3. 8 sas2n cos B = - 17 Зл sina - 5 2 Given: cOs a = Sin'a -Cos a 1. sin B = 2. Sio33/% - Cos'a=1 を 2. cosa 25 9. Sin るこcos'a*! Sina Tosa tan B = 3. tan a = 4. %3D

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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### Trigonometric Identities and Angle Calculation

**Given:**

- \(\sin \alpha = \frac{3\sqrt{5}}{5}, \quad \frac{\pi}{2} \leq \alpha \leq 2\pi\)
- \(\cos \beta = \frac{8}{17}, \quad \pi \leq \beta \leq \frac{3\pi}{2}\)

---

**1. Calculate \(\cos \alpha\):**

Using the identity \(\cos^2 \alpha = 1 - \sin^2 \alpha\),

\[
\sin^2 \alpha = \left(\frac{3\sqrt{5}}{5}\right)^2 = \frac{9}{5}
\]

\[
\cos^2 \alpha = 1 - \frac{9}{5} = \frac{16}{25}
\]

Thus, \(\cos \alpha = \frac{4}{5}\).

---

**2. Calculate \(\sin \beta\):**

Using the identity \(\sin^2 \beta + \cos^2 \beta = 1\),

\[
\cos^2 \beta = \left(\frac{8}{17}\right)^2 = \frac{64}{289}
\]

\[
\sin^2 \beta = 1 - \frac{64}{289} = \frac{225}{289}
\]

Thus, \(\sin \beta = \frac{15}{17}\).

---

**3. Calculate \(\tan \alpha\):**

Using the identity \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\),

\[
\tan \alpha = \frac{-3\sqrt{5}/5}{-4/5} = \frac{3\sqrt{5}}{4}
\]

---

**4. Calculate \(\tan \beta\):**

Using the identity \(\tan \beta = \frac{\sin \beta}{\cos \beta}\),

\[
\tan \beta = \frac{3\sqrt{17}/17}{8/17} = \frac{3\sqrt{17}}{8}
\]

---

**5. Calculate \(\sin(\alpha + \beta)\):**

Using the identity \(\sin(\alpha + \beta) = \sin \alpha \cos \
Transcribed Image Text:### Trigonometric Identities and Angle Calculation **Given:** - \(\sin \alpha = \frac{3\sqrt{5}}{5}, \quad \frac{\pi}{2} \leq \alpha \leq 2\pi\) - \(\cos \beta = \frac{8}{17}, \quad \pi \leq \beta \leq \frac{3\pi}{2}\) --- **1. Calculate \(\cos \alpha\):** Using the identity \(\cos^2 \alpha = 1 - \sin^2 \alpha\), \[ \sin^2 \alpha = \left(\frac{3\sqrt{5}}{5}\right)^2 = \frac{9}{5} \] \[ \cos^2 \alpha = 1 - \frac{9}{5} = \frac{16}{25} \] Thus, \(\cos \alpha = \frac{4}{5}\). --- **2. Calculate \(\sin \beta\):** Using the identity \(\sin^2 \beta + \cos^2 \beta = 1\), \[ \cos^2 \beta = \left(\frac{8}{17}\right)^2 = \frac{64}{289} \] \[ \sin^2 \beta = 1 - \frac{64}{289} = \frac{225}{289} \] Thus, \(\sin \beta = \frac{15}{17}\). --- **3. Calculate \(\tan \alpha\):** Using the identity \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\), \[ \tan \alpha = \frac{-3\sqrt{5}/5}{-4/5} = \frac{3\sqrt{5}}{4} \] --- **4. Calculate \(\tan \beta\):** Using the identity \(\tan \beta = \frac{\sin \beta}{\cos \beta}\), \[ \tan \beta = \frac{3\sqrt{17}/17}{8/17} = \frac{3\sqrt{17}}{8} \] --- **5. Calculate \(\sin(\alpha + \beta)\):** Using the identity \(\sin(\alpha + \beta) = \sin \alpha \cos \
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