Fundamental Trigonometric Identities

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

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

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

By [L1] with $x$ replaced by $-x$, and then [L3], $\sin (\pi \mathrm{/}2-x)=\cos x$ and $\cos (\pi \mathrm{/}2-x)=\sin x$; [L1] also gives the displayed quarter-turn formulas.

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

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.

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

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

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

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

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.

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

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

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

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

Use ${\sin }^{2}x=1-{\cos }^{2}x$ and ${\cos }^{2}x=1-{\sin }^{2}x$ to collect the stated cubic forms.

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

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

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

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

Add and subtract the addition and subtraction formulas for cosine.

Add the addition and subtraction formulas for sine.

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

Divide the double-angle identities by ${\cos }^2(\theta \mathrm{/}2)$ to obtain the rational formulas; $1+t^2>0$.

For a point with $x\mathrm{\ne }-1$, put $t=y\mathrm{/}(1+x)$ and use $x^2+y^2=1$ to simplify both rational expressions to $x$ and $y$.

The same formula $t=y\mathrm{/}(1+x)$ proves uniqueness.

The Pythagorean identity puts every image point on $S^1$.

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

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

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

Otherwise $R>0$ and $(a\mathrm{/}R{)}^2+(b\mathrm{/}R{)}^2=1$, so the unit-circle parametrization gives the stated unique $\varphi$.

Expand $R\cos (x-\varphi )$ and substitute the two defining coordinates of $\varphi$.

For each $k\ge 1$, the power-difference factorization gives $x^k-a^k=(x-a){\sum }_{j<k}{a}^j{x}^{k-1-j}$.

Since $p(a)=0$, write $p(x)={\sum }_{k<n}{a}_k(x^k-a^k)$.

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

At degree $0$ the polynomial is a nonzero constant and has no root.

Assume the claim at degree $n$.

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

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

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

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

In each recurrence of [L1], $2x$ times the degree-$n$ term has degree $n+1$, whereas the subtracted predecessor has degree $n-1$. Thus no leading-term cancellation is possible, and the next leading coefficients are $2\cdot 2^{n-1}=2^n$ for $T_{n+1}$ and $2\cdot 2^n=2^{n+1}$ for $U_{n+1}$.

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

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

Assume the identities at $n$ and $n-1$.

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

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

At $n=0$ the two displayed finite sums give $1$ and $0$.

Assume the two formulas at $n$.

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

By [L2], those adjacent sums are exactly $\left(\genfrac{}{}{0px}{}{n+1}{r}\right)$; even $r$ contribute to cosine with sign $(-1{)}^{r\mathrm{/}2}$ and odd $r$ contribute to sine with sign $(-1{)}^{(r-1)\mathrm{/}2}$. This proves both formulas at $n+1$ without using complex numbers.

By [L1], $P_n=2^{1-n}{T}_n$ is monic of degree $n$. Put $y_j=\cos ((n-j)\pi \mathrm{/}n)$ for $0\le j\le n$. By [L3], $y_0<\cdots <y_n$, and [L2] gives $P_n(y_j)=(-1{)}^{n-j}{2}^{1-n}$.

For each $x\in [-1,1]$, [L3] supplies $\theta$ with $x=\cos \theta$; [L2] then gives $\mathrm{\mid }{P}_n(x)\mathrm{\mid }=2^{1-n}\mathrm{\mid }\cos (n\theta )\mathrm{\mid }\le 2^{1-n}$. Equality holds at every $y_j$, so ${\max }_{[-1,1]}\mathrm{\mid }{P}_n\mathrm{\mid }=2^{1-n}$.

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

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