Prove \(\displaystyle{4}{\sin{{\left({2}\frac{\pi}{{5}}\right)}}}+{\tan{{\left({2}\frac{\pi}{{5}}\right)}}}={5}{\cot{{\left(\frac{\pi}{{5}}\right)}}}\) Attempt: Let the side length be

Beryneingmk39

Beryneingmk39

Answered question

2022-04-01

Prove 4sin(2π5)+tan(2π5)=5cot(π5)
Attempt: Let the side length be 1. Then by breaking down the pentagon into 5 congruent triangles, it is easy to show that the area is A=54cot(π5)=52R2sin(2π5)=5r2tan(π5) where R is the circumradius and r is the inradius.
Then it suffices to prove that the area can also be written as sin(2π5)+14tan(2π5) I tried to break down the area of a regular pentagon as a trapezoid and triangle, but the algebra got messy: A=(1+cos(2π5))sin(2π5)+(12)(1+2cos(2π5))(R+rsin(2π5))

Answer & Explanation

Nathanial Carey

Nathanial Carey

Beginner2022-04-02Added 12 answers

If you cut the pentagon ABCDE into three triangles ABC, ACD, ADE then
AreaABC=12sin3π5=12sin2π5
and ABC=AED, so it remains to show AreaACD=14tan2π5 or equivalently, the perpendicular distance of A from CD is 12tan2π5 which is immediate by looking at the base angles ACD=ADC=d2π5

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