Prove that $ABC$ is right-angled

Question

Prove that if $\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$, then $ABC$ is right-angled.

I only found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$, but I have no idea what to do next.

Thank you in advance for your answers!

Answer 1

$$\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$$ Use $\cos^2{A} =\frac{1+\cos2A}2$. Then $$\cos2A+\cos2B+\cos2C=-1$$ Use $\cos2A+\cos2B+\cos2C=-1-4\cos A\cos B\cos C$

Then $$-1-4\cos A\cos B\cos C=-1$$ $$4\cos A\cos B\cos C=0$$ Then $\angle A$ or $\angle B$ or $\angle C$ is $\frac{\pi}{2}$

Answer 2

HINT:

$$F=\cos^2A+\cos^2B+\cos^2C=\cos^2A-\sin^2B+\cos^2C+1$$

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

$$F=\cos(A+B)\cos(A-B)+\cos^2C+1$$

Now as $\cos(A+B)=\cos(\pi-C)=-\cos C,$

$$F=-\cos C\cos(A-B)+\cos C\{-\cos(A+B)\}=1-\cos C(2\cos A\cos B)$$

Answer 3

By $½$-angle relations, $cos²(A)+cos²(B)+cos² (C) = 1 + ½{cos(2A)+cos(2B)} + cos²(C)$

$= 1 + cos(A+B)cos(A−B) + cos²(C) = 1 – cos (C)cos(A−B) + cos²(C),$ using $A+B=π−C$

$= 1 + cos(C){cos(C) −cos(A−B)} = 1 − 2cos (C)sin(½(C−A+B))sin(½(C+A−B))$

$= 1 – 2cos(C)sin(½π−A)sin( ½π−B) = 1 – 2cos(A)cos(B)cos(C)$

$∴ cos²(A)+cos²(B)+cos²(C) = 1 → cos(A)cos (B)cos(C) = 0 → A,B or C = ½π$

Answer 4

As you found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$ note that $\sin^2 \theta + \cos^2 \theta =1$

Then $1-\cos^2 A + 1- \cos^2 B + 1 - \cos^2 C = 2$

$ 3 -(\cos^2 A + \cos^2 B + \cos^2 C)= 2$

$3 - 2 = \cos^2 A + \cos^2 B + \cos^2 C$

$\cos^2 A + \cos^2 B + \cos^2 C =1$