Show that \tan 18\degree =\tan 6\degree \cdotp \tan(60\degree - 6\degree )\cdotp \tan(60\degree + 6\degree )

Collected in the board: Trigonometry

Steven Zheng posted 1 hour ago


First let's prove the identity

\tan3θ = \tanθ\tan(60\degree−\tanθ )\tan(60\degree +θ)

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 }


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

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

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°

Steven Zheng posted 1 hour ago

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