Question Number 46007 by Meritguide1234 last updated on 19/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18
![T_k =1−tan^4 ((π/2^k )) ={1−tan^2 ((π/2^k ))}{1+tan^2 ((π/2^k ))} ={((cos^2 ((π/2^k ))−sin^2 ((π/2^k )))/(cos^2 ((π/2^k ))))}×(1/(cos^2 ((π/2^k )))) T_k =((cos(((2π)/2^k )))/(cos^4 ((π/2^k ))))=((cos((π/2^(k−1) )))/({cos((π/2^k ))}^4 )) now T_3 ×T_4 ×T_5 ×T_6 ...×T_n =((cos((π/2^(3−1) ))cos((π/2^(4−1) ))cos((π/2^(5−1) ))...cos((π/2^(n−1) )))/([cos((π/2^3 ))cos((π/2^4 ))cos((π/2^5 ))...cos((π/2^n ))]^4 )) now see the first term of N_r =cos((π/2^2 )) see the nth of D_r =(1/([cos((π/2^n ))]^4 )) both remain unchanged... =((cos((π/2^2 )))/(cos^4 ((π/2^n ))))×(1/([cos((π/2^3 ))cos((π/2^4 ))cos((π/2^5 ))...cos((π/2^(n−1) ))]^3 )) =the red marked value to be calculated... wait... =((cos((π/4)))/(cos((π/2^n ))))×(1/([cos((π/2^3 ))cos((π/2^4 ))...cos((π/2^(n−1) ))cos((π/2^n ))]^3 )) sin((π/2^2 ))=2cos((π/2^3 ))sin((π/2^3 )) =2^2 cos((π/2^3 ))cos((π/2^4 ))sin((π/2^4 )) =2^3 cos((π/2^3 ))cos((π/2^4 ))cos((π/2^5 ))sin((π/2^5 )) cos((π/2^3 ))cos((π/2^4 ))cos((π/2^5 ))...cos((π/2^n ))=((sin((π/2^2 )))/(2^(f(n)) sin((π/2^n ))))=((sin((π/4)))/(2^(f(n)) ×sin((π/2^n )))) the value of f(n)to find...wait and condition n→∞ to put pls wait... now putting thevalue ((cos((π/4)))/(cos((π/2^n ))))×[((2^(f(n)) sin((π/2^n )))/(sin((π/4))))]^3 when n→∞ cos((π/2^n ))→1 but sin((π/2^n ))→0 so answer is zero pls check...](Q46132.png)
$${T}_{{k}} =\mathrm{1}−{tan}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}^{{k}} }\right) \\ $$$$\:\:\:\:\:=\left\{\mathrm{1}−{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\right\}\left\{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\right\} \\ $$$$\:\:\:\:\:\:=\left\{\frac{{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}^{{k}} }\right)−{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}^{{k}} }\right)}{{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}^{{k}} }\right)}\right\}×\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}^{{k}} }\right)} \\ $$$${T}_{{k}} =\frac{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{2}^{{k}} }\right)}{{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}^{{k}} }\right)}=\frac{{cos}\left(\frac{\pi}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{\left\{{cos}\left(\frac{\pi}{\mathrm{2}^{{k}} }\right)\right\}^{\mathrm{4}} } \\ $$$${now} \\ $$$${T}_{\mathrm{3}} ×{T}_{\mathrm{4}} ×{T}_{\mathrm{5}} ×{T}_{\mathrm{6}} ...×{T}_{{n}} \\ $$$$=\frac{{cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{3}−\mathrm{1}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{4}−\mathrm{1}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{5}−\mathrm{1}} }\right)...{cos}\left(\frac{\pi}{\mathrm{2}^{{n}−\mathrm{1}} }\right)}{\left[{cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{3}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{4}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{5}} }\right)...{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\right]^{\mathrm{4}} } \\ $$$${now}\:{see}\:{the}\:{first}\:{term}\:{of}\:{N}_{{r}} ={cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{2}} }\right) \\ $$$${see}\:{the}\:{nth}\:{of}\:{D}_{{r}} =\frac{\mathrm{1}}{\left[{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\right]^{\mathrm{4}} } \\ $$$${both}\:{remain}\:{unchanged}... \\ $$$$=\frac{{cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{2}} }\right)}{{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}×\frac{\mathrm{1}}{\left[{cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{3}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{4}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{5}} }\right)...{cos}\left(\frac{\pi}{\mathrm{2}^{{n}−\mathrm{1}} }\right)\right]^{\mathrm{3}} } \\ $$$$={the}\:{red}\:{marked}\:{value}\:{to}\:{be}\:{calculated}... \\ $$$${wait}... \\ $$$$=\frac{{cos}\left(\frac{\pi}{\mathrm{4}}\right)}{{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}×\frac{\mathrm{1}}{\left[{cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{3}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{4}} }\right)...{cos}\left(\frac{\pi}{\mathrm{2}^{{n}−\mathrm{1}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\right]^{\mathrm{3}} } \\ $$$${sin}\left(\frac{\pi}{\mathrm{2}^{\mathrm{2}} }\right)=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{3}} }\right){sin}\left(\frac{\pi}{\mathrm{2}^{\mathrm{3}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{2}} {cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{3}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{4}} }\right){sin}\left(\frac{\pi}{\mathrm{2}^{\mathrm{4}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{3}} {cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{3}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{4}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{5}} }\right){sin}\left(\frac{\pi}{\mathrm{2}^{\mathrm{5}} }\right) \\ $$$${cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{3}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{4}} }\right){cos}\left(\frac{\pi}{\mathrm{2}^{\mathrm{5}} }\right)...{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)=\frac{{sin}\left(\frac{\pi}{\mathrm{2}^{\mathrm{2}} }\right)}{\mathrm{2}^{{f}\left({n}\right)} {sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}=\frac{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{2}^{{f}\left({n}\right)} ×{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)} \\ $$$${the}\:{value}\:{of}\:{f}\left({n}\right){to}\:{find}...{wait}\:\:\:{and}\:{condition}\: \\ $$$${n}\rightarrow\infty\:{to}\:{put}\:\:\:{pls}\:{wait}... \\ $$$${now}\:{putting}\:{thevalue} \\ $$$$\frac{{cos}\left(\frac{\pi}{\mathrm{4}}\right)}{{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}×\left[\frac{\mathrm{2}^{{f}\left({n}\right)} {sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)}{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\right]^{\mathrm{3}} \\ $$$${when}\:{n}\rightarrow\infty\:\:{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\rightarrow\mathrm{1}\:{but}\:{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\rightarrow\mathrm{0} \\ $$$${so}\:\boldsymbol{{answer}}\:\boldsymbol{{is}}\:\boldsymbol{{zero}}\:\boldsymbol{{pls}}\:\boldsymbol{{check}}... \\ $$$$ \\ $$
Commented by Meritguide1234 last updated on 22/Oct/18
