Chapter 9: Problem 14

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

Short Answer

Expert verified

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: $$\sqrt{29}\cos{\left[3t+\arctan{\left(\frac{2}{5}\right)}\right]}$$

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

Short Answer

Step by step solution

Identify the amplitude and angular frequency

Calculate the amplitude A

Find the phase difference

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

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