1
$\begingroup$

I derived the following trigonometrical equation from a real triangle, knowing that the angle $\alpha$ is an acute one:

$\sin 3\alpha = 2\sin\alpha$

Just by eye-balling the equation, and remembering the trigonometric unit circle, I know that:

$\sin 90 = 1 = 2\sin 30 = 2 \cdot 0.5 = 1$

Therefore, $\alpha = 30^\circ$ is a possible solution.
I am unaware of any trigonometric identity to help me simplify this equation in order to get all possible solutions for $\alpha$, and this intuitive solution is the best I can come up with. I plugged this equation into symbolab.com, but their solution seems very long-winded, and I am hoping for the possibility that a simpler one exists.

How can I solve this type of problem when the intuitive approach fails?

$\endgroup$
2
$\begingroup$

If triple angle relation for sine is known, letting $ \sin \alpha =s$,

$$3 s -4 s^3= 2 s\rightarrow s =(0, \pm \frac12), \quad \alpha=(0, \pm 30^{\circ}, 150^{\circ}\pm30^{\circ} ) ..$$

$\endgroup$
  • $\begingroup$ Forgive my ignorance, but I don't understand what this syntax is meant to represent. $\endgroup$ – daedsidog Dec 8 '18 at 15:48
  • $\begingroup$ Sorry I should have given it fully, shall edit it in the answer. $\endgroup$ – Narasimham Dec 8 '18 at 16:14
1
$\begingroup$

Here is a lead.

$$\sin(3x) = \sin(2x + x) = \sin(2x)\cos(x) + \cos(2x)\sin(x).$$ Now invoke some double angle magig to get $$\sin(3x) = 2\sin(x)\cos^2(x) + (1 - 2\sin^2(x))\sin(x).$$ Next, use the pythagorean identities to gete $$\sin(3x) = 2\sin(x)(1 - \sin^2(x)) + (1 - 2\sin^2(x))\sin(x). $$ Can you use this?

$\endgroup$
  • $\begingroup$ This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there? $\endgroup$ – daedsidog Dec 8 '18 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.