prove that: cos α×cos 2α×cos 2^2 α...cos 2^n α =((sin 2^(n+1) α)/(2^(n+1) sin α))


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Question Number 72145 by Maclaurin Stickker last updated on 25/Oct/19

$p r o v e t h a t :$ $\cos α \times \cos 2 α \times c o s 2^{2} α . . . c o s 2^{n} α$ $= \frac{\sin 2^{n + 1} α}{2^{n + 1} \sin α}$
Commented by kaivan.ahmadi last updated on 25/Oct/19

$f i r s t w a y : w e s o l v e b y i n d u c t i o n$ $n = 1 \Rightarrow \frac{s i n 2^{2} α}{2^{2} s i n α} = \frac{s i n 2 α c o s 2 α}{2 s i n α} = c o s α c o s 2 α$ $i n d u c t i o n h y p o t h e s i s$ $n = k - 1 \Rightarrow c o s α \times c o s 2 α \times . . . \times c o s 2^{k - 1} α = \frac{s i n 2^{k} α}{2^{k} s i n α}$ $n o w$ $n = k \Rightarrow c o s α \times c o s 2 α \times . . . \times c o s 2^{k - 1} α \times c o s 2^{k} α =$ $\frac{s i n 2^{k} α c o s 2^{k} α}{2^{k} s i n α} = \frac{\frac{1}{2} s i n 2^{k + 1} α}{2^{k} s i n α} = \frac{s i n 2^{k + 1} α}{2^{k + 1} s i n α} .$ $s e c o n d w a y :$ $\frac{s i n 2^{n + 1} α}{2^{n + 1} s i n α} = \frac{s i n 2^{n} α c o s 2^{n} α}{2^{n} s i n α} = \frac{s i n 2^{n - 1} α c o s 2^{n - 1} α c o s 2^{n} α}{2^{n - 1} s i n α} =$ $. . . = \frac{s i n 2^{0} α c o s 2^{0} α c o s 2 α . . . c o s 2^{n} α}{s i n α} =$ $c o s α c o s 2 α . . . c o s 2^{n} α$
Commented by Maclaurin Stickker last updated on 25/Oct/19

$T h a n k y o u f o r y o u r t i m e .$
Commented by mathmax by abdo last updated on 25/Oct/19
$let A_n =cos(α)cos(2α)....cos(2^n α) ⇒A_n =Π_(k=0) ^n cos(2^k α) let B_n =sin(α)sin(2α)....sin(2^n α)=Π_(k=0) ^n sin(2^k α) ⇒ A_n .B_n =Π_(k=0) ^n cos(2^k α)sin(2^k α)=Π_(k=0) ^n {(1/2)sin(2^(k+1) α)} =(1/2^(n+1) )Π_(k=0) ^n sin(2^(k+1) α) =_(k+1=i) (1/2^(n+1) )Π_(i=1) ^(n+1) sin(2^i α) =(1/2^(n+1) )Π_(i=0) ^n sin(2^i α)((sin(2^(n+1) α))/(sinα)) =((sin(2^(n+1) α))/(2^n sinα)) ×B_n but B_n ≠0 ⇒ A_n =((sin(2^(n+1) α))/(2^(n+1) sin(α)))$
$l e t A_{n} = c o s (α) c o s (2 α) . . . . c o s (2^{n} α) \Rightarrow A_{n} = \prod_{k = 0}^{n} c o s (2^{k} α)$ $l e t B_{n} = s i n (α) s i n (2 α) . . . . s i n (2^{n} α) = \prod_{k = 0}^{n} s i n (2^{k} α) \Rightarrow$ $A_{n} . B_{n} = \prod_{k = 0}^{n} c o s (2^{k} α) s i n (2^{k} α) = \prod_{k = 0}^{n} {\frac{1}{2} s i n (2^{k + 1} α)}$ $= \frac{1}{2^{n + 1}} \prod_{k = 0}^{n} s i n (2^{k + 1} α) =_{k + 1 = i} \frac{1}{2^{n + 1}} \prod_{i = 1}^{n + 1} s i n (2^{i} α)$ $= \frac{1}{2^{n + 1}} \prod_{i = 0}^{n} s i n (2^{i} α) \frac{s i n (2^{n + 1} α)}{s i n α} = \frac{s i n (2^{n + 1} α)}{2^{n} s i n α} \times B_{n} b u t B_{n} \neq 0 \Rightarrow$ $A_{n} = \frac{s i n (2^{n + 1} α)}{2^{n + 1} s i n (α)}$
Commented by mathmax by abdo last updated on 25/Oct/19

$r e c u r e n c e m e t h o d l e t A_{n} = c o s (α) c o s (2 α) . . . c o s (2^{n} α)$ $f o r n = 0 A_{0} = c o s (α) a n d c o s α = \frac{s i n (2 α)}{2 s i n α} (t r u e) l e t s u p p o s e$ $A_{n} = \frac{s i n (2^{n + 1} α)}{2^{n + 1} s i n α} \Rightarrow A_{n + 1} = c o s (α) c o s (2 α) . . . c o s (2^{n + 1} α)$ $= c o s (α) c o s (2 α) . . . . c o s (2^{n} α) c o s (2^{n + 1} α)$ $= \frac{s i n (2^{n + 1} α)}{2^{n + 1} s i n (α)} \times c o s (2^{n + 1} α) = \frac{1}{2^{n + 1} s i n (α)} \times \frac{1}{2} s i n (2^{n + 2} α)$ $= \frac{s i n (2^{n + 2} α)}{2^{n + 2} s i n (α)} t h e r e l a t i o n i s t r u e f o r (n + 1)$
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