Telescopic Trigonometry!

Algebra Level 3

∏ n = 1 ∞ 1 1 − tan ⁡ 2 ( 2 − n ) = tan ⁡ k \large \prod _{ n=1 }^{ \infty }{ \frac { 1 }{ 1-\tan ^{ 2 }{( { 2 }^{ -n }) } } } = \tan k ,

The equation above holds true for some integer k k .

Find k k − k ! k^k - k! .


The answer is 0.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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2 solutions

P = lim ⁡ N → ∞ ∏ n = 1 N 1 1 − tan ⁡ 2 1 2 n = lim ⁡ N → ∞ ∏ n = 1 N 2 tan ⁡ 1 2 n 1 − tan ⁡ 2 1 2 n ⋅ 1 2 tan ⁡ 1 2 n = lim ⁡ N → ∞ ∏ n = 1 N tan ⁡ 1 2 n − 1 tan ⁡ 1 2 n = lim ⁡ N → ∞ tan ⁡ 1 2 tan ⁡ 1 2 ⋅ tan ⁡ 1 2 2 tan ⁡ 1 4 ⋅ tan ⁡ 1 4 2 tan ⁡ 1 8 ⋯ tan ⁡ 1 2 N − 1 2 tan ⁡ 1 2 N = lim ⁡ N → ∞ tan ⁡ 1 2 N tan ⁡ 1 ∘ 2 N = lim ⁡ N → ∞ tan ⁡ 1 tan ⁡ 1 2 N 1 2 N = tan ⁡ 1 \begin{aligned} P & = \lim_{N \to \infty}\prod_{n=1}^N \frac 1{1-\tan^2 \frac 1{2^n}} \\ & = \lim_{N \to \infty}\prod_{n=1}^N {\color{#3D99F6}\frac {2\tan \frac 1{2^n}}{1-\tan^2 \frac 1{2^n}}}\cdot \frac 1{2\tan \frac 1{2^n}} \\ & = \lim_{N \to \infty}\prod_{n=1}^N \frac {\color{#3D99F6} \tan \frac 1{2^{n-1}}}{\tan \frac 1{2^n}} \\ & = \lim_{N \to \infty} \frac {\tan 1}{2 \cancel {\tan \frac 12}}\cdot \frac {\cancel{\tan \frac 12}}{2 \cancel{\tan \frac 14}}\cdot \frac {\cancel{\tan \frac 14}}{2 \cancel{\tan \frac 18}}\cdots \frac {\cancel{\tan \frac 1{2^{N-1}}}}{2 \tan \frac 1{2^N}} \\ & = \lim_{N \to \infty} \frac {\tan 1}{2^N \tan \frac {1^\circ}{2^N}} = \lim_{N \to \infty} \frac {\tan 1}{\frac {\tan \frac 1{2^N}}{\frac 1{2^N}}} \\ & = \tan 1 \end{aligned}

Therefore, k k − k ! = 1 1 − 1 = 0 k^k - k! = 1^1 - 1 = \boxed{0} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Priyanshu Mishra
May 26, 2018

We have

tan ⁡ 2 x = 2 tan ⁡ x 1 − tan ⁡ 2 x ∏ n = 1 N 1 1 − tan ⁡ 2 ( 2 − n ) = ∏ n = 1 N tan ⁡ ( 2 − n + 1 ) 2 tan ⁡ ( 2 − n ) = 2 − N tan ⁡ ( 2 − N ) tan ⁡ 1 \large\ { \displaystyle \tan { 2x } = \frac { 2\tan { x } }{ 1 - \tan ^{ 2 }{ x } } \\ \displaystyle \prod _{ n=1 }^{ N }{ \frac { 1 }{ 1 - \tan ^{ 2 }{ \left( { 2 }^{ -n } \right) } } } = \displaystyle \prod _{ n=1 }^{ N }{ \frac { \tan { \left( { 2 }^{ -n+1 } \right) } }{ 2\tan { \left( { 2 }^{ -n } \right) } } } = \displaystyle \frac { { 2 }^{ -N } }{ \tan { \left( { 2 }^{ -N } \right) } } \tan { 1 } } .

And the rest follows.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

In the problem, tan ⁡ k ∘ \tan k^\circ should just be tan ⁡ k \tan k . You specifically say that k k is in radians. Likewise, k ∈ ( 0 , 180 ) k \in (0,180) should say k ∈ ( 0 , π ) k \in (0,\pi) .

Jon Haussmann - 2 years, 11 months ago

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