If \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]

1 \[\left( 2^n + 1 \right)x = \pi \left( \text{ Given } \right)\] \[ \Rightarrow 2^n x + x = \pi\] \[ \Rightarrow 2^n x = \pi - x\] \[ \Rightarrow \sin 2^n x = \sin\left( \pi - x \right)\] \[ \Rightarrow \sin 2^n x = \sin x . . . (1) \] \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 2^n \times \frac{\sin 2^n x}{2^n \sin x}\]\[ = \frac{\sin 2^n x}{\sin x}\]\[ = \frac{\sin x}{\sin x} \left[ \text{ From } (1) \right]\]

If ( 2 N + 1 ) X = π , Then 2 N Cos X Cos 2 X Cos 2 2 X . . . Cos 2 N − 1 X = 1 - Mathematics

प्रश्न

If $(2^{n} + 1) x = π,$ then $2^{n} \cos x \cos 2 x \cos 2^{2} x . . . \cos 2^{n - 1} x = 1$

पर्याय

-1
1
1/2
None of these

MCQ

उत्तर

$(2^{n} + 1) x = π (Given)$

$\Rightarrow 2^{n} x + x = π$

$\Rightarrow 2^{n} x = π - x$

$\Rightarrow \sin 2^{n} x = \sin (π - x)$

$\Rightarrow \sin 2^{n} x = \sin x . . . (1)$

$2^{n} \cos x \cos 2 x \cos 2^{2} x . . . \cos 2^{n - 1} x = 2^{n} \times \frac{\sin 2^{n} x}{2^{n} \sin x}$
$= \frac{\sin 2^{n} x}{\sin x}$
$= \frac{\sin x}{\sin x} [From (1)]$

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 27 Q 26 Q 28

पाठ 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.5 [पृष्ठ ४५]

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पाठ 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.5 | Q 27 | पृष्ठ ४५

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If ( 2 N + 1 ) X = π , Then 2 N Cos X Cos 2 X Cos 2 2 X . . . Cos 2 N − 1 X = 1 - Mathematics

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If ( 2 N + 1 ) X = π , Then 2 N Cos X Cos 2 X Cos 2 2 X . . . Cos 2 N − 1 X = 1 - Mathematics

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