Question

Express \(5 \cos 3 t+2 \sin 3 t\) in the form \(A \cos (\omega t+\alpha), \alpha \geq 0\)

Short answer

Question: Express the function \(5 \cos 3t + 2 \sin 3t\) in the form of \(A \cos(\omega t + \alpha)\). Answer: The function \(5 \cos 3t + 2 \sin 3t\) can be expressed as \(\sqrt{29}\cos{\left[3t+\arctan{\left(\frac{2}{5}\right)}\right]}\).

Step-by-step solution

Identify the amplitude and angular frequency

The given function is $$5\cos{3t} + 2 \sin{3t}$$ From this function, we can identify the angular frequency \(\omega\) as 3. To find the amplitude A, we will use the amplitude formula: $$A = \sqrt{a^2 + b^2}$$ Where \(a\) and \(b\) are the coefficients of \(\cos{\omega t}\) and \(\sin{\omega t}\) respectively. In this case, \(a = 5\) and \(b = 2\).

Calculate the amplitude A

Using the formula \(A = \sqrt{a^2 + b^2}\), we have: $$A = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}$$ So, the amplitude A is \(\sqrt{29}\).

Find the phase difference

To find the phase difference \(\alpha\), we use the formula: $$\alpha = \arctan{\frac{b}{a}}$$ In this case, we have: $$\alpha = \arctan{\frac{2}{5}}$$

Express the given function as \(A \cos (\omega t+\alpha)\)

Now we can express the given function in the form \(A \cos(\omega t + \alpha)\) as: $$\sqrt{29}\cos{[3t+\arctan{(\frac{2}{5})}]}$$ Thus, the given function \(5 \cos 3t + 2 \sin 3t\) can be expressed as: $$\sqrt{29}\cos{\left[3t+\arctan{\left(\frac{2}{5}\right)}\right]}$$

Key concepts

Amplitude Formula

Understanding the amplitude formula is crucial to converting trigonometric expressions.
The amplitude is essentially the maximum value of a wave. It tells how tall or powerful the wave is.
To find the amplitude in a wave expression involving both cosine and sine, we use a specific formula. Here is the formula to calculate amplitude:

• The expression needs to be in the form of: \(a \cos(\omega t) + b \sin(\omega t)\).
• The amplitude \(A\) is given by \(A = \sqrt{a^2 + b^2}\).
• Here, \(a\) is the coefficient of cosine, and \(b\) is the coefficient of sine.

By applying this formula, the amplitude of our function, \(5\cos{3t} + 2\sin{3t}\), is \(\sqrt{29}\). This means this function's wave reaches just over 5.38 units high or low from the center, making it an important value when analyzing wave functions.

Phase Difference

Phase difference is a concept describing the shift from one wave to the next. It's often measured as an angle.
To find the phase difference for trigonometric identities, we use the phase angle formula.
This helps determine how much the wave lags or leads its normal position.Let's break down finding the phase difference:

• For the expression \(a \cos(\omega t) + b \sin(\omega t)\), the phase difference \(\alpha\) is found using the formula: \(\alpha = \arctan{\left(\frac{b}{a}\right)}\).
• This formula is derived from trigonometric identities involving the tangent function.
• The result, \(\alpha\), determines the horizontal shift of the wave on a graph.

For our function \(5\cos{3t} + 2\sin{3t}\), the phase difference is \(\arctan{\left(\frac{2}{5}\right)}\). This tells us how different the current wave is from a baseline we might compare it to.

Angular Frequency

Angular frequency relates to how fast something cycles through its phases. It's key in understanding wave behavior in trigonometry.
This is the speed at which the wave oscillates, often described in radians per time unit.
Understanding angular frequency involves:

• In the expression of \(a \cos(\omega t) + b \sin(\omega t)\), \(\omega\) represents the angular frequency.
• It tells us how many cycles occur in a certain time frame.
• Measured in radians per second, it offers a glimpse into the wave's speed without direct reference to time.

In our function, \(5\cos{3t} + 2\sin{3t}\), the angular frequency \(\omega\) is 3. This indicates that the wave completes 3 cycles in the time it takes for \(t\) to rise by 2\(\pi\) radians, giving insights into how quickly the wave moves.