nn, maxx[1,1]q(x)21n=maxx[1,1]Pn(x).\max_{x\in[-1,1]}|q(x)|\ge2^{1-n}=\max_{x\in[-1,1]}|P_n(x)|. Equality is attained by PnP_n. The conventions and prerequisite facts used below are recorded in Chebyshev polynomials of the first and second kinds by their three-term recurrences, Degrees and leading coefficients of the Chebyshev polynomials, Tn(cosθ)=cos(nθ)T_n(\cos\theta)=\cos(n\theta) and Un(cosθ)sinθ=sin((n+1)θ)U_n(\cos\theta)\sin\theta=\sin((n+1)\theta) for every nNn\in\mathbb N, Signs, monotonicity intervals, and ranges of sine and cosine, A nonzero real polynomial of degree nn has no more than nn distinct real roots, Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials, Extreme value theorem: a continuous real function on a nonempty compact subset of R\mathbb{R} attains a greatest and a least value, Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, and Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on [a,b][a,b] takes every value between f(a)f(a) and f(b)f(b).

Facts & Assumptions

Given: A natural n1n\ge1 and a monic polynomial qq of degree nn.

[L1]

Degrees and leading coefficients of the Chebyshev polynomials says that TnT_n has degree nn and leading coefficient 2n12^{n-1}.

[L2]

Tn(cosθ)=cos(nθ)T_n(\cos\theta)=\cos(n\theta) and Un(cosθ)sinθ=sin((n+1)θ)U_n(\cos\theta)\sin\theta=\sin((n+1)\theta) for every nNn\in\mathbb N gives Tn(cosθ)=cos(nθ)T_n(\cos\theta)=\cos(n\theta) and Tn(cos(jπ/n))=(1)jT_n(\cos(j\pi/n))=(-1)^j for 0jn0\le j\le n.

[L3]

Signs, monotonicity intervals, and ranges of sine and cosine says that cosine has range [1,1][-1,1] and is strictly decreasing on [0,π][0,\pi].

[L5]

A nonzero real polynomial of degree nn has no more than nn distinct real roots bounds the number of distinct real roots by the degree.

[L6]

Extreme value theorem: a continuous real function on a nonempty compact subset of R\mathbb{R} attains a greatest and a least value makes the maximum of q|q| on the nonempty compact interval [1,1][-1,1] exist once qq is continuous.

Proof

technique · contradiction
1.1

By [L1], Pn=21nTnP_n=2^{1-n}T_n is monic of degree nn. Put yj=cos((nj)π/n)y_j=\cos((n-j)\pi/n) for 0jn0\le j\le n. By [L3], y0<<yny_0<\cdots<y_n, and [L2] gives Pn(yj)=(1)nj21nP_n(y_j)=(-1)^{n-j}2^{1-n}.

L1L2L3
2.1

For each x[1,1]x\in[-1,1], [L3] supplies θ\theta with x=cosθx=\cos\theta; [L2] then gives Pn(x)=21ncos(nθ)21n|P_n(x)|=2^{1-n}|\cos(n\theta)|\le2^{1-n}. Equality holds at every yjy_j, so max[1,1]Pn=21n\max_{[-1,1]}|P_n|=2^{1-n}.

L2L3step 1.1
2.2

By [L7], qq and hence q|q| are continuous, so [L6] makes the displayed maximum well-defined. Suppose, for contradiction, that it is <21n<2^{1-n}. Then r:=qPnr:=q-P_n has degree at most n1n-1. At the successive points yjy_j, the values of rr have the opposite alternating signs to PnP_n, hence are nonzero and alternate.

L6L7assume-contrastep 1.1
3.1

By [L7], rr is continuous, so [L4] gives a root of rr in each disjoint interval (yj,yj+1)(y_j,y_{j+1}) (0j<n)(0\le j<n). Thus rr has at least nn distinct roots, contradicting [L5] because step 2.2 makes rr nonzero of degree at most n1n-1. The contradiction proves the lower bound, while step 2.1 proves equality for PnP_n.

L4L5L7step 2.1step 2.2discharge-contradiction

5 · Examples, counterexamples and false statements

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