# When is $4 ab \sin^2 θ = (a+b)^2$ ?

I know that by trial and error it is only possible when $a=b$, but what is the actual solution process?

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–  lab bhattacharjee Apr 16 at 3:07

By Pythagoras

$$4ab\sin^2(\theta)=4ab(1-\cos^2(\theta))$$

Now subtract $4ab$ on both sides of your equation to see

$$-4ab\cos^2(\theta)=(a+b)^2-4ab=(a-b)^2\ge 0$$

Therefore we must have either $a=b$ or $a,b$ must have different signs. But in the latter case the original equation implies $a=-b$. If $(a,b)\not=(0,0)$, then this also determines the angle $\theta$.

So the only solutions are

$$\left\{\begin{array}{l} a=b, \theta=(2k+1)\frac{\pi}{2}, k\in\mathbb{Z}\\ a=-b, \theta=k\pi, k\in\mathbb{Z}\\ a=b=0, \theta\;\text{arbitrary} \end{array}\right.$$

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