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Use the following polar equation to answer the following questions. \( r=f(\theta)=3 \sin (2 \theta) \) name: \( R \) R -0\( ) \) 10. State the values of \( \theta \) where \( f(\theta)=0 \)

Ask by Moreno King. in the United States
Mar 11,2025

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Answer

The values of \(\theta\) where \( f(\theta)=0 \) are \(\theta=\frac{k\pi}{2}\) for any integer \(k\).

Solution

1. We start with the equation \[ f(\theta)=3 \sin(2 \theta)=0. \] 2. Divide both sides by 3 (since 3 is not zero) to get \[ \sin(2\theta)=0. \] 3. The sine function \(\sin x\) is zero when \[ x=k\pi \quad \text{for } k\in \mathbb{Z}. \] So, set \[ 2\theta=k\pi. \] 4. Solve for \(\theta\) by dividing both sides by 2: \[ \theta=\frac{k\pi}{2} \quad \text{for } k\in \mathbb{Z}. \] Thus, the values of \(\theta\) where \( f(\theta)=0 \) are \[ \theta=\frac{k\pi}{2} \quad \text{for } k\in \mathbb{Z}. \]

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Beyond the Answer

To find where \( f(\theta) = 0 \) in the polar equation \( r = 3 \sin(2\theta) \), we set \( 3 \sin(2\theta) = 0 \). This leads to solving \( \sin(2\theta) = 0 \). The sine function equals zero at multiples of \( \pi \), so we have \( 2\theta = n\pi \), where \( n \) is any integer. Therefore, the values of \( \theta \) can be expressed as \( \theta = \frac{n\pi}{2} \), resulting in angles like \( 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \), and continuing infinitely in both directions! The beauty of this polar plot is that it creates a visually striking pattern. In this specific case, the function describes a rose curve with petals that appear at specific angles, and knowing where the sine function equals zero helps in graphically locating where these petals intersect the origin. So, grab a graphing tool and see those vibrant petals dance around the pole! 🌸📈

Related Questions

\( \varphi=\arctan \left(\frac{y}{x}\right) \) Penulisan system koordinat polar adalah \( \boldsymbol{r} \angle \boldsymbol{\varphi} \) Sebaliknya, untuk mengkonversi dari system koordinat polar menuju system koordinat cartesian seperti pada persamaan berikut: \( x=r \cdot \cos (\varphi) \) \( y=r \cdot \sin (\varphi) \) Untuk operasi penjumlahan dan pengurangan harus diubah ke system koordinat cartesian. Sedangkan untuk operasi perkalian dan pembagian dapat dilakukan dengan sangat mudah. Misal \( A=4 \angle 60^{\circ} \) dan \( B=2 \angle 20^{\circ} \) Perkalian: \( A \cdot B=\left(4 \angle 60^{\circ}\right) \cdot\left(2 \angle 20^{\circ}\right) \) Dalam proses perkalian, jari-jari (r) dikalikan, sedangkan sudut \( (\varphi) \) dijumlahkan. \( A \cdot B=(4.2) \angle\left(60^{\circ}+20^{\circ}\right) \) A. \( B=8 \angle 80^{\circ} \) Pembagian: \[ \frac{A}{B}=\frac{4 \angle 60^{\circ}}{2 \angle 20^{\circ}} \] Dalam proses pembagian, jari-jari (r) dibagi seperti biasa, sedangkan sudut \( (\varphi) \) pembilangdikurangi sudut penyebut. \[ \begin{array}{l} \frac{A}{B}=\frac{4}{2} \angle\left(60^{\circ}-20^{\circ}\right) \\ \frac{A}{B}=2 \angle 40^{\circ} \end{array} \] Tugas 1. (harus diseertai dengan cara) 1. Tentukan hasil dari \( \sqrt{-13.69} \) Diketahui \( A=1.5+0.5 i \) dan \( B=1+2.5 i \), maka tentukan: 2. \( A+B \) 3. \( A-B \) 4. \( A * B \) 5. \( \frac{A}{B} \) 6. \( \frac{A^{2}-A B}{B} \) 7. Bentuk polar \( A \) 8. Bentuk polar \( B \) 9. \( A * B \) dengan cara polar 10. \( \frac{A}{B} \) dengan cara polar 11. Bentuk cartesian dari jawaban no. 9 dan bandinglan hasilnya dengan jawaban no. 4 12. Bentuk cartesian dari jawaban no. 10 dan bandingkan hasilnya dengan jawaban no.5. Materi selanjutnya -> Euler dan Teorema de Moivre.

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