MHB Week #36 - Simplifying Trigonometric Expressions | December 3rd, 2012

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The discussion focuses on simplifying the expression \((a \cos x + b \sin x)^2 + (b \cos x - a \sin x)^2\). Two members provided solutions that both arrive at the same result, \(a^2 + b^2\). The first solution uses direct expansion and trigonometric identities, while the second employs linear combination identities and the Pythagorean identity. Both approaches confirm the simplification is consistent and accurate. The thread highlights the collaborative effort in problem-solving within the community.
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Simplify [math]\left({a \cos x + b \sin x}\right)^2 + \left({b \cos x - a \sin x}\right)^2[/math]
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) Sudharaka
3) soroban
4) BAdhi
5) Siron

Solution #1 (from soroban):[sp] Simplify: $(a\cos x + b\sin x)^2 + (b\cos x - a\sin x)^2$

We have: .[/color]$a^2\cos^2\!x + 2ab\sin x\cos x + b^2\sin^2\!x + b^2\cos^2\!x - 2ab\sin x\cos x + a^2\sin^2\!x $

. . . . . [/color]$=\;a^2\sin^2\!x + a^2\cos^2\!x + b^2\sin^2\!x + b^2\cos^2\!x $

. . . . . [/color]$=\;a^2\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}} + b^2\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}}$

. . . . . [/color]$=\;a^2 + b^2$
[/size][/sp]

Solution #2 (from MarkFL):[sp]Using linear combination identities, we have:

$\displaystyle (a^2+b^2)\sin^2\left(x+\tan^{-1}\left(\frac{a}{b} \right) \right)+(a^2+b^2)\sin^2\left(x-\cot^{-1}\left(\frac{a}{b} \right)+\pi \right)$

Using the identity:

$\displaystyle \tan^{-1}(\theta)+\cot^{-1}(\theta)=\frac{\pi}{2}$ we may write:

$\displaystyle (a^2+b^2)\sin^2\left(x+\tan^{-1}\left(\frac{a}{b} \right) \right)+(a^2+b^2)\sin^2\left(x+\tan^{-1}\left(\frac{a}{b} \right)+\frac{\pi}{2} \right)$

Using the identity $\displaystyle \sin\left(\theta+\frac{\pi}{2} \right)=\cos(\theta)$, we have:

$\displaystyle (a^2+b^2)\sin^2\left(x+\tan^{-1}\left(\frac{a}{b} \right) \right)+(a^2+b^2)\cos^2\left(x+\tan^{-1}\left(\frac{a}{b} \right) \right)$

$\displaystyle (a^2+b^2)\left(\sin^2\left(x+\tan^{-1}\left(\frac{a}{b} \right) \right)+\cos^2\left(x+\tan^{-1}\left(\frac{a}{b} \right) \right) \right)$

Using the Pythagorean identity $\displaystyle \sin^2(\theta)+\cos^2(\theta)=1$ we are left with:

$\displaystyle a^2+b^2$[/sp]
 
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