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 4 of 6: Factor the equation into a product
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In plain words

A sum of two terms becomes a single product exactly when the two cosines are opposite, which is precisely what step 3 established.

cos⁡(α+γ2)(acos⁡β−bcos⁡α)=0\cos\left(\alpha+\frac{\gamma}{2}\right)\bigl(a\cos\beta - b\cos\alpha\bigr) = 0
Detailed analysis

Substituting cos⁡(β+γ2)=−cos⁡(α+γ2)\cos(\beta+\frac{\gamma}{2})=-\cos(\alpha+\frac{\gamma}{2}) from step 3 into the identity of step 2 turns acos⁡βcos⁡(α+γ2)−bcos⁡αcos⁡(α+γ2)=0a\cos\beta\cos(\alpha+\frac{\gamma}{2}) - b\cos\alpha\cos(\alpha+\frac{\gamma}{2}) = 0, i.e. the product shown.