Is \( n \) Equal to 23 in the Given Trigonometric Product Equation?

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SUMMARY

The equation $$(1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots(1+\tan 45^{\circ})=2^n$$ has been solved, revealing that \( n \) equals 23. The derivation utilizes the identity \( \tan(45-k) = \frac{1 - \tan k}{1 + \tan k} \) to establish pairs of terms that simplify to 2. Specifically, the products \( (1+\tan k)(1+\tan(45-k)) \) yield 2 for \( k = 1, 2, \ldots, 22 \), leading to a total of 22 pairs, and including \( 1 + \tan 45 = 2 \) results in \( 2^{23} \).

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anemone
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Given that $$(1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots(1+\tan 45^{\circ})=2^n$$, find $n$.
 
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anemone said:
Given that $$(1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots(1+\tan 45^{\circ})=2^n$$, find $n$.

Note k in degrees and degree symbol not mentioned

tan (45-k)) = (tan 45- tank )/( 1+ tan 45 tan k)
= ( 1- tan k)/ (1 + tan k)
So 1+ tan (45-k) = ( 1+ tan k + 1 – tan k) / ( 1+ tan k) = 2/(1+ tan k)
Or (1 + tan (45-k))(1+ tan k) = 2

So (1 + tan 1) ( 1+ tan 44) = 2
(1+ tan 2)(1 + tan 43) = 2
( 1 + tan 22)(1+ tan 23) = 2

Hence (1+tan 1)( 1+ tan 2) … ( 1 + tan 44) = 2^22

As 1 + tan 45 = 2 so multiplying we get
(1+tan 1)( 1+ tan 2) … ( 1 + tan 44)( 1+ tan 45) = 2^23
hence n = 23
 

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