The value of cos-22-cos-23-cos-210-sin-210 is -

The correct option is C 1512A=cosπ2^2·cosπ2^3· ... ·cosπ2^10·sinπ2^10 Put π2^10=x⇒π=2^10x ⇒ nbsp; A= (cos x·cos2x· ... ·cos 2^8x)sin x As we know sin 2^nθ2^n ...

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Standard XII

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

The value of ...

Question

The value of $cos \frac{π}{2^{2}} \cdot cos \frac{π}{2^{3}} \cdot . . . \cdot cos \frac{π}{2^{10}} \cdot sin \frac{π}{2^{10}}$ is :

\frac{1}{1024}

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\frac{1}{2}

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\frac{1}{512}

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\frac{1}{256}

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Solution

The correct option is C $\frac{1}{512}$
$A = cos \frac{π}{2^{2}} \cdot cos \frac{π}{2^{3}} \cdot . . . \cdot cos \frac{π}{2^{10}} \cdot sin \frac{π}{2^{10}}$
Put $\frac{π}{2^{10}} = x \Rightarrow π = 2^{10} x$
$\Rightarrow A = (cos x \cdot cos 2 x \cdot . . . \cdot cos 2^{8} x) sin x$
As we know
$\frac{sin 2^{n} θ}{2^{n} sin θ} = cos θ \cdot cos 2 θ \cdot . . . cos 2^{n - 1} θ \cdot$
Here, $n - 1 = 8 \Rightarrow n = 9$
$\Rightarrow A = \frac{sin 2^{9} x}{2^{9} sin x} \cdot sin x$
$\Rightarrow A = \frac{sin 2^{9} \frac{π}{2^{10}}}{2^{9}}$
$\Rightarrow A = \frac{1}{512}$

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The value of cosπ22⋅cosπ23⋅ ... ⋅cosπ210⋅sinπ210 is :

The correct option is C 1512A=cosπ22⋅cosπ23⋅ ... ⋅cosπ210⋅sinπ210 Put π210=x⇒π=210x ⇒A=(cosx⋅cos2x⋅ ... ⋅cos28x)sinx As we know sin2nθ2nsinθ=cosθ⋅cos2θ⋅ ... cos2n−1θ⋅ Here, n−1=8⇒n=9 ⇒A=sin29x29sinx⋅sinx ⇒A=sin29π21029 ⇒A=1512

The value of $cos \frac{π}{2^{2}} \cdot cos \frac{π}{2^{3}} \cdot . . . \cdot cos \frac{π}{2^{10}} \cdot sin \frac{π}{2^{10}}$ is :