Double Angle Summation

Calculus Level 3

Find the value of $z$ , where $z$ is a positive integer given that

$\large \sum _{ k=0 }^{ \infty }{ \frac { { (-1) }^{ k }(\frac {\pi} { z})^{ 2k } }{ (2k)! } } =\frac { \sqrt { \sqrt { 2 } +2 } }{ 2 }$

1 solution

Will McGlaughlin
May 10, 2018

Firstly we observe that $\sum _{ k=0 }^{ \infty }{ \frac { { (-1) }^{ k }(\frac {\pi} { z})^{ 2k } }{ (2k)! } } =cos(\frac {\pi} { z})$

Now we are solving for $cos(\frac {\pi} { z})=\frac { \sqrt { \sqrt { 2 } +2 } }{ 2 }$

Squaring both sides we achieve $\cos ^{ 2 }{ \frac { \pi }{ z } } =\frac { \sqrt { 2 } +2 }{ 4 }$

Applying the double angle formula for cosine $\frac { cos(\frac { 2\pi }{ z } )+1 }{ 2 } =\frac { \sqrt { 2 } +2 }{ 4 }$

$cos(\frac { 2\pi }{ z } )+1=\frac { \sqrt { 2 } }{ 2 } +1$

$cos(\frac { 2\pi }{ z } )=\frac { \sqrt { 2 } }{ 2 }$

$\frac { 2\pi }{ z } =\frac { \pi }{ 4 }$

$\frac { \pi }{ z } =\frac { \pi }{ 8}$

therefore $z=\boxed{8}$

z actually has other solutions, for instance 8/15 and 8/17. Should your question ask for only integer solutions?

Joseph Newton - 3 years, 1 month ago

yes my bad, its been corrected

Will McGlaughlin - 3 years, 1 month ago