Find the value of $\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$

Question

My Attempt: $$\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$$ $$\sin \dfrac {180}{10} - \sin \dfrac {3\times 180}{10}$$ $$\sin 18^\circ - \sin 54^\circ$$

Now,

Let $A=18^\circ$. $$5A=90^\circ$$ $$2A+3A=90^\circ$$ $$3A=90^\circ - 2A$$ Taking 'sin' on the both sides, $$\sin 3A=\sin (90^\circ - 2A)$$ $$3\sin A-4\sin^3 A=\cos 2A$$

What should I do further?

See https://math.stackexchange.com/questions/827540/proving-trigonometric-equation-cos36-circ-cos72-circ-1-2 — lab bhattacharjee, Apr 15 '17 at 11:15

score 2 · Answer 1 · answered Apr 15 '17 at 10:52

2

$\cos 2A=1-2\sin^2 A$, then substitute $x=\sin A$.

answered Apr 15 '17 at 10:52

szw1710

8,102

Michael Rozenberg · Accepted Answer · 2017-04-15T13:49:24.283

2

$$\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}=\cos72^{\circ}+\cos144^{\circ}=$$ $$=\frac{2\sin36^{\circ}\cos72^{\circ}+2\sin36^{\circ}\cos144^{\circ}}{2\sin36^{\circ}}=$$ $$=\frac{\sin108^{\circ}-\sin36^{\circ}+\sin180^{\circ}-\sin108^{\circ}}{2\sin36^{\circ}}=-\frac{1}{2}.$$ Done!

edited Apr 15 '17 at 13:49

answered Apr 15 '17 at 11:04

Michael Rozenberg

194,933

How did you get $\dfrac {-1}{2}$? Could you show the calculations? – pi-π Apr 15 '17 at 13:18
@Ramanujan It was typo. See now, please. – Michael Rozenberg Apr 15 '17 at 13:50

Find the value of $\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$

2 Answers2