MathLabs

Problem 2

If aa, bb, and cc are the sides and α\alpha, β\beta, and γ\gamma are the respective angles of a triangle for which a+b=tan⁡γ2(atan⁡α+btan⁡β)a+b=\tan\frac{\gamma}{2}\left(a\tan\alpha+b\tan\beta\right), prove that the triangle is isosceles.
Step 3 of 6: Use α+β+γ=π\alpha+\beta+\gamma=\pi to relate the two cosines
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In plain words

This is a triangle-angle identity that holds for every triangle, not just this one — it is what lets the two cosine terms in step 2 cancel into a single factor.

cos⁡(β+γ2)=−cos⁡(α+γ2)\cos\left(\beta+\frac{\gamma}{2}\right) = -\cos\left(\alpha+\frac{\gamma}{2}\right)
Detailed analysis

Since γ=π−α−β\gamma=\pi-\alpha-\beta, we get α+γ2=π2−β−α2\alpha+\frac{\gamma}{2}=\frac{\pi}{2}-\frac{\beta-\alpha}{2} and β+γ2=π2+β−α2\beta+\frac{\gamma}{2}=\frac{\pi}{2}+\frac{\beta-\alpha}{2}. Because cos⁡(π2−x)=sin⁡x\cos(\frac{\pi}{2}-x)=\sin x and cos⁡(π2+x)=−sin⁡x\cos(\frac{\pi}{2}+x)=-\sin x, the two cosines are negatives of each other for any triangle, independently of the equation from step 2.