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Problem of the week #36 - December 3rd, 2012

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Jameson

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Jan 26, 2012
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Jameson

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Jan 26, 2012
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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):
Simplify: $(a\cos x + b\sin x)^2 + (b\cos x - a\sin x)^2$

We have: .$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 $

. . . . . $=\;a^2\sin^2\!x + a^2\cos^2\!x + b^2\sin^2\!x + b^2\cos^2\!x $

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

. . . . . $=\;a^2 + b^2$


Solution #2 (from MarkFL):
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:

$\displaystyle a^2+b^2$
 
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