If n-1-2-3- - then cos-cos 2 -cos 4 -cos 2n-1-A- sin 2 n 0-1 sin 2 n-C- 5114-1-4n-1sinD- sin 2 n-2 n sin-

The correct option is D sin 2^n α/2^n sin α ∵ α cos 2 α cos 4 α⋯ cos 2^n 1α =sin 2^n α/2^n sin α ...

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General Solution of Trigonometric Equation

If n=1,2,3, …...

Question

If n = 1, 2, 3, . . ., then $c o s α c o s 2 α$

$c o s 4 α \dots c o s 2^{n - 1} α$

$\frac{s i n 2 n α}{2 n s i n α}$

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$\frac{s i n 2^{n} α}{2^{n} s i n^{n - 1} α}$

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$\frac{s i n 4^{n - 1} α}{4^{n - 1} s i n α}$

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$\frac{s i n 2^{n} α}{2^{n} s i n α}$

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Solution

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Similar questions

n = 1, 2, 3, . . ., then \cos α \cos 2 α \cos 4 α . . . \cos 2^{n - 1} α

\frac{\sin 2 n α}{2 n \sin α}

\frac{\sin 2^{n} α}{2^{n} \sin 2^{n - 1} α}

\frac{\sin 4^{n - 1} α}{4^{n - 1} \sin α}

\frac{\sin 2^{n} α}{2^{n} \sin α}

R = {cos}^{2} (\frac{n - 1}{n}) π + {cos}^{2} (\frac{n - 2}{n}) π + \dots + {cos}^{2} (\frac{2 π}{n}) + {cos}^{2} (\frac{π}{n})

R

t_{n} = \frac{1}{4} (n + 2) (n + 3)

n = 1, 2, 3, . . .

\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + . . . \frac{1}{t_{2003}} =

t_{n} = \frac{1}{4} (n + 2) (n + 3)

n = 1, 2, 3....

\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + . . . . . . . . . + \frac{1}{t_{2003}} =

t_{n} = \frac{1}{4} (n + 2) (n + 3)

n = 1, 2, 3

\frac{1}{t_{I}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + \dots \dots . . + \frac{1}{t_{20}} =

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