Three simple harmonic motions with equal amplitudes A and identical...

The three simple harmonic motions can be represented as phasors of length $A$ with angles $0^{{}^{{}^{\circ}}}$, $60^{{}^{{}^{\circ}}}$, and $120^{{}^{{}^{\circ}}}$ respectively.

The resultant amplitude $R$ can be found by resolving the phasors into horizontal and vertical components.

$R_x = A \cos(0^{{}^{{}^{\circ}}}) + A \cos(60^{{}^{{}^{\circ}}}) + A \cos(120^{{}^{{}^{\circ}}})$

$R_x = A \left(1 + \dfrac{1}{2} - \dfrac{1}{2}\right) = A$

$R_y = A \sin(0^{{}^{{}^{\circ}}}) + A \sin(60^{{}^{{}^{\circ}}}) + A \sin(120^{{}^{{}^{\circ}}})$

$R_y = A \left(0 + \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2}\right) = \sqrt{3}A$

The resultant amplitude is given by $R = \sqrt{R_x^2 + R_y^2}$

$R = \sqrt{A^2 + (\sqrt{3}A)^2} = \sqrt{A^2 + 3A^2} = \sqrt{4A^2} = 2A$

Alternatively, the resultant of the first and third phasors ($0^{{}^{{}^{\circ}}}$ and $120^{{}^{{}^{\circ}}}$) has a magnitude of $\sqrt{A^2 + A^2 + 2A^2 \cos(120^{{}^{{}^{\circ}}})} = A$ and bisects the angle, directing it at $60^{{}^{{}^{\circ}}}$. Since the second phasor is also directed at $60^{{}^{{}^{\circ}}}$, their magnitudes add up directly to give $A + A = 2A$.