Trig problem or limit problem?

Geometry Level 3

$\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \tan { \left( \frac { \pi }{ 4 \times { 2 }^{ r } } \right) } \sec { \left( \frac { \pi }{ 4\times { 2 }^{ r-1 } } \right) } } } = \ ?$

The answer is 1.

1 solution

Vishnu Kadiri
Apr 11, 2019

Let $\frac { \pi }{ 4 } =A$ . Therefore, $\tan { \frac { A }{ 2 } } \sec { A } =\frac { \sin { \frac { A }{ 2 } } }{ \cos { \frac { A }{ 2 } } \cos { A } } =\frac { \sin { A } \cos { \frac { A }{ 2 } } -\cos { A } \sin { \frac { A }{ 2 } } }{ \cos { \frac { A }{ 2 } } \cos { A } } =\tan { A } -\tan { \frac { A }{ 2 } }$ $\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \tan { \left( \frac { \pi }{ 4*{ 2 }^{ r } } \right) } \sec { \left( \frac { \pi }{ 4*{ 2 }^{ r-1 } } \right) } } } =\lim _{ n\rightarrow \infty }{ \tan { A } -\tan { \frac { A }{ { 2 }^{ n+1 } } } } =\tan { \frac { \pi }{ 4 } } =1$

Very nice! Although the property you prove holds for all $A$ (apart from undefined points of the tangent and secant functions). What you can show is that the sum telescopes for any $A$ , then plug in $A=\frac{\pi}{4}$ at the end (at the moment, it looks as though the identity is a special property of $A=\frac{\pi}{4}$ )

Chris Lewis - 2 years, 1 month ago

Yes, thanks for pointing out!

Vishnu Kadiri - 2 years, 1 month ago

Trig problem or limit problem?

The answer is 1.

1 solution

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