How to prove this identity without using the actual values of tan⁡54∘ and sin⁡24∘tan⁡54∘=sin⁡24∘1−3sin⁡24∘

I have a proof.Note that 2⋅36∘=180∘−3⋅36∘ thereforesin⁡2⋅36∘=sin⁡3⋅36∘and thus2sin⁡36∘cos⁡36∘=3sin⁡36∘−4sin336∘and therefore2cos⁡36∘=3−4sin236∘so2cos⁡36∘=3cos236∘−sin236∘=(3cos⁡36∘−sin⁡36∘)(3cos⁡36∘+sin⁡36∘)Nowsin⁡24∘=sin⁡(60∘−36∘)=32cos⁡36∘−12sin⁡36∘and therefore with the last equation we havecos⁡36∘=sin⁡24∘(3cos⁡36∘+sin⁡36∘)dividing by sin⁡36∘ we getcot⁡36∘=sin⁡24∘(3cot⁡36∘+1)Note that tan⁡54∘=tan⁡(90∘−36∘)=cot⁡36∘tan⁡54∘=sin⁡24∘(3tan⁡54∘+1)from whichtan⁡54∘=sin⁡24∘1−3sin⁡24∘follows.

Expert answers to "How to prove this identity without using the..."

Bryant Jacobs

Answered question

2022-04-06

How to prove this identity without using the actual values of

{\tan 54}^{\circ}

and

{\sin 24}^{\circ}

{\tan 54}^{\circ} = \frac{{\sin 24}^{\circ}}{1 - \sqrt{3} {\sin 24}^{\circ}}

Answer & Explanation

Coupewopmergorlpt

Beginner2022-04-07Added 13 answers

I have a proof.
Note that

2 \cdot 36^{\circ} = 180^{\circ} - 3 \cdot 36^{\circ}

therefore

\sin 2 \cdot 36^{\circ} = \sin 3 \cdot 36^{\circ}

and thus

2 {\sin 36}^{\circ} {\cos 36}^{\circ} = 3 {\sin 36}^{\circ} - 4 \sin^{3} 36^{\circ}

and therefore

2 {\cos 36}^{\circ} = 3 - 4 \sin^{2} 36^{\circ}

2 {\cos 36}^{\circ} = 3 \cos^{2} 36^{\circ} - \sin^{2} 36^{\circ}

= (\sqrt{3} {\cos 36}^{\circ} - {\sin 36}^{\circ}) (\sqrt{3} {\cos 36}^{\circ} + {\sin 36}^{\circ})

Now

{\sin 24}^{\circ} = \sin (60^{\circ} - 36^{\circ}) = \frac{\sqrt{3}}{2} {\cos 36}^{\circ} - \frac{1}{2} {\sin 36}^{\circ}

and therefore with the last equation we have

{\cos 36}^{\circ} = {\sin 24}^{\circ} (\sqrt{3} {\cos 36}^{\circ} + {\sin 36}^{\circ})

dividing by

{\sin 36}^{\circ}

we get

{\cot 36}^{\circ} = {\sin 24}^{\circ} (\sqrt{3} {\cot 36}^{\circ} + 1)

Note that

{\tan 54}^{\circ} = \tan (90^{\circ} - 36^{\circ}) = {\cot 36}^{\circ}

{\tan 54}^{\circ} = {\sin 24}^{\circ} (\sqrt{3} {\tan 54}^{\circ} + 1)

from which

{\tan 54}^{\circ} = \frac{{\sin 24}^{\circ}}{1 - \sqrt{3} {\sin 24}^{\circ}}

follows.

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