A generalization of the Pythagorean trigonometric identity

"Discovery is seeing what everybody else has seen, and thinking what nobody else has thought."

-- Albert Szent-Györgyi (1893 - 1986)

I don't know how I missed it, but the Pythagorean trigonometric identity is a special case of the generalized half-angle formulas. 

The following is a generalization of the half-angle formulas presented at Nabla - Applications of Trigonometry for a triangle.

Generalization. Let $a$, $b$, $c$, $d$ be the sides of a general convex quadrilateral, $s$ is the semiperimeter, and  $\alpha$ and $\gamma$ are opposite angles, then

$$ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}=(s-a)(s-d).\tag{1}$$

For a proof of $(1)$ see pp. 8 in MATINF.

The Pythagorean identity as a special case

 In $(1)$, consider the case when $a=c$, $b=d$ and $\alpha=\gamma$ so that the quadrilateral is a parallelogram . Then

$$ab\sin^2{\frac{\alpha}{2}}+ab\cos^2{\frac{\alpha}{2}}=\frac{-a+b+c+d}{2}\cdot{\frac{a+b+c-d}{2}}=ab.$$

Dividing both sides by $ab$ you get

$$\sin^2{\frac{\alpha}{2}}+\cos^2{\frac{\alpha}{2}}=1.$$

Or by making $\frac{\alpha}{2}=\theta$,

$$\sin^2{\theta}+\cos^2{\theta}=1,$$

which is the Pythagorean trigonometric identity.

An interesting discussion about this generalization is available at MathSE.

For more implications of $(1)$ I invite you to see The theoretical importance of the half-angle formulas.

Another proof of Euler inequality via the half-angle formulas

The Euler's inequality is an immediate consequence of Euler's identity in a triangle,
$$OI^2=R^2−2Rr.$$
An additional proof of Euler's inequality is given at Elias Lampakis, Am Math Monthly, 122 (9), November 2015, p 892. Now, continuing with this madness of proving everything from the half-angle formulas, we give an alternative proof of Euler's inequality.

Theorem (Euler): If $R$ and $r$ are the circumradius and, respectively, the inradius of a triangle, then
$$R\geq2r.\tag{1}$$

Proof. Let $a$, $b$ and $c$ be the sides of $\triangle{ABC}$. Denote $s$ its semiperimeter and $\angle{BAC}=\alpha$. It is easy to show that 

$$(s-a)\tan{\frac{\alpha}{2}}=r.\tag{2}$$

Also, it is well-known that

$$\frac{abc}{4\Delta}=R\qquad and \qquad\Delta=\frac{bc\sin{\alpha}}{2},\tag{3}$$

where $\Delta$ is the area of $\triangle{ABC}$. Substituting $(2)$ and $(3)$ in $(1)$ we have

$$\begin{aligned}\frac{abc}{4\Delta}&\geq(-a+b+c)\tan{\frac{\alpha}{2}}\\\frac{abc}{4bc\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}}&\geq(-a+b+c)\tan{\frac{\alpha}{2}}\\abc&\geq(-a+b+c)(4bc\sin^2{\frac{\alpha}{2}})\\abc&\geq(-a+b+c)(a-b+c)(a+b-c)\end{aligned}$$

This is Padoa's inequality, so the proof is complete.

$\sum_{cyc}\tan\frac\alpha2\tan\frac\beta2\geq4$ for a cyclic quadrilateral

 Let $ABCD$ be a cyclic quadrilateral with sides $a$, $b$, $c$ and $d$. Denote $s$ the semiperimeter and let $\angle{DAB}=\alpha$, $\angle{ABC}=\beta$, $\angle{BCD}=\gamma$ and $\angle{CDA}=\delta$. Then the following inequality holds

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}\geq4.\tag{1}$$

Proof. Substituting from the half-angle formula for the tangent we have that

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}=\sqrt{\frac{(s-a)(s-d)}{(s-b)(s-c)}}\cdot{\sqrt{\frac{(s-a)(s-b)}{(s-c)(s-d)}}}=\frac{s-a}{s-c}.$$

Similarly, 

$$\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=\frac{s-b}{s-d}\qquad\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}=\frac{s-c}{s-a}\qquad\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}=\frac{s-d}{s-b}$$

Thus, the left-hand side of $(1)$ can be rewritten as follows

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}=\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}.$$

But, 

$$\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}=\frac{a-c}{s-a}+\frac{b-d}{s-b}+\frac{c-a}{s-c}+\frac{d-b}{s-d}+4.\tag{2}$$

Since $\frac{a-c}{s-a}+\frac{c-a}{s-c}=\frac{4(a-c)^2}{(-a+b+c+d)(a+b-c+d)}$, and similarly for $\frac{b-d}{s-b}+\frac{d-b}{s-d}$, then $\frac{a-c}{s-a}+\frac{b-d}{s-b}+\frac{c-a}{s-c}+\frac{d-b}{s-d}$ is positive. Hence,

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}\geq4.$$

$\square$

Notice equality holds when $ABCD$ is rectangular. 

From $(2)$ it also follows that

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}>\frac{a-c}{s-a}-\frac{a-c}{s-c}+\frac{b-d}{s-b}-\frac{ b-d}{s-d}.$$

A huge list of inequalities can be seen at Cut-the-knot.org.