Question

Show that \(A \cos \omega_{0} t+B \sin \omega_{0} t\) can be written in the form \(r \sin \left(\omega_{0} t-\theta\right) .\) Determine \(r\) and \(\theta\) in terms of \(A\) and \(B\). If \(R \cos \left(\omega_{0} t-\delta\right)=r \sin \left(\omega_{0} t-\theta\right),\) determine the relationship among \(R, r, \delta,\) and \(\theta .\)

Short answer

Answer: The relationship among R, r, δ, and θ is: $$ \begin{cases} R = \pm r \\ δ = \theta - \frac{\pi}{2} \end{cases} $$

Step-by-step solution

Rewrite the given expression using the sine addition formula

Use the sine addition formula \(\sin(x - y) = \sin x \cos y - \cos x \sin y\) to rewrite the given expression \(A\cos\omega_0t + B\sin\omega_0t\) into the form of \(r\sin(\omega_0t - \theta)\). The sine addition formula can be transformed into our desired form: $$r\sin(\omega_0t - \theta) = r\sin\omega_0t\cos\theta - r\cos\omega_0t\sin\theta$$ We want to match the terms with the given expression, which can be done by setting: $$r\cos\theta = A\quad(1)$$ $$r\sin\theta = B\quad(2)$$

Find the relationship between r and θ

Now that we have the expressions for r and θ in terms of A and B, we need to solve for r and θ. Squaring both equations (1) and (2) and adding them, we get: $$(r\cos\theta)^2 + (r\sin\theta)^2 = A^2 + B^2$$ $$r^2(\cos^2\theta + \sin^2\theta) = A^2 + B^2$$ Since \(\cos^2\theta + \sin^2\theta = 1\), we have: $$r^2 = A^2 + B^2$$ Taking the square root on both sides: $$r = \sqrt{A^2 + B^2}$$ Now, to find θ, divide equation (2) by equation (1): $$\frac{r\sin\theta}{r\cos\theta} = \frac{B}{A}$$ Simplifying: $$\tan\theta = \frac{B}{A}$$ $$\theta = \arctan\left(\frac{B}{A}\right)$$

Determine the relationship among R, r, δ, and θ

We are given that: $$R\cos(\omega_0t - \delta) = r\sin(\omega_0t - \theta)$$ Use the cosine addition formula \(\cos(x - y) = \cos x \cos y + \sin x \sin y\) to rewrite the left side of the equation: $$R\cos\omega_0t\cos\delta + R\sin\omega_0t\sin\delta = r\sin\omega_0t\cos\theta - r\cos\omega_0t\sin\theta$$ The two expressions will be equal if: $$R\cos\delta = -r\sin\theta\quad(3)$$ $$R\sin\delta = r\cos\theta\quad(4)$$ Now, to find the relationship between R and r, square both equations (3) and (4) and add them: $$(R\cos\delta)^2 + (R\sin\delta)^2 = r^2\sin^2\theta + r^2\cos^2\theta$$ $$R^2(\cos^2\delta + \sin^2\delta) = r^2(\sin^2\theta + \cos^2\theta)$$ Since \(\cos^2\delta + \sin^2\delta = 1\) and \(\sin^2\theta + \cos^2\theta = 1\), we have: $$R^2 = r^2$$ Thus, the relationship between R and r is: $$R = \pm r$$ To find the relationship between δ and θ, divide equation (4) by equation (3): $$\frac{R\sin\delta}{-R\cos\delta} = \frac{r\cos\theta}{r\sin\theta}$$ Simplifying: $$\tan_δ = -\cot_θ$$ $$δ = \arctan\left(\frac{-1}{\tan\theta}\right)$$ $$δ = \theta - \frac{\pi}{2}$$ So, the relationship among R, r, δ, and θ is: $$ \begin{cases} R = \pm r \\ δ = \theta - \frac{\pi}{2} \end{cases} $$