Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 18003 by Tinkutara last updated on 13/Jul/17

The value of cosA∙cos2A∙cos2^2 A ..... cos(2^(n − 1) A),  where A ∈ R may be  (1) 1  (2) 2  (3) −1  (4) ((sin 2^n  A)/(2^n  sin A))

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}{A}\centerdot\mathrm{cos2}{A}\centerdot\mathrm{cos2}^{\mathrm{2}} {A}\:.....\:\mathrm{cos}\left(\mathrm{2}^{{n}\:−\:\mathrm{1}} {A}\right), \\ $$$$\mathrm{where}\:{A}\:\in\:{R}\:\mathrm{may}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:−\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{sin}\:\mathrm{2}^{{n}} \:{A}}{\mathrm{2}^{{n}} \:\mathrm{sin}\:{A}} \\ $$

Answered by alex041103 last updated on 13/Jul/17

Let E=cos(A)cos(2A)...=Π_(k=0) ^(n−1) cos(2^k A)  Because sin(2^(k+1) A)=2sin(2^k A)cos(2^k A)  ⇒cos(2^k A)=((sin(2^(k+1) A))/(2sin(2^k A)))  ⇒E=Π_(k=0) ^(n−1) ((sin(2^(k+1) A))/(2sin(2^k A)))=  =[Π_(k=0) ^(n−1) sin(2^(k+1) A)]×[Π_(k=0) ^(n−1) (1/(2sin(2^k A)))]  =((sin(2^n A))/(2sin(A)))[Π_(k=0) ^(n−2) sin(2^(k+1) A)][Π_(k=1) ^(n−1) (1/(2sin(2^k A)))]  =((sin(2^n A))/(2sin(A)))[Π_(k=1) ^(n−1) sin(2^k A)][Π_(k=1) ^(n−1) (1/(2sin(2^k A)))]  =((sin(2^n A))/(2sin(A)))[Π_(k=1) ^(n−1) ((sin(2^k A))/(2sin(2^k A)))]  =((sin(2^n A))/(2sin(A)))[Π_(k=1) ^(n−1) (1/2)]  =((sin(2^n A))/(2sin(A)))×(1/2^(n−1) )  =((sun(2^n A))/(2^n sin(A))) = cos(A)cos(2A)...

$${Let}\:{E}={cos}\left({A}\right){cos}\left(\mathrm{2}{A}\right)...=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}{cos}\left(\mathrm{2}^{{k}} {A}\right) \\ $$$${Because}\:{sin}\left(\mathrm{2}^{{k}+\mathrm{1}} {A}\right)=\mathrm{2}{sin}\left(\mathrm{2}^{{k}} {A}\right){cos}\left(\mathrm{2}^{{k}} {A}\right) \\ $$$$\Rightarrow{cos}\left(\mathrm{2}^{{k}} {A}\right)=\frac{{sin}\left(\mathrm{2}^{{k}+\mathrm{1}} {A}\right)}{\mathrm{2}{sin}\left(\mathrm{2}^{{k}} {A}\right)} \\ $$$$\Rightarrow{E}=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{{sin}\left(\mathrm{2}^{{k}+\mathrm{1}} {A}\right)}{\mathrm{2}{sin}\left(\mathrm{2}^{{k}} {A}\right)}= \\ $$$$=\left[\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}{sin}\left(\mathrm{2}^{{k}+\mathrm{1}} {A}\right)\right]×\left[\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\mathrm{1}}{\mathrm{2}{sin}\left(\mathrm{2}^{{k}} {A}\right)}\right] \\ $$$$=\frac{{sin}\left(\mathrm{2}^{{n}} {A}\right)}{\mathrm{2}{sin}\left({A}\right)}\left[\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\prod}}{sin}\left(\mathrm{2}^{{k}+\mathrm{1}} {A}\right)\right]\left[\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\mathrm{1}}{\mathrm{2}{sin}\left(\mathrm{2}^{{k}} {A}\right)}\right] \\ $$$$=\frac{{sin}\left(\mathrm{2}^{{n}} {A}\right)}{\mathrm{2}{sin}\left({A}\right)}\left[\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}{sin}\left(\mathrm{2}^{{k}} {A}\right)\right]\left[\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\mathrm{1}}{\mathrm{2}{sin}\left(\mathrm{2}^{{k}} {A}\right)}\right] \\ $$$$=\frac{{sin}\left(\mathrm{2}^{{n}} {A}\right)}{\mathrm{2}{sin}\left({A}\right)}\left[\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{{sin}\left(\mathrm{2}^{{k}} {A}\right)}{\mathrm{2}{sin}\left(\mathrm{2}^{{k}} {A}\right)}\right] \\ $$$$=\frac{{sin}\left(\mathrm{2}^{{n}} {A}\right)}{\mathrm{2}{sin}\left({A}\right)}\left[\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$=\frac{{sin}\left(\mathrm{2}^{{n}} {A}\right)}{\mathrm{2}{sin}\left({A}\right)}×\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$=\frac{{sun}\left(\mathrm{2}^{{n}} {A}\right)}{\mathrm{2}^{{n}} {sin}\left({A}\right)}\:=\:{cos}\left({A}\right){cos}\left(\mathrm{2}{A}\right)... \\ $$

Commented by Tinkutara last updated on 13/Jul/17

But which options are correct? It can  have more than one option(s).

$$\mathrm{But}\:\mathrm{which}\:\mathrm{options}\:\mathrm{are}\:\mathrm{correct}?\:\mathrm{It}\:\mathrm{can} \\ $$$$\mathrm{have}\:\mathrm{more}\:\mathrm{than}\:\mathrm{one}\:\mathrm{option}\left(\mathrm{s}\right). \\ $$

Commented by alex041103 last updated on 13/Jul/17

you′re right  it can be 1 when A=2π  and  −1 for A=π and n=1  but i′m still thinking about the 2

$${you}'{re}\:{right} \\ $$$${it}\:{can}\:{be}\:\mathrm{1}\:{when}\:{A}=\mathrm{2}\pi \\ $$$${and}\:\:−\mathrm{1}\:{for}\:{A}=\pi\:{and}\:{n}=\mathrm{1} \\ $$$${but}\:{i}'{m}\:{still}\:{thinking}\:{about}\:{the}\:\mathrm{2} \\ $$

Commented by Tinkutara last updated on 13/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com