Ltn -cosx-2-cosx-4-cosx-8-cosx-2n-

The correct option is B sin x/xcos θ. cos 2θ…… cos 2^n 1θ = sin 2^n θ/2^n sin θ Cos x/2. cos x/4…… cos x/2^n=sin x/2^n sin ( x/2^n ) ⇒Ltn →∞sin x/2^n sin ( x/2 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

Ltn →∞cosx/2·...

Question

$L t n \to \infty {c o s \frac{x}{2} . c o s \frac{x}{4} . c o s \frac{x}{8} \dots \dots c o s \frac{x}{2^{n}}}$ =

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s i n x

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\frac{s i n x}{x}

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Solution

The correct option is C $\frac{s i n x}{x}$
$c o s θ . c o s 2 θ \dots \dots c o s 2^{n - 1} θ = \frac{s i n 2^{n} θ}{2^{n} s i n θ}$
$C o s \frac{x}{2} . c o s \frac{x}{4} \dots \dots c o s \frac{x}{2^{n}} = \frac{s i n x}{2^{n} s i n (\frac{x}{2^{n}})}$
$\Rightarrow L t n \to \infty \frac{s i n x}{2^{n} s i n (\frac{x}{2^{n}})} = \frac{s i n x}{x}$

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

Q. Evaluate:

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Ltn→∞{cos x2.cos x4.cos x8……cos x2n}=

The correct option is C sin xxcos θ.cos 2θ……cos 2n−1θ=sin 2nθ2n sin θ Cos x2.cos x4……cos x2n=sin x2n sin(x2n) ⇒Ltn→∞sin x2n sin(x2n)=sin xx

$L t n \to \infty {c o s \frac{x}{2} . c o s \frac{x}{4} . c o s \frac{x}{8} \dots \dots c o s \frac{x}{2^{n}}}$ =