Trigonometry help (1 Viewer)

cookiez69 · Dec 28, 2012

a) Prove that

provided that

b) Hence find the exact value of

ANSWER:

I've already answered the first question, I just need help with the second one... Please and thank you :]

RealiseNothing · Dec 28, 2012

Let $\theta =\frac{\pi}{8}$ and take the square root of both sides, then simplify.

cookiez69 · Dec 29, 2012

Thanks, so far I've gotten

, which when plugged into the calculator is also sqrt(2)-1, I just need help simplifying that please.

Carrotsticks · Dec 29, 2012

cookiez69 said:
Thanks, so far I've gotten , which when plugged into the calculator is also sqrt(2)-1, I just need help simplifying that please.

$\\ \sqrt{3-2\sqrt{2}} = \sqrt{2-2\sqrt{2}+1} = \sqrt{(\sqrt{2})^2-2\sqrt{2}+(1)^2} = \sqrt{(\sqrt{2}-1)^2} = \pm (\sqrt{2}-1) \\\\ $But since $ \sqrt{3-2 \sqrt{2}} > 0 $ , we must take the positive case and therefore $ \sqrt{3-2\sqrt{2}} = \sqrt{2}-1$

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Trigonometry help (1 Viewer)

cookiez69

What a stupid name, Nat.

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RealiseNothing

what is that?It is Cowpea

cookiez69

What a stupid name, Nat.

Carrotsticks

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