Question

Express \(\frac{1}{2} \cos t+\sin t\) in the form \(A \sin (\omega t-\alpha), \alpha \geq 0\)

Short answer

Question: Rewrite the sinusoidal trigonometric expression \(\frac{1}{2} \cos t + \sin t\) in the form \(A \sin (\omega t - \alpha)\), with \(\alpha \geq 0\). Answer: \(\frac{1}{2} \cos t + \sin t = \frac{\sqrt{5}}{2} \sin (t - (\pi - \arctan\left(\frac{1}{2}\right)))\).

Step-by-step solution

Use the sine addition formula

Recall the sine addition formula: \(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta\). We want to rewrite \(\frac{1}{2} \cos t + \sin t\) as a sine with a phase shift: \(A \sin (\omega t - \alpha)\). Let's compare the given expression with the sine addition formula: \(\frac{1}{2} \cos t + \sin t = A\sin(\omega t)\cos(-\alpha) + A\cos(\omega t)\sin(-\alpha)\). It becomes clear that we should take \(\omega =1\), so that coefficients of \(\cos t\) and \(\sin t\) match on both sides of the equation above.

Match coefficients

Now, our task is to find \(A\) and \(\alpha\): \(\frac{1}{2} \cos t + \sin t = A\sin t\cos(-\alpha) + A\cos t\sin(-\alpha)\). For the coefficients to match, we need: \(\frac{1}{2} = A\sin(-\alpha) \quad(1)\) \(1 = A\cos(-\alpha) \quad(2)\)

Find A and α

Square both equations (1) and (2) and add them: \((\frac{1}{2})^2 + 1^2 = (A\sin(-\alpha))^2 + (A\cos(-\alpha))^2\) \(\frac{1}{4} + 1 = A^2 (\sin^2(-\alpha) + \cos^2(-\alpha))\) \(\frac{5}{4} = A^2\). Taking the square root of both sides, we obtain: \(A = \frac{\sqrt{5}}{2}\). Now, let's solve for \(\alpha\). Divide equation (1) by (2): \(\frac{1/2}{1} = \frac{\sin(-\alpha)}{\cos(-\alpha)}\) \(\tan(-\alpha) = \frac{1}{2}\) \(\alpha = -\arctan\left(\frac{1}{2}\right)\). For \(\alpha\) to satisfy \(\alpha \geq 0\), we can add \(\pi\) to the obtained value: \(\alpha = \pi - \arctan\left(\frac{1}{2}\right)\). Finally, we found all the necessary values and can express our original function in the desired form:

Write the final expression

The function \(\frac{1}{2} \cos t + \sin t\) can be expressed in the form \(A \sin (\omega t - \alpha)\) with the found values for \(A\), \(\omega\), and \(\alpha\): \(\frac{1}{2} \cos t + \sin t = \frac{\sqrt{5}}{2} \sin (t - (\pi - \arctan\left(\frac{1}{2}\right)))\).

Key concepts

Sine Addition Formula

The Sine Addition Formula is a key identity in trigonometry, allowing us to express the sine of a sum of angles as a product involving sine and cosine functions. The formula is given by:

• \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)

In this particular exercise, we want to rewrite the expression \( \frac{1}{2} \cos t + \sin t \) using the sine addition formula. We aim to express it in the form \( A \sin(\omega t - \alpha) \). This involves transforming the given linear combination into a single sine term with a specific amplitude \( A \), frequency \( \omega \), and phase shift \( \alpha \). By doing so, we make it more comparable to harmonic sine waves, offering a clearer understanding of its oscillatory behavior.
Initially, comparing our expression with the sine addition formula shows us how sine and cosine terms mix with coefficients on each side of the equation. Recognizing the patterns and mappings in the formula is crucial, as it sets the stage for further transformations, paving the way to represent our initial equation effectively with matched coefficients.

Phase Shift

Phase shift is a concept that describes the horizontal shift of a sinusoidal function on a graph. In trigonometry, when we rewrite a combination of sine and cosine functions as a pure sine function, understanding phase shifts becomes very important. The variable \( \alpha \) represents the phase shift in the transformation \( A \sin(\omega t - \alpha) \). A positive \( \alpha \) denotes a left shift, while a negative one shifts it to the right.
In the exercise, we calculate the phase shift \( \alpha \) after matching terms so that \( \alpha \geq 0 \). Initially obtaining \( \alpha = -\arctan\left(\frac{1}{2}\right) \), we need to adjust it by adding \( \pi \) since arcsine functions can result in negative angles, which aren't suitable for our constraints. This gives us \( \alpha = \pi - \arctan\left(\frac{1}{2}\right) \), ensuring that the phase is represented in a permissible format while maintaining the integrity of the function's orientation on the graph.

Coefficient Matching

Coefficient matching is a vital step when transforming trigonometric expressions into a standard form. In this process, we equate the coefficients of sine and cosine on both sides of their respective equations to find the unknowns. This technique requires careful analysis and algebraic manipulation to ensure accurate results.
For the given problem, the steps involve setting \( \frac{1}{2} \) and \( 1 \) equal to \( A\sin(-\alpha) \) and \( A\cos(-\alpha) \) respectively. This is done separately for terms associated with \( \cos t \) and \( \sin t \), leading to a system of equations:

• \( \frac{1}{2} = A\sin(-\alpha) \)
• \( 1 = A\cos(-\alpha) \)

After solving these equations, we determined that \( A = \frac{\sqrt{5}}{2} \). Square and sum the equations to solve for \( A \), as the sum of squares involves the Pythagorean identity \( \sin^2\alpha + \cos^2\alpha = 1 \). This elegantly simplifies the system, confirming the correct calculation of both amplitude \( A \) and angle \( \alpha \).