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Question Number 72145 by Maclaurin Stickker last updated on 25/Oct/19
prove that: cos α×cos 2α×cos 2^2 α...cos 2^n α =((sin 2^(n+1) α)/(2^(n+1) sin α))
$${prove}\:{that}: \\ $$$$\mathrm{cos}\:\alpha×\mathrm{cos}\:\mathrm{2}\alpha×{cos}\:\mathrm{2}^{\mathrm{2}} \alpha…{cos}\:\mathrm{2}^{{n}} \alpha \\ $$$$=\frac{\mathrm{sin}\:\mathrm{2}^{{n}+\mathrm{1}} \alpha}{\mathrm{2}^{{n}+\mathrm{1}} \mathrm{sin}\:\alpha} \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 25/Oct/19
first way :we solve by induction n=1⇒((sin2^2 α)/(2^2 sinα))=((sin2αcos2α)/(2sinα))=cosαcos2α induction hypothesis n=k−1⇒cosα×cos2α×...×cos2^(k−1) α=((sin2^k α)/(2^k sinα)) now n=k⇒cosα×cos2α×...×cos2^(k−1) α×cos2^k α= ((sin2^k αcos2^k α)/(2^k sinα))=(((1/2)sin2^(k+1) α)/(2^k sinα))=((sin2^(k+1) α)/(2^(k+1) sinα)). second way: ((sin2^(n+1) α)/(2^(n+1) sinα))=((sin2^n αcos2^n α)/(2^n sinα))=((sin2^(n−1) αcos2^(n−1) αcos2^n α)/(2^(n−1) sinα))= ...=((sin2^0 αcos2^0 αcos2α...cos2^n α)/(sinα))= cosαcos2α...cos2^n α
$${first}\:{way}\::{we}\:{solve}\:{by}\:{induction} \\ $$$${n}=\mathrm{1}\Rightarrow\frac{{sin}\mathrm{2}^{\mathrm{2}} \alpha}{\mathrm{2}^{\mathrm{2}} {sin}\alpha}=\frac{{sin}\mathrm{2}\alpha{cos}\mathrm{2}\alpha}{\mathrm{2}{sin}\alpha}={cos}\alpha{cos}\mathrm{2}\alpha \\ $$$${induction}\:{hypothesis} \\ $$$${n}={k}−\mathrm{1}\Rightarrow{cos}\alpha×{cos}\mathrm{2}\alpha×…×{cos}\mathrm{2}^{{k}−\mathrm{1}} \alpha=\frac{{sin}\mathrm{2}^{{k}} \alpha}{\mathrm{2}^{{k}} {sin}\alpha} \\ $$$${now} \\ $$$${n}={k}\Rightarrow{cos}\alpha×{cos}\mathrm{2}\alpha×…×{cos}\mathrm{2}^{{k}−\mathrm{1}} \alpha×{cos}\mathrm{2}^{{k}} \alpha= \\ $$$$\frac{{sin}\mathrm{2}^{{k}} \alpha{cos}\mathrm{2}^{{k}} \alpha}{\mathrm{2}^{{k}} {sin}\alpha}=\frac{\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}^{{k}+\mathrm{1}} \alpha}{\mathrm{2}^{{k}} {sin}\alpha}=\frac{{sin}\mathrm{2}^{{k}+\mathrm{1}} \alpha}{\mathrm{2}^{{k}+\mathrm{1}} {sin}\alpha}. \\ $$$$ \\ $$$$ \\ $$$${second}\:{way}: \\ $$$$ \\ $$$$\frac{{sin}\mathrm{2}^{{n}+\mathrm{1}} \alpha}{\mathrm{2}^{{n}+\mathrm{1}} {sin}\alpha}=\frac{{sin}\mathrm{2}^{{n}} \alpha{cos}\mathrm{2}^{{n}} \alpha}{\mathrm{2}^{{n}} {sin}\alpha}=\frac{{sin}\mathrm{2}^{{n}−\mathrm{1}} \alpha{cos}\mathrm{2}^{{n}−\mathrm{1}} \alpha{cos}\mathrm{2}^{{n}} \alpha}{\mathrm{2}^{{n}−\mathrm{1}} {sin}\alpha}= \\ $$$$…=\frac{{sin}\mathrm{2}^{\mathrm{0}} \alpha{cos}\mathrm{2}^{\mathrm{0}} \alpha{cos}\mathrm{2}\alpha…{cos}\mathrm{2}^{{n}} \alpha}{{sin}\alpha}= \\ $$$${cos}\alpha{cos}\mathrm{2}\alpha…{cos}\mathrm{2}^{{n}} \alpha \\ $$$$ \\ $$
Commented by Maclaurin Stickker last updated on 25/Oct/19
Thank you for your time.
$${Thank}\:{you}\:{for}\:{your}\:{time}. \\ $$
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)))
$${let}\:{A}_{{n}} ={cos}\left(\alpha\right){cos}\left(\mathrm{2}\alpha\right)….{cos}\left(\mathrm{2}^{{n}} \alpha\right)\:\Rightarrow{A}_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}} {cos}\left(\mathrm{2}^{{k}} \alpha\right) \\ $$$${let}\:{B}_{{n}} ={sin}\left(\alpha\right){sin}\left(\mathrm{2}\alpha\right)….{sin}\left(\mathrm{2}^{{n}} \alpha\right)=\prod_{{k}=\mathrm{0}} ^{{n}} \:{sin}\left(\mathrm{2}^{{k}} \alpha\right)\:\Rightarrow \\ $$$${A}_{{n}} .{B}_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}} \:{cos}\left(\mathrm{2}^{{k}} \alpha\right){sin}\left(\mathrm{2}^{{k}} \alpha\right)=\prod_{{k}=\mathrm{0}} ^{{n}} \left\{\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}^{{k}+\mathrm{1}} \alpha\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\prod_{{k}=\mathrm{0}} ^{{n}} \:{sin}\left(\mathrm{2}^{{k}+\mathrm{1}} \alpha\right)\:=_{{k}+\mathrm{1}={i}} \:\:\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\prod_{{i}=\mathrm{1}} ^{{n}+\mathrm{1}} \:{sin}\left(\mathrm{2}^{{i}} \alpha\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\prod_{{i}=\mathrm{0}} ^{{n}} \:{sin}\left(\mathrm{2}^{{i}} \alpha\right)\frac{{sin}\left(\mathrm{2}^{{n}+\mathrm{1}} \alpha\right)}{{sin}\alpha}\:=\frac{{sin}\left(\mathrm{2}^{{n}+\mathrm{1}} \alpha\right)}{\mathrm{2}^{{n}} {sin}\alpha}\:×{B}_{{n}} \:\:{but}\:{B}_{{n}} \neq\mathrm{0}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{{sin}\left(\mathrm{2}^{{n}+\mathrm{1}} \alpha\right)}{\mathrm{2}^{{n}+\mathrm{1}} {sin}\left(\alpha\right)} \\ $$
Commented by mathmax by abdo last updated on 25/Oct/19
recurence method let A_n =cos(α)cos(2α)...cos(2^n α) for n=0 A_0 =cos(α) and cosα=((sin(2α))/(2sinα)) (true) let suppose A_n =((sin(2^(n+1) α))/(2^(n+1) sinα)) ⇒ A_(n+1) =cos(α)cos(2α)...cos(2^(n+1) α) =cos(α)cos(2α)....cos(2^n α)cos(2^(n+1) α) =((sin(2^(n+1) α))/(2^(n+1) sin(α)))×cos(2^(n+1) α) =(1/(2^(n+1) sin(α)))×(1/2)sin(2^(n+2) α) =((sin(2^(n+2) α))/(2^(n+2) sin(α))) the relation is true for (n+1)
$${recurence}\:{method}\:{let}\:{A}_{{n}} ={cos}\left(\alpha\right){cos}\left(\mathrm{2}\alpha\right)…{cos}\left(\mathrm{2}^{{n}} \alpha\right) \\ $$$${for}\:{n}=\mathrm{0}\:\:{A}_{\mathrm{0}} ={cos}\left(\alpha\right)\:{and}\:{cos}\alpha=\frac{{sin}\left(\mathrm{2}\alpha\right)}{\mathrm{2}{sin}\alpha}\:\:\left({true}\right)\:{let}\:{suppose} \\ $$$${A}_{{n}} =\frac{{sin}\left(\mathrm{2}^{{n}+\mathrm{1}} \alpha\right)}{\mathrm{2}^{{n}+\mathrm{1}} {sin}\alpha}\:\Rightarrow\:{A}_{{n}+\mathrm{1}} ={cos}\left(\alpha\right){cos}\left(\mathrm{2}\alpha\right)…{cos}\left(\mathrm{2}^{{n}+\mathrm{1}} \alpha\right) \\ $$$$={cos}\left(\alpha\right){cos}\left(\mathrm{2}\alpha\right)….{cos}\left(\mathrm{2}^{{n}} \alpha\right){cos}\left(\mathrm{2}^{{n}+\mathrm{1}} \alpha\right) \\ $$$$=\frac{{sin}\left(\mathrm{2}^{{n}+\mathrm{1}} \alpha\right)}{\mathrm{2}^{{n}+\mathrm{1}} {sin}\left(\alpha\right)}×{cos}\left(\mathrm{2}^{{n}+\mathrm{1}} \alpha\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} {sin}\left(\alpha\right)}×\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}^{{n}+\mathrm{2}} \alpha\right) \\ $$$$=\frac{{sin}\left(\mathrm{2}^{{n}+\mathrm{2}} \alpha\right)}{\mathrm{2}^{{n}+\mathrm{2}} {sin}\left(\alpha\right)}\:{the}\:{relation}\:{is}\:{true}\:\:{for}\:\left({n}+\mathrm{1}\right) \\ $$

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