Another Verify the identity that I can't get:
$$\sec^2 \frac{x}{2} = \frac{2}{1+\cos x}$$
$$ = \frac{1 + \left(\frac{1}{\cos x}\right)}{2}$$
$$ = \frac{\cos x + 1}{2 \cos x}$$
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This is confusing. $(\sec^2x)/2$ or $\sec^2{x/2}$? – Yiyuan Lee Mar 19 '14 at 08:03
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It's the second one. sec^2 (x/2) – KKendall Mar 19 '14 at 08:03
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Sorry, just changed the title – KKendall Mar 19 '14 at 08:04
1 Answers
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Again, both ways are possible :
Going backwards (which is simpler, coincidentally)
$$\frac{2}{1 + \cos x} = \frac{2}{1 + (2\cos^2\frac{x}{2} - 1)}\\ = \frac{2}{2\cos^2\frac{x}{2}}\\ = \sec^2\frac{x}{2}$$
Going forward, start by rewriting as:
$$\sec^2\frac{x}{2} = \frac{1}{\cos^2\frac{x}{2}}$$
By the double angle formula for $\cos$, $\cos 2\theta = 2\cos^2\theta - 1$ we have (by letting $\theta = \frac{x}{2}$):
$$\cos x = 2\cos^2\frac{x}{2} - 1$$
Rearrange to get $$\cos^2\frac{x}{2} = \frac{1 + \cos x}{2}$$
Simply substitute this back, to deduce
$$\sec^2\frac{x}{2} = \frac{1}{\frac{1 + \cos x}{2}} = \frac{2}{1 + \cos x}$$
It's probably worth mentioning that both approaches are equivalent, differing only by which direction you choose to prove the identity.
Yiyuan Lee
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@KKendall No problem! Back in my days (actually, just a year ago), our teachers gave us countless drills on trigo questions (in fact, questions on probably all other topics as well). I've probably seen this approach somewhere, so the problem really is familiar to me. – Yiyuan Lee Mar 19 '14 at 08:19