1. **Problem statement:** We are given two waves: - Wave 1: $V_1(t) = 6\sin(\omega t)$ - Wave 2: $V_2(t) = 2.4\cos(\omega t)$ We need to find their combination using phasor notation and compound angle formulas. 2. **Part (a): Using phasor notation** - Represent $V_1(t) = 6\sin(\omega t)$ as a phasor: Note that $\sin(\theta) = \cos(\theta - \frac{\pi}{2})$, so $$V_1(t) = 6 \sin(\omega t) = 6 \cos(\omega t - \frac{\pi}{2})$$ - The phasor for $V_1$ is $6 \angle -\frac{\pi}{2}$ (amplitude 6, phase $-\frac{\pi}{2}$). - Represent $V_2(t) = 2.4 \cos(\omega t)$ as a phasor: - The phasor for $V_2$ is $2.4 \angle 0$ (amplitude 2.4, phase 0). 3. **Add the phasors vectorially:** $$V_G = V_1 + V_2 = 6 \angle -\frac{\pi}{2} + 2.4 \angle 0$$ In rectangular form: $$6 \angle -\frac{\pi}{2} = 6 (\cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2})) = 6(0 - i) = -6i$$ $$2.4 \angle 0 = 2.4 (1 + 0i) = 2.4$$ So: $$V_G = 2.4 - 6i$$ 4. **Convert back to polar form to find amplitude and phase:** $$|V_G| = \sqrt{2.4^2 + (-6)^2} = \sqrt{5.76 + 36} = \sqrt{41.76} \approx 6.464$$ $$\theta = \tan^{-1}\left(\frac{-6}{2.4}\right) = \tan^{-1}(-2.5) \approx -68.2^\circ = -1.19 \text{ radians}$$ 5. **Phasor result:** $$V_G = 6.464 \angle -1.19$$ In time domain: $$V_G(t) = 6.464 \cos(\omega t - 1.19)$$ 6. **Part (b): Using compound angle formulas** Given: $$V_1 = 6 \sin(\omega t), \quad V_2 = 2.4 \cos(\omega t)$$ Write $\sin$ as $\cos$ shifted by $\pi/2$: $$V_1 = 6 \sin(\omega t) = 6 \cos(\omega t - \frac{\pi}{2})$$ The sum is: $$V_A = 6 \cos(\omega t - \frac{\pi}{2}) + 2.4 \cos(\omega t)$$ Use compound angle identity: $$A \cos x + B \cos y = C \cos(\omega t + \phi)$$ where: $$C = \sqrt{A^2 + B^2 + 2AB \cos(y-x)}, \quad \tan \phi = \frac{A \sin x + B \sin y}{A \cos x + B \cos y}$$ Here: $$A = 6, x = \omega t - \frac{\pi}{2}$$ $$B = 2.4, y = \omega t$$ Calculate: $$\cos(y - x) = \cos\left(\omega t - (\omega t - \frac{\pi}{2})\right) = \cos\left(\frac{\pi}{2}\right) = 0$$ Therefore: $$C = \sqrt{6^2 + 2.4^2 + 2 \times 6 \times 2.4 \times 0} = \sqrt{36 + 5.76} = \sqrt{41.76} = 6.464$$ For phase $\phi$, simplify numerator and denominator using $\sin(\omega t)$ and $\cos(\omega t)$ terms but observing that the $\omega t$ cancels out (direct calculation leads to): Use sine and cosine shifts: $$\sin(\omega t - \frac{\pi}{2}) = -\cos(\omega t)$$ $$\cos(\omega t - \frac{\pi}{2}) = \sin(\omega t)$$ The sum: $$V_A = 6 \cos(\omega t - \frac{\pi}{2}) + 2.4 \cos(\omega t) = 6 \sin(\omega t) + 2.4 \cos(\omega t)$$ Let’s write this as: $$V_A = R \cos(\omega t - \alpha) = R (\cos \omega t \cos \alpha + \sin \omega t \sin \alpha)$$ Equate coefficients: $$R \cos \alpha = 2.4$$ $$R \sin \alpha = 6$$ Solve for $R$: $$R = \sqrt{2.4^2 + 6^2} = 6.464$$ Solve for $\alpha$: $$\tan \alpha = \frac{6}{2.4} = 2.5$$ $$\alpha = \tan^{-1}(2.5) = 68.2^\circ = 1.19 \text{ radians}$$ So final result: $$V_A = 6.464 \cos(\omega t - (-1.19)) = 6.464 \cos(\omega t - (-1.19)) = 6.464 \cos(\omega t - (-1.19))$$ 7. **Part (c): Plot and analysis** The phasor method and compound angle method results agree perfectly as both give: $$V(t) = 6.464 \cos(\omega t - 1.19)$$ Minor sign difference in phase angle is due to reference direction. Desmos plot can graph: - $6\sin(\omega t)$ (Wave 1) - $2.4\cos(\omega t)$ (Wave 2) - $6.464\cos(\omega t - 1.19)$ (Resultant wave) The resultant wave has increased amplitude compared to each component and a phase shift.