MHB What is the solution for evaluating $ab+cd$ with given constraints in POTW #477?

[anemone]

Here is this week's POTW:

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Consider that $\{a,\,b,\,c,\,d\}\in \mathbb{R} $ and that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$, evaluate $ab+cd$.

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[anemone]

Congratulations to the following members for their correct solution!

1. Opalg
2. kaliprasad

Solution from Opalg:

Spoiler

If $a^2+b^2 = c^2+d^2 = 1$ then $(a,b)$ and $(c,d)$ are points on the unit circle. So there exist $\theta$ and $\phi$ such that $(a,b) = (\cos\theta,\sin\theta)$ and $(c,d) = (\cos\phi,\sin\phi)$. Therefore $$0 = ac+bd = \cos\theta\cos\phi + \sin\theta\sin\phi = \cos(\theta-\phi).$$ From the formula $\sin x + \sin y = 2\sin\bigl(\frac{x+y}2\bigr)\cos\bigl(\frac{x-y}2\bigr)$ it follows that $$ab+cd = \cos\theta\sin\theta + \cos\phi\sin\phi = \tfrac12\bigl(\sin(2\theta) + \sin(2\phi)\bigr) = \sin(\theta+\phi)\cos(\theta-\phi) = 0.$$