# Show that $x=y=z$ where $\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi$

Given that $$\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi.$$ Also given that $$x+y+z=\frac{3}{2}.$$ Then prove that $$x=y=z.$$
My attempt: Let us assume $$\cos^{-1} x=a,\> \cos^{-1} y =b, \> \cos^{-1} z=c.$$ Then we have $$a+b+c=\pi \implies a+b = \pi - c.$$ This follows that \begin{align*} \cos(a+b)=\cos(\pi - c) & \implies \cos a \cos b - \sin a \sin b = - \cos c \\ & \implies xy-\sqrt{1-x^2} \sqrt{1-y^2}=z.\end{align*} Now i am not able to proceed from here. Please help me to solve this.

Another way to solve the same could be $$\cos^{-1}x=A$$, $$\cos^{-1}y=B$$ and $$\cos^{-1}z=C$$

and thus $$A+B+C=\pi$$ and the condition becomes $$\cos A+\cos B+\cos C=\frac{3}{2}$$,

which can be simplified to $$2\cos\frac{A+B}{2}\frac{A-B}{2}+1-2\sin^2{\frac{c}{2}}=\frac{3}{2}$$, and we know $$\cos\frac{A+B}{2}=\cos\frac{\pi-C}{2}=\sin\frac{C}{2}$$

Rearranging as quadratic in $$\sin\frac{c}{2}$$ we can write the equation as $$2\sin^2\frac{c}{2}-2\cos\frac{A-B}{2}\sin\frac{c}{2}+\frac{1}{2}=0$$

For real roots $$D\ge0$$, so we get $$4\cos^2\frac{A-B}{2}-4\ge0$$

i.e. $$\sin^2\frac{A-B}{2}\le0$$ which is only possible when $$\sin^2\frac{A-B}{2}=0$$

Therefore, $$A=B=C=\frac{\pi}{3}\Rightarrow x=y=z=\frac{1}{2}$$

With your substitution you have $$a,b,c\in [0,2\pi],\ a+b+c=\pi$$ and

$$\cos a+\cos b+\cos c=\frac{3}{2}$$

We have that:

$$\cos c=1-2\sin^2 \frac{c}{2},\ \cos a+\cos b=2\cos \frac{a+b}{2}\cos\frac{a-b}{2}$$

Therefore:

$$\cos a+\cos b+\cos c=2\cos \frac{a+b}{2}\cos\frac{a-b}{2}+\cos c\\ =2\sin \frac{c}{2}\cos \frac{a-b}{2}+1-2\sin^2\frac{c}{2}\leq 2\sin \frac{c}{2}+1-2\sin^2\frac{c}{2}$$

Therefore:

$$\frac{3}{2}\leq 2\sin \frac{c}{2}+1-2\sin^2\frac{c}{2}\Rightarrow 2\left(\sin \frac{c}{2}-\frac{1}{2}\right)^2\leq 0$$

This implies $$c=\frac{\pi}{3}$$. Similarly we can prove that $$a=b=\frac{\pi}{3}$$, which gives $$x=y=z=\frac{1}{2}$$.

• Sir this solution is fine, but I want a basic algebraic solution without Jensem's inequality. – abcdmath Mar 8 '20 at 7:53
• @abcdmath, I have edited the solution. – LHF Mar 8 '20 at 8:01

This is equivalent to the equation of the sum of cosines of the tree angles of triangle,

\begin{align} \cos \alpha+\cos\beta+\cos\gamma &=\frac{3}{2} , \end{align}

it is well-known that this sum can be expressed in terms of $$r$$ and $$R$$, the radii or inscribed and circumscribed circle, respectively as

\begin{align} \cos\alpha+\cos\beta+\cos\gamma &=\frac rR+1=\frac{3}{2} , \end{align}

which gives \begin{align} R&=2\,r , \end{align} and this can be true only for the equilateral triangle.