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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

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प्रश्न

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

 

विकल्प

  • -1

  • 1

  • 1/2

  • None of these

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उत्तर

1

\[\left( 2^n + 1 \right)x = \pi \left( \text{ Given } \right)\]

\[ \Rightarrow 2^n x + x = \pi\]

\[ \Rightarrow 2^n x = \pi - x\]

\[ \Rightarrow \sin 2^n x = \sin\left( \pi - x \right)\]

\[ \Rightarrow \sin 2^n x = \sin x . . . (1) \]

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

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.5 [पृष्ठ ४५]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.5 | Q 27 | पृष्ठ ४५

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