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Integration of Trigonometric Functions

Prove that: ...

Question

Prove that: $c o s θ . c o s 2 θ . c o s 2^{2} θ . c o s 2^{3} θ \dots c o s 2^{n - 1} θ = \frac{s i n 2^{n} θ}{2^{n} s i n θ}$

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Solution

$s i n 2 x = 2 s i n x c o s x$ [double angle formula]
$s i n 4 x$ = $s i n (2 (2 x)) = 2 s i n 2 x c o s 2 x = 4 s i n x c o s x c o s 2 x$
Similarly
$s i n 16 x = 2 s i n (8 x) c o s (8 x)$
$= 2.2 s i n (4 x) c o s (4 x) c o s (8 x)$
$= 2.2.2. s i n (2 x) c o s (2 x) c o s (4 x) c o s (8 x)$
$= 2.2.2.2. c o s (x) s i n (x) c o s (2 x) c o s (4 x) c o s (8 x)$
Now similarly of we generalise this
$s i n (2.2.2.... (n + 1) t i m e s) = 2 c o s (2.2.... (n) t i m e s) s i n (2.2... n t i m e s)$
$2^{2} c o s (2^{n} x) s i n (2^{n} - 1 x) c o s (2^{n} - 1 x)$
$2^{n} c o s (2^{n} x) c o s (2^{n} - 1 x) . . . . . . c o s (4 x) c o s (2 x) s i n (2 x)$
$2^{n} + 1 c o s (2^{n} x) c o s (2^{n} - 1 x) . . . . . . . c o s (2 x) c o s (x) s i n (x)$
$c o s (x) c o s (2 x) c o s (4 x) c o s (8 x) . . . . . . . c o s (2^{n} x) = s i n (2^{n} + 1 x) / (2^{n} + 1 s i n (x))$

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