Very complicated solution. There's a much easier one expressing a in terms of cot(Pi/24) and then cot(Pi/24)=cos(Pi/24) / sin(Pi/24). Then the numerator is equal to sqr(3) * cos(Pi/24) - sin(Pi/24), which is 2 * sin(7*Pi/24) And the denominator is cos(Pi/24) + sin(Pi/24) which is sqr(2) * sin(7*Pi/24) Hense a=sqr(2)
Note ∏/6+∏/24 = ∏/3-∏/8 and tan ∏/3 = √3 and tan ∏/8 = √2-1 (using tan∏/4=1). So, tan [ ∏/6+∏/24] = tan[∏/3-∏/8] = [√3-( √2-1)]/[1+√6-√3]. But tan( ∏/6+∏/24) simplifies to [4-(√3-1)a]/[(√3+1)a]. So, [4-(√3-1)a]/[(√3+1)a] = [√3-( √2-1)]/[1+√6-√3] which yields, after some algebra, a= √2.
Με ολο το σεβασμο θελω να σας παρακαλεσω κατι. Να γραφετε το 3,14...σαν π και οχι σαν ^και απο πανω μια παυλα _. ελληνικο π. Και παλι μπραβο για τις ασκησεις σας. Ευχαριστω.
let a=pi/24, c=cos(a),s=sin(a), r3 = sqrt(3). Problem is rewritten (1+x)/(r3-x) = c/s => x= (r3*c-s)/(s+c) = r3 - (r3+1)s/(s+c) or x = r3 - (r3+1)/(1+cot(a)) ~1.4142. (OK, maybe not as nice as your method)
Very complicated solution.
There's a much easier one expressing a in terms of cot(Pi/24) and then cot(Pi/24)=cos(Pi/24) / sin(Pi/24).
Then the numerator is equal to sqr(3) * cos(Pi/24) - sin(Pi/24), which is 2 * sin(7*Pi/24)
And the denominator is cos(Pi/24) + sin(Pi/24) which is sqr(2) * sin(7*Pi/24)
Hense a=sqr(2)
This is both a nice problem and a nice presentation.
Note ∏/6+∏/24 = ∏/3-∏/8 and tan ∏/3 = √3 and tan ∏/8 = √2-1 (using tan∏/4=1). So, tan [ ∏/6+∏/24] = tan[∏/3-∏/8] = [√3-( √2-1)]/[1+√6-√3]. But tan( ∏/6+∏/24) simplifies to [4-(√3-1)a]/[(√3+1)a]. So, [4-(√3-1)a]/[(√3+1)a] = [√3-( √2-1)]/[1+√6-√3] which yields, after some algebra, a= √2.
a= √2
A=√2
Με ολο το σεβασμο θελω να σας παρακαλεσω κατι. Να γραφετε το 3,14...σαν π και οχι σαν ^και απο πανω μια παυλα _. ελληνικο π. Και παλι μπραβο για τις ασκησεις σας. Ευχαριστω.
let a=pi/24, c=cos(a),s=sin(a), r3 = sqrt(3). Problem is rewritten (1+x)/(r3-x) = c/s => x= (r3*c-s)/(s+c) = r3 - (r3+1)s/(s+c)
or x = r3 - (r3+1)/(1+cot(a)) ~1.4142. (OK, maybe not as nice as your method)
cotPi/24 cot Pi/4^6 cotPi/2^23^2 cotPi/1^23^1 23 (cotPi ➖ 3cotPi+2). { 2a+2a ➖}=2a^2/(3)^2 ➖ (a)^2= 2a^2/{9 ➖ a^2}={ 2a^2/7}= 3 .1a^2 3^1.1^1a^2^1 3a^2 (a ➖ 3a+2) .
All this algebra is not needed cot(pi/24) is a constant, just solve for a in terms of cot(pi/24) simplification is not always needed.
Yes, that will give the solution for x if cot(pi/24) is given.
@@ericerpelding2348 cot(pi/24) is not a variable it is given open your eyes
Made a mistake at 3.30