Evaluate- limn-cosx-2cosx-4cosx-8-cosx-2n

The correct option is C sin xx cos(x2)cos(x4)… cos(x2^n) #160; = sin(x)2sin(x2)sin(x2)2sin(x4)sin(x4)2sin(x8)…sin(x2^n 1)2sin(x2^n) ......Using for ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

Evaluate: l...

Question

Evaluate: $lim n \to \infty cos \frac{x}{2} cos \frac{x}{4} cos \frac{x}{8} \dots cos \frac{x}{2^{n}}$

1

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\frac{sin x}{x}

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\frac{x}{sin x}

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2

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Solution

The correct option is C $\frac{sin x}{x}$
$c o s (\frac{x}{2}) c o s (\frac{x}{4}) \dots c o s (\frac{x}{2^{n}}) = \frac{s i n (x)}{2 s i n (\frac{x}{2})} \frac{s i n (\frac{x}{2})}{2 s i n (\frac{x}{4})} \frac{s i n (\frac{x}{4})}{2 s i n (\frac{x}{8})} \dots \frac{s i n (\frac{x}{2^{n - 1}})}{2 s i n (\frac{x}{2^{n}})}$ ......Using formula $(s i n 2 x = 2 s i n x . c o s x)$

$c o s (\frac{x}{2}) c o s (\frac{x}{4}) \dots c o s (\frac{x}{2^{n}}) = \frac{s i n (x)}{2^{n} s i n (\frac{x}{2^{n}})}$

$lim n \to \infty c o s (\frac{x}{2}) c o s (\frac{x}{4}) \dots c o s (\frac{x}{2^{n}}) = lim n \to \infty \frac{s i n (x)}{2^{n} s i n (\frac{x}{2^{n}})}$

$lim n \to \infty \frac{s i n (x)}{x \cdot \frac{s i n (\frac{x}{2^{n}})}{\frac{x}{2^{n}}}} = \frac{s i n x}{x}$

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Similar questions

Q. Evaluate:

lim n \to \infty (cos \frac{x}{2} cos \frac{x}{4} cos \frac{x}{8} . . . cos \frac{x}{2^{n}})

${lim}_{n \to \infty} {c o s \frac{x}{2} c o s \frac{x}{4} c o s \frac{x}{8} . . . . . c o s \frac{x}{2^{n}}}$ =

Q. The value of

lim n \to \infty cos (\frac{x}{2}) cos (\frac{x}{4}) cos (\frac{x}{8}) . . . cos (\frac{x}{2^{n}})

Q. Given that

cos \frac{x}{2} cos \frac{x}{4} cos \frac{x}{8} \dots = \frac{sin x}{x}

, then expression

\frac{1}{2^{2}} {sec}^{2} \frac{x}{2} + \frac{1}{2^{4}} {sec}^{2} \frac{x}{4} + \dots

equals

Q. The value of:

lim n \to 0 cos (\frac{x}{2}) cos (\frac{x}{4}) cos (\frac{x}{8}) . . . . . . cos (\frac{x}{2^{n}})

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Evaluate: limn→∞cosx2cosx4cosx8…cosx2n

Evaluate: $lim n \to \infty cos \frac{x}{2} cos \frac{x}{4} cos \frac{x}{8} \dots cos \frac{x}{2^{n}}$