Let triangle ABC satisfy cos A:cos B:cos C = 1:1:2, then find the value of sin A.

So first thing I noted is for the ratio to make sense that triangle ABC is an acute isosceles triangle. Then B=A and C=pi-2A. Note that both cos A and sin A will be positive.

Then cos A/cos(pi-2A) = 1/2, and then apply basic trig identities to reduce this to 2*cos A = 1-2*cos^2 A.

This is quadratic in cos A: 2(cos A)^2 + 2*cos A - 1 = 0, then cos A = (-1+/-sqrt(3))/2. But the negative root must be rejected. That leaves cos A = (-1+sqrt(3))/2.

Then sin A = sqrt[1 - ((-1+sqrt(3))/2)^2] = **sqrt(sqrt(3)/2)** = 0.9306.