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: $$ \begin{cases} R = \pm r \\ δ = \theta - \frac{\pi}{2} \end{cases} $$

Key concepts

Sine Addition Formula

The Sine Addition Formula is a key identity in trigonometry that helps in simplifying expressions. It states that for any angles \(x\) and \(y\), the sine of their difference is given by:
\[ \sin(x - y) = \sin x \cos y - \cos x \sin y \]
This formula is very useful when you want to convert a sum of sine and cosine functions into a singular sine function with a phase shift.

For example, consider the expression \( A \cos \omega_0 t + B \sin \omega_0 t \). To express it as a sine function using the Sine Addition Formula, it can be rewritten in the form \( r \sin(\omega_0 t - \theta) \).

By matching terms in the expression, we deduce that:

• \( r \cos \theta = A \)
• \( r \sin \theta = B \)

Both equations help in finding the specific values of \( r \) and \( \theta \), giving a cleaner and more unified trigonometric expression.

Cosine Addition Formula

The Cosine Addition Formula is another fundamental trigonometric identity that allows the expressing of cosine of the sum or difference of two angles. It is given as:
\[ \cos(x - y) = \cos x \cos y + \sin x \sin y \]
This identity is highly beneficial when working with problems that involve converting complex trigonometric expressions.

In the exercise solution, the formula is used to express a cosine function in terms of a sine function. Specifically, if you are comparing expressions like \( R \cos(\omega_0 t - \delta) \) with \( r \sin(\omega_0 t - \theta) \), this formula helps establish the equivalence by expanding the cosine expression.

Using this identity, we can measure the relationships between different parameters in trigonometric equations. For example:

• \( R \cos \delta = -r \sin \theta \)
• \( R \sin \delta = r \cos \theta \)

Consequently, solving these can help to determine the relationships of angles \( \delta \) and \( \theta \).

Tangent Function

The Tangent Function is a basic trigonometric function that relates the sine and cosine of an angle. It is defined as:
\( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
In the context of the exercise, the tangent is used to find the angle \( \theta \) in terms of \( A \) and \( B \).

When you are given \( r \sin \theta = B \) and \( r \cos \theta = A \), you can express the tangent of \( \theta \) as:
\( \tan \theta = \frac{B}{A} \)
This is straightforward but incredibly powerful for determining angles in equations.

In practical applications, it allows you to solve for \( \theta \) using the arctangent function:
\( \theta = \arctan\left(\frac{B}{A}\right) \)

Additionally, while comparing expressions involving phase shifts, the relationship \( \tan \delta = -\cot \theta \) is used to find \( \delta \). Remember that \( \cot \theta \) represents the reciprocal of \( \tan \theta \), helping in alternating between trigonometric forms. This can lead to various interesting equations and solutions.