Prove that \(\displaystyle{\tan{{\left({6}^{\circ}\right)}}}{\tan{{\left({42}^{\circ}\right)}}}={\tan{{\left({12}^{\circ}\right)}}}{\tan{{\left({24}^{\circ}\right)}}}\)

Pasegeabe85xy

Pasegeabe85xy

Answered question

2022-03-28

Prove that tan(6)tan(42)=tan(12)tan(24)

Answer & Explanation

memantangti17

memantangti17

Beginner2022-03-29Added 13 answers

Here is a proof that I got after back calculation. Let θ=36 Then, one can show that θ satisfies the following equation
(cosθ+1)(4cos2θ2cosθ1)=0
since cos36>0, θ satisfies
4cos2θ2cosθ1=0
2cos2θ2cosθ=1
cos2θcosθ=12
Let ϕ=12. Then θ=3ϕ and
cos6ϕcos3ϕ=12
cos6ϕ+1=cos3ϕ+12=cos3ϕ+cos5ϕ
2cos23ϕ=2cos4ϕcosϕ
cos6cos42+sin6sin42cos6cos42sin6sin42
tan6tan42=tan24tan12

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