Prove identity in a triangle

Question

I want to show that if $ABC$ is a triangle then $$\sin^2(A/2)+ \sin^2(B/2) + \sin^2(C/2) =1-2\sin(A/2) \sin(B/2) \sin(C/2)$$

Well I eventually got it after much algebra, but I am looking for a shorter solution, or maybe even a geometric one?

Answer 1

Let $\alpha=\pi-2A$, $\beta=\pi-2B$ and $\gamma=\pi-2C$.

Thus, $\alpha+\beta+\gamma=\pi$ and we need to prove that $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1,$$ which is obvious for acute-angled triangle $ABC$ (it's just law of cosines for new triangle).

In the general case we obtain: $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=$$ $$=\cos^2\alpha+\cos^2\beta+\cos^2(\alpha+\beta)-2\cos\alpha\cos\beta\cos(\alpha+\beta)=$$ $$=\cos^2\alpha+\cos^2\beta+\cos^2\alpha\cos^2\beta+\sin^2\alpha\sin^2\beta-2\sin\alpha\sin\beta\cos\alpha\cos\beta-$$ $$-2\cos\alpha\cos\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=$$ $$=\cos^2\alpha+\cos^2\beta-\cos^2\alpha\cos^2\beta+\sin^2\alpha\sin^2\beta=$$ $$=\cos^2\alpha\sin^2\beta+\cos^2\beta+\sin^2\alpha\sin^2\beta=\sin^2\beta+\cos^2\beta=1.$$ Done!

Answer 2

use the so called half angle formulas: $$\sin(\alpha/2)=\sqrt{\frac{(s-b)(s-a)}{b c}}$$ etc then we get $$\frac{(s-b)(s-c)}{ac}+\frac{(s-a)(s-c)}{ac}+\frac{(s-a)(s-b)}{ab}=1-2\frac{(s-a)(s-b)(s-c)}{abc}$$ with $$s=\frac{a+b+c}{2}$$ simplifying both sides we get $$1/4\,{\frac {{a}^{3}-{a}^{2}b-{a}^{2}c-a{b}^{2}+6\,bca-a{c}^{2}+{b}^{3 }-{b}^{2}c-b{c}^{2}+{c}^{3}}{bca}} $$

Answer 3

A different way of the two precedent.

Because of $\cos^2(X)=1-\sin^2(X)$, the proposed equality is equivalent to $$►\sin^2(A/2)+ \sin^2(B/2)=\cos^2(C/2)-2\sin(A/2) \sin(B/2)\sin(C/2)$$ and, by complementary angles, $$►\cos^2(C/2)=\sin^2(\frac{A+B}{2})=[\sin(A/2)\cos(B/2)+\cos(A/2)\sin(B/2)]^2$$ expanding the square,

$$►\sin^2(A/2)+\sin^2(B/2)-2\sin^2(A/2)\cos^2(B/2)+2\sin(A/2)\cos(A/2)\sin(B/2)\cos(B/2)$$ Finally one has $$0=-\sin(A/2)\sin(B/2)+\cos(A/2)\cos(B/2)-\cos(A/2+B/2)$$ which is the very well-known formula $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$.

Answer 4

As $\dfrac A2+\dfrac B2=\dfrac\pi2-\dfrac C2,\sin\dfrac C2=\cos\dfrac{A+B}2$

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$\sin^2\dfrac A2+\sin^2\dfrac B2+\sin^2\dfrac C2$$

$$=1-\left(\cos^2\dfrac A2-\sin^2\dfrac B2\right)+\sin^2\dfrac C2$$

$$=1-\cos\dfrac{A+B}2\cos\dfrac{A-B}2+\cos^2\dfrac{A+B}2$$

$$=1-\cos\dfrac{A+B}2\left(\cos\dfrac{A-B}2-\cos\dfrac{A+B}2\right)$$

Can you take it from here?