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

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The equation \( (1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots(1+\tan 45^{\circ})=2^n \) leads to the determination of \( n \). By using the identity \( 1 + \tan(45^\circ - k) = \frac{2}{1 + \tan k} \), pairs of terms can be simplified to yield \( (1 + \tan k)(1 + \tan(45^\circ - k)) = 2 \). This results in 22 pairs contributing to the product, giving \( (1+\tan 1)(1+\tan 2)\cdots(1+\tan 44) = 2^{22} \). Including \( 1 + \tan 45 = 2 \) leads to the final product being \( 2^{23} \). Therefore, \( n \) is determined to be 23.
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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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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