Answered: "Compute limx→π2(1−sin⁡x)(1−sin2x)…(1−sinnx)cos2nx"

Addison Fuller

Answered question

2022-03-25

Compute

lim_{x \to \frac{π}{2}} \frac{(1 - \sin x) (1 - \sin^{2} x) \dots (1 - \sin^{n} x)}{\cos^{2 n} x}

Chelsea Chen

Beginner2022-03-26Added 5 answers

lim_{x \to \frac{π}{2}} \frac{(1 - \sin x) (1 - \sin^{2} x) \dots (1 - \sin^{n} x)}{\cos^{2 n} x} = lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{\cos^{2} x} \cdot \frac{1 - \sin^{2} x}{\cos^{2} x} \cdot \frac{1 - \sin^{n} x}{\cos^{2} x}

= \frac{1}{2} \cdot \frac{2}{2} \dots \frac{n}{2} = \frac{n!}{2^{n}}

because, near

\frac{π}{2}, 1 - \sin^{k} x

beahaves like

\frac{k}{2} {(x - \frac{π}{2})}^{2}

whereas

\cos^{2} x

behaves like

{(x - \frac{π}{2})}^{2}

cineworld93uowb

Beginner2022-03-27Added 16 answers

Note that

lim_{x \to \frac{π}{2}} \frac{1 - \sin^{k} (x)}{1 - \sin (x)} = lim_{x \to \frac{π}{2}} \sum_{j = 0}^{k - 1} \sin^{j} (x)

=k
and

lim_{x \to \frac{π}{2}} \frac{\cos^{2 n} (x)}{{(1 - \sin (x))}^{n}} = lim_{x \to \frac{π}{2}} {(1 + \sin (x))}^{n}

= 2^{n}

Thus, after dividing the numerator and denominator by

{(1 - \sin (x))}^{n}

we get

lim_{x \to \frac{π}{2}} \frac{\prod_{k = 1}^{n} (1 - \sin^{k} (x))}{\cos^{2 n} (x)} = \frac{n!}{2^{n}}

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