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?
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This graphical explanation easily generalizes to the situation where lenghts $a$ and $b$ are not $1$. – hmakholm left over Monica Jun 04 '19 at 21:04
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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 – HFKy Jun 04 '19 at 21:04
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$a \sin(x) + b\cos(x) = r \sin(x + \theta)$ 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.
Robert Israel
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