1
$\begingroup$

For $$\sin (x)+\sqrt{3} \cos (x)$$, we can rewrite it as $$2 \sin \left(x+\frac{\pi }{3}\right)$$.

Is there a formula to represent $$a\sin(x)+b\cos(x)$$ just by using sine or cosine function just as the aforementioned example?

$\endgroup$
  • $\begingroup$ This graphical explanation easily generalizes to the situation where lenghts $a$ and $b$ are not $1$. $\endgroup$ – hmakholm left over Monica Jun 4 '19 at 21:04
  • $\begingroup$ I would guess is $/sqrt(a^2 + b^2)sen(x + atan (b/a))$. I cant remember very well but I suppose maybe you can prove it using $e^{ix} = cos x + i sen x $ and taking real and imaginary parts and somethings like this $\endgroup$ – HFKy Jun 4 '19 at 21:04
5
$\begingroup$

$a \sin(x) + b\cos(x) = r \sin(x + \theta)$ where $r = \sqrt{a^2 + b^2}$, $a/r = \cos(\theta)$ and $b/r = \sin(\theta)$. Thus if $a > 0$, $\theta = \arcsin(b/r)$ while if $a < 0$, $\theta = \pi - \arcsin(b/r)$ will do.

| cite | improve this answer | |
$\endgroup$

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