Try this beautiful problem from the PRMO, 2018 based on Nearest value.
If x=cos1cos2cos3.....cos89 and y=cos2cos6cos10....cos86, then what is the integer nearest to \(\frac{2}{7}log_2{\frac{y}{x}}\)?
Algebra
Numbers
Multiples
But try the problem first...
Answer: is 19.
PRMO, 2018, Question 14
Higher Algebra by Hall and Knight
First hint
\(\frac{y}{x}\)=\(\frac{cos2cos6cos10.....cos86}{cos1cos2cos3....cos89}\)
=\(2^{44}\times\sqrt{2}\frac{cos2cos6cos10...cos86}{sin2sin4...sin88}\)
[ since cos\(\theta\)=sin(90-\(\theta\)) from cos 46 upto cos 89 and 2sin\(\theta\)cos\(\theta\)=sin2\(\theta\)]
Second Hint
=\(\frac{2^{\frac{89}{2}}sin4sin8sin12...sin88}{sin2sin4sin6...sin88}\)
[ since sin\(\theta\)=cos(90-\(\theta\))]
=\(\frac{2^{\frac{89}{2}}}{cos4cos8cos12..cos88}\)
[ since cos\(\theta\)=sin(90-\(\theta\))]
Final Step
=\(\frac{2^\frac{89}{2}}{\frac{1}{2}^{22}}\)
[since \(cos4cos8cos12...cos88\)
\(=(cos4cos56cos64)(cos8cos52cos68)(cos12cos48cos72)(cos16cos44cos76)(cos20cos40cos80)(cos24cos36cos84)(cos28cos32cos88)cos60\)
\(=(1/2)^{15}(cos12cos24cos36cos48cos60cos72cos84)\)
\(=(1/2)^{16}(cos12cos48cos72)(cos24cos36cos84)\)
\(=(1/2)^{20}(cos36cos72)\)
\(=(1/2)^{20}(cos36sin18)\)
\(=(1/2)^{22}(4sin18cos18cos36/cos18)\)
\(=(1/2)^{22}(sin72/cos18)\)
\(=(1/2)^{22}\)]
=\(2^\frac{133}{2}\)
\(\frac{2}{7}log_2{\frac{y}{x}}\)=\(\frac{2}{7} \times \frac{133}{2}\)=19.
Try this beautiful problem from the PRMO, 2018 based on Nearest value.
If x=cos1cos2cos3.....cos89 and y=cos2cos6cos10....cos86, then what is the integer nearest to \(\frac{2}{7}log_2{\frac{y}{x}}\)?
Algebra
Numbers
Multiples
But try the problem first...
Answer: is 19.
PRMO, 2018, Question 14
Higher Algebra by Hall and Knight
First hint
\(\frac{y}{x}\)=\(\frac{cos2cos6cos10.....cos86}{cos1cos2cos3....cos89}\)
=\(2^{44}\times\sqrt{2}\frac{cos2cos6cos10...cos86}{sin2sin4...sin88}\)
[ since cos\(\theta\)=sin(90-\(\theta\)) from cos 46 upto cos 89 and 2sin\(\theta\)cos\(\theta\)=sin2\(\theta\)]
Second Hint
=\(\frac{2^{\frac{89}{2}}sin4sin8sin12...sin88}{sin2sin4sin6...sin88}\)
[ since sin\(\theta\)=cos(90-\(\theta\))]
=\(\frac{2^{\frac{89}{2}}}{cos4cos8cos12..cos88}\)
[ since cos\(\theta\)=sin(90-\(\theta\))]
Final Step
=\(\frac{2^\frac{89}{2}}{\frac{1}{2}^{22}}\)
[since \(cos4cos8cos12...cos88\)
\(=(cos4cos56cos64)(cos8cos52cos68)(cos12cos48cos72)(cos16cos44cos76)(cos20cos40cos80)(cos24cos36cos84)(cos28cos32cos88)cos60\)
\(=(1/2)^{15}(cos12cos24cos36cos48cos60cos72cos84)\)
\(=(1/2)^{16}(cos12cos48cos72)(cos24cos36cos84)\)
\(=(1/2)^{20}(cos36cos72)\)
\(=(1/2)^{20}(cos36sin18)\)
\(=(1/2)^{22}(4sin18cos18cos36/cos18)\)
\(=(1/2)^{22}(sin72/cos18)\)
\(=(1/2)^{22}\)]
=\(2^\frac{133}{2}\)
\(\frac{2}{7}log_2{\frac{y}{x}}\)=\(\frac{2}{7} \times \frac{133}{2}\)=19.
