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

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    2017
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The discussion revolves around evaluating the expression $\sin^2 (A+B) +\pi\sin (A+B)\cos (A+B) +\sqrt{2}\cos^2 (A+B)$, given that $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2+\pi x+\sqrt{2}=0$. Participants are encouraged to follow the guidelines for the Problem of the Week (POTW) and submit their solutions. Members Opalg and kaliprasad successfully provided correct solutions to the problem. The thread emphasizes the importance of understanding trigonometric identities and relationships between angles A and B. Engaging with such problems enhances mathematical problem-solving skills.
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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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Congratulations to the following members for their correct solution::)

1. Opalg
2. kaliprasad

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