ric identity:

$$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$

Letting

$$A = \omega t + \phi_1 \quad \text{and} \quad B = \omega t + \phi_2,$$

the sum becomes:

$$y = y_1 + y_2 = 2a \cos\left(\frac{\phi_1 - \phi_2}{2}\right) \sin \left(\omega t + \frac{\phi_1 + \phi_2}{2}\right)$$

This represents a single sinusoidal wave with:

Amplitude: $$A' = 2a \left|\cos\left(\frac{\phi_1 - \phi_2}{2}\right)\right|$$

Angular frequency: $$\omega$$ (and hence the same time period)

The problem states that the amplitude of the resultant wave is the same as that of the individual waves. Thus, we require:

$$2a \left|\cos\left(\frac{\phi_1 - \phi_2}{2}\right)\right| = a$$

Dividing both sides by $$a$$ (assuming $$a \neq 0$$), we get:

$$2\left|\cos\left(\frac{\phi_1 - \phi_2}{2}\right)\right| = 1$$

This simplifies to:

$$\left|\cos\left(\frac{\phi_1 - \phi_2}{2}\right)\right| = \frac{1}{2}$$

The cosine function equals $$\frac{1}{2}$$ when its argument is $$\frac{\pi}{3}$$ (or $$-\frac{\pi}{3}$$, but the absolute value takes care of the sign). Similarly, $$\cos\left(\frac{\phi_1 - \phi_2}{2}\right)$$ equals $$-\frac{1}{2}$$ when the argument is $$\frac{2\pi}{3}$$ (or $$-\frac{2\pi}{3}$$). In either case, the magnitude is $$\frac{1}{2}$$.

Choosing the positive difference for simplicity, we set:

$$\frac{\phi_1 - \phi_2}{2} = \frac{\pi}{3} \quad \Rightarrow \quad |\phi_1 - \phi_2| = \frac{2\pi}{3}$$

This result can be written as:

$$\phi_1 - \phi_2 = \cos^{-1}\left(-\frac{1}{2}\right)$$

since

$$\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}.$$

Thus, the correct answer is Option A:

$$\cos^{-1}\left(\frac{-1}{2}\right).$$

Waves

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