Prove $\tan3°\tan63°\tan69°=\tan15°$
And assuming we don't know that $\tan15^{\circ}$ part, how to just evaluate $\tan3^{\circ} tan63^{\circ} tan69^{\circ}$?
Asked
Active
Viewed 422 times
2
-
1https://math.stackexchange.com/questions/477364/prove-that-tan-a-tan-b-tan-c-tan-a-tan-b-tan-c-abc-180-circ?noredirect=1&lq=1 – Darío A. Gutiérrez Jun 07 '17 at 13:10
2 Answers
2
We need to prove that $$\sin3^{\circ}\sin63^{\circ}\sin69^{\circ}\cos15^{\circ}=\cos3^{\circ}\cos63^{\circ}\cos69^{\circ}\sin15^{\circ}$$ or $$(\cos60^{\circ}-\cos66^{\circ})(\sin84^{\circ}+\sin54^{\circ})=(\cos60^{\circ}+\cos66^{\circ})(\sin84^{\circ}-\sin54^{\circ})$$ or $$\sin84^{\circ}\cos66^{\circ}=\sin54\cos60^{\circ}$$ or $$\sin150^{\circ}+\sin18^{\circ}=\sin54^{\circ}$$ or $$\frac{1}{2}+\frac{\sqrt5-1}{4}=\frac{\sqrt5+1}{4},$$ which is obvious.
Done!
Michael Rozenberg
- 194,933
-
Thank you for this concise proving. What if it's just an evaluation of $\tan3^{\circ} \tan63^{\circ} \tan69^{\circ}$? Can that be solved by a similar method? – mo-han Jun 08 '17 at 00:50
-
@mo-han Yes. Why no? The value $2-\sqrt3$ would be say about a similar method. – Michael Rozenberg Jun 08 '17 at 02:40
-
Need more details, it's just too difficult for me to fill up the gap. I suppose those transformation above would not be intact any more. – mo-han Jun 08 '17 at 03:13
0
Though I hate reverse engineering, I'm yet to find how the problem came being.
Here is one approach of validating the proposition:
We need $$\dfrac{\sin3^\circ\sin63^\circ}{\cos3^\circ\cos63^\circ}=\dfrac{\sin15^\circ\sin21^\circ}{\cos15^\circ\cos21^\circ}$$
Using componendo and dividendo, $$\dfrac{\cos(63-3)^\circ}{\cos(63+3)^\circ}=\dfrac{\cos(21-15)^\circ}{\cos(21+15)^\circ}$$
$$\iff\cos36^\circ=2\cos6^\circ\cos66^\circ=\cos(66-6)^\circ+\cos(66+6)^\circ$$
which has been established here: Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
lab bhattacharjee
- 274,582