Solution for POTW #259: Finding the Value of a Trigonometric Expression

[anemone]

Here is this week's POTW:

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Suppose $\tan A$ and $\tan B$ are the roots of $x^2+\pi x+\sqrt{2}=0$. Evaluate

$\sin^2 (A+B) +\pi\sin (A+B)\cos (A+B) +\sqrt{2}\cos^2 (A+B)$

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[anemone]

Congratulations to the following members for their correct solution::)

1. Opalg
2. kaliprasad

Solution from Opalg:

Spoiler

Let $f(x) = x^2 + \pi x + \sqrt2$.

The sum of the roots of $f(x)$ is $\tan A+ \tan B = -\pi$, and their product is $\tan A\tan B = \sqrt2$. Therefore $$\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A\tan B} = \frac{-\pi}{1-\sqrt2} = \frac{\pi}{\sqrt2 - 1}.$$

It follows that $$\begin{aligned} \sin^2 (A+B) +\pi\sin (A+B)\cos (A+B) +\sqrt{2}\cos^2 (A+B) &= \cos^2 (A+B)\bigl(\tan^2 (A+B) +\pi\tan (A+B) +\sqrt{2}\bigr) \\ &= \cos^2 (A+B) \, f\bigl(\tan(A+B)\bigr) \\ &= \frac{f\bigl(\tan(A+B)\bigr)}{\sec^2(A+B)} \\ &= \frac{f\bigl(\tan(A+B)\bigr)}{1 + \tan^2(A+B)} \\ &= \frac{\frac{\pi^2}{(\sqrt2-1)^2} + \pi\frac{\pi}{\sqrt2-1} + \sqrt2}{1 + \frac{\pi^2}{(\sqrt2-1)^2}} \\ &= \frac{\pi^2 + \pi^2(\sqrt2-1) + \sqrt2(\sqrt2-1)^2}{(\sqrt2-1)^2 + \pi^2} \\ &= \frac{\sqrt2 \bigl(\pi^2 + (\sqrt2-1)^2\bigr)}{(\sqrt2-1)^2 + \pi^2} \\ &= \sqrt2. \end{aligned} $$