Shorter solution to this trig question? (1 Viewer)

calamebe · Sep 9, 2015

Hey, I did a question and I ended up taking a whole page of working out, and I'm guessing I did it the complicated way. I'll post my working and the question, any shorter methods would be appreciated, thanks!
http://imgur.com/3W1niXM
http://imgur.com/85sr1Z4

kawaiipotato · Sep 9, 2015

I've found a way that's shorter but i'm not sure if Mathematics students learn this.
$$ Convert $ cos^{2}\theta $ into $ \frac{1+cos2\theta}{2}$

$(\sqrt3 + 1)(\frac{1+cos2\theta}{2}) - 1 - ( \sqrt3 - 1)sin\theta cos\theta = 0$

$(\sqrt3 + 1)(1+cos2\theta) - 2 - 2(\sqrt3 - 1)sin\theta cos\theta = 0$

$(\sqrt3 + 1)(cos2\theta) + \sqrt3 - 1 - (\sqrt3 - 1)sin2\theta = 0$

$(\sqrt3 + 1)(cos2\theta) - (\sqrt3 - 1)sin2\theta = 1 - \sqrt3$

$$ Consider $ (\sqrt3 + 1)(cos2\theta) - (\sqrt3 - 1)sin2\theta \equiv Rcos(2\theta + \alpha)$

$(\sqrt3 + 1)(cos2\theta) - (\sqrt3 - 1)sin2\theta \equiv Rcos2\theta cos\alpha - Rsin2\theta sin\alpha$

$\therefore (\sqrt3 + 1) = Rcos\alpha$

$\therefore (\sqrt3 - 1) = Rsin\alpha$

$$ Squaring and adding, $ R^2 = 8$

$R = 2\sqrt2 , R>0$

$$ Dividing the two equations, $ \tan\alpha = \frac{\sqrt3 - 1}{\sqrt3 + 1}$

$$ Note that $ \frac{\sqrt3 - 1}{\sqrt3 + 1} = \frac{\tan(\frac{\pi}{3}) - \tan(\frac{\pi}{4})}{\tan(\frac{\pi}{3}) + \tan( \frac{\pi}{4})}$

$= \tan(\frac{\pi}{3} - \frac{\pi}{4}) = \tan(\frac{\pi}{12})$

$$ Hence, $ \tan\alpha = \tan(\frac{\pi}{12})$

$\alpha = \frac{\pi}{12} , $for $ 0 \leq \alpha \leq \frac{\pi}{2}$

$\therefore 2\sqrt2 cos(2\theta + \frac{\pi}{12}) = 1 - \sqrt3$

$Then just solve for 2\theta$

Maybe someone else has a diff. method

calamebe · Sep 9, 2015

kawaiipotato said:
I've found a way that's shorter but i'm not sure if Mathematics students learn this.
$$ Convert $ cos^{2}\theta $ into $ \frac{1+cos2\theta}{2}$

$(\sqrt3 + 1)(\frac{1+cos2\theta}{2}) - 1 - ( \sqrt3 - 1)sin\theta cos\theta = 0$

$(\sqrt3 + 1)(1+cos2\theta) - 2 - 2(\sqrt3 - 1)sin\theta cos\theta = 0$

$(\sqrt3 + 1)(cos2\theta) + \sqrt3 - 1 - (\sqrt3 - 1)sin2\theta = 0$

$(\sqrt3 + 1)(cos2\theta) - (\sqrt3 - 1)sin2\theta = 1 - \sqrt3$

$$ Consider $ (\sqrt3 + 1)(cos2\theta) - (\sqrt3 - 1)sin2\theta \equiv Rcos(2\theta + \alpha)$

$(\sqrt3 + 1)(cos2\theta) - (\sqrt3 - 1)sin2\theta \equiv Rcos2\theta cos\alpha - Rsin2\theta sin\alpha$

$\therefore (\sqrt3 + 1) = Rcos\alpha$

$\therefore (\sqrt3 - 1) = Rsin\alpha$

$$ Squaring and adding, $ R^2 = 8$

$R = 2\sqrt2 , R>0$

$$ Dividing the two equations, $ \tan\alpha = \frac{\sqrt3 - 1}{\sqrt3 + 1}$

$$ Note that $ \frac{\sqrt3 - 1}{\sqrt3 + 1} = \frac{\tan(\frac{\pi}{3}) - \tan(\frac{\pi}{4})}{\tan(\frac{\pi}{3}) + \tan(\frac{\pi}{4}}$

$= \tan(\frac{\pi}{3} - \frac{\pi}{4}) = \tan(\frac{\pi}{12})$

$$ Hence, $ \tan\alpha = \tan(\frac{\pi}{12})$

$\alpha = \frac{\pi}{12} , $for $ 0 \leq \alpha \leq \frac{\pi}{2}$

$\therefore 2\sqrt2 cos(2\theta + \frac{\pi}{12}) = 1 - \sqrt3$

$Then just solve for 2\theta$

Maybe someone else has a diff. method

Well I get all of it so Ext 1 students definitely learnt all of that, and I don't think that any of that is extension stuff.

psyc1011 · Sep 9, 2015

Another possiblemethod

$(\sqrt{3} +1) \cos ^2 \theta - 1 = (\sqrt{3} - 1) \sin \theta \cos \theta$

1 = s^2 + c^2

$\sqrt{3} \cos ^2 \theta - \sin ^2 \theta = (\sqrt{3} - 1) \sin \theta \cos \theta$

Divide by c^2 and note that cos(theta) != 0, i.e. theta != ...

$\sqrt{3} - \tan ^2 \theta = (\sqrt{3} -1)\tan \theta$

$\tan ^2 \theta + (\sqrt{3} - 1)\tan \theta - \sqrt{3}=0$

Then solve quadratic but continuation is brutal lol

$\tan \theta = \frac{1-\sqrt{3} \pm \sqrt{4+2\sqrt{3}}}{2}$

By calculator,

$\theta = \tan ^{-1} \frac{1-\sqrt{3} - \sqrt{4+2\sqrt{3}}}{2} = -60 + k(180) = 120, 300$

$\theta = \tan ^{-1} \frac{1-\sqrt{3} + \sqrt{4+2\sqrt{3}}}{2} = 45 + k(180) = 45, 225$

VBN2470 · Sep 9, 2015

Bring everything to the LHS and factorise to get $(\sqrt{3}\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=0 \Rightarrow \theta=\frac{\pi}{4}, \frac{5\pi}{4}, \frac{2\pi}{3}, \frac{5\pi}{3}$ where $0\leq{\theta}\leq{2\pi}$ .

Shorter solution to this trig question? (1 Viewer)

calamebe

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kawaiipotato

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calamebe

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psyc1011

#truth

VBN2470

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