Question
Show that \tan 18\degree =\tan 6\degree \cdotp \tan(60\degree - 6\degree )\cdotp \tan(60\degree + 6\degree )
Answer
First let's prove the identity
\tan3θ = \tanθ\tan(60\degree−\tanθ )\tan(60\degree +θ)
(1)
Using triple angle identity for tangent
\tanθ\tan(60\degree−\tanθ )\tan(60\degree +θ)
=\tanθ⋅\dfrac{\tan60\degree- \tanθ}{1+\tan60\degree\tanθ }\cdot \dfrac{\tan60\degree+\tanθ }{1-\tan60\degree\tanθ}
=\tanθ⋅\dfrac{\sqrt{3} - \tanθ}{1+\sqrt{3} \degree\tanθ }\cdot \dfrac{\sqrt{3} +\tanθ }{1-\sqrt{3} \degree\tanθ}
=\dfrac{3\tanθ -\tan^3θ }{1-3\tan^2θ }
=\tan3θ
Then let's plug θ =6\degree and θ =18\degree to the equation (1), respectively
Plug θ =6\degree
\tan 18\degree =\tan 6\degree \cdotp \tan(60\degree - 6\degree )\cdotp \tan(60\degree + 6\degree )
Then we get the following equation
\tan 18\degree =\tan 6\degree \cdotp \tan54\degree\cdotp \tan66\degree
(2)
Plug θ =18\degree
\tan 54\degree =\tan 18\degree \cdotp \tan(60\degree - 18\degree )\cdotp \tan(60\degree + 18\degree )
\tan54\degree =\tan 18\degree \cdotp \tan42\degree\cdotp \tan78\degree
(3)
Multiplying (2) and (3) gives
\tan 6\degree \cdotp \tan66\degree \cdotp \tan42\degree\cdotp \tan78\degree=1
Mutiply both sides by \tan 12°\tan24°\tan48° . Now we have proved the equality
\tan6° =\tan12°\tan24°\tan48°