Type or upload your work and solution. Your work should include right triangles in the appropriate quadrants with appropriate labels If \( \cos \alpha=-3 / 5 \) and \( \alpha \) is in Quad III and \( \tan \beta=-5 / 12 \) and \( \beta \) is in Quad II. Find the following: a) \( \cos (\alpha+\beta) \) b) \( \sin (\alpha+\beta) \) c) \( (\alpha+\beta) \) is in Quadrant

Sure! Let's work through each part of the problem step by step. We'll use right triangles in the appropriate quadrants and apply trigonometric identities to find the solutions. --- ## Given: 1. \( \cos \alpha = -\frac{3}{5} \) and \( \alpha \) is in **Quadrant III**. 2. \( \tan \beta = -\frac{5}{12} \) and \( \beta \) is in **Quadrant II**. ## Objectives: a) Find \( \cos (\alpha + \beta) \) b) Find \( \sin (\alpha + \beta) \) c) Determine the quadrant in which \( (\alpha + \beta) \) lies. --- ## Part a) \( \cos (\alpha + \beta) \) ### Step 1: Recall the Cosine Addition Formula \[ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \] ### Step 2: Find \( \sin \alpha \) and \( \cos \beta \), \( \sin \beta \) #### For \( \alpha \) in Quadrant III: - \( \cos \alpha = -\frac{3}{5} \) - In Quadrant III, both sine and cosine are negative. **Constructing the Right Triangle for \( \alpha \):** - Adjacent side: 3 (negative) - Hypotenuse: 5 - Opposite side: \( y \) (to be found) Using Pythagoras: \[ y = -\sqrt{5^2 - 3^2} = -\sqrt{25 - 9} = -\sqrt{16} = -4 \] \[ \sin \alpha = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{-4}{5} \] #### For \( \beta \) in Quadrant II: - \( \tan \beta = -\frac{5}{12} \) - In Quadrant II, sine is positive and cosine is negative. **Constructing the Right Triangle for \( \beta \):** - Opposite side: 5 (since tangent is opposite/adjacent, and tangent is negative in Quadrant II, adjacent side is negative) - Adjacent side: -12 - Hypotenuse: \( r \) (to be found) Using Pythagoras: \[ r = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \] \[ \cos \beta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{-12}{13} \] \[ \sin \beta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{13} \] ### Step 3: Apply the Cosine Addition Formula \[ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \] \[ = \left(-\frac{3}{5}\right) \left(-\frac{12}{13}\right) - \left(\frac{-4}{5}\right) \left(\frac{5}{13}\right) \] \[ = \frac{36}{65} - \left(-\frac{20}{65}\right) \] \[ = \frac{36}{65} + \frac{20}{65} = \frac{56}{65} \] **Answer for Part a):** \[ \cos (\alpha + \beta) = \frac{56}{65} \] --- ## Part b) \( \sin (\alpha + \beta) \) ### Step 1: Recall the Sine Addition Formula \[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] ### Step 2: Use the Previously Found Values \[ \sin (\alpha + \beta) = \left(\frac{-4}{5}\right) \left(-\frac{12}{13}\right) + \left(-\frac{3}{5}\right) \left(\frac{5}{13}\right) \] \[ = \frac{48}{65} + \left(-\frac{15}{65}\right) \] \[ = \frac{48}{65} - \frac{15}{65} = \frac{33}{65} \] **Answer for Part b):** \[ \sin (\alpha + \beta) = \frac{33}{65} \] --- ## Part c) Determine the Quadrant of \( (\alpha + \beta) \) To determine the quadrant where \( (\alpha + \beta) \) lies, we consider the signs of \( \cos (\alpha + \beta) \) and \( \sin (\alpha + \beta) \). From Parts a) and b): \[ \cos (\alpha + \beta) = \frac{56}{65} > 0 \] \[ \sin (\alpha + \beta) = \frac{33}{65} > 0 \] ### Step 1: Analyze the Signs - Both sine and cosine of \( (\alpha + \beta) \) are positive. ### Step 2: Recall Quadrant Sign Rules - **Quadrant I:** Both sine and cosine are positive. - **Quadrant II:** Sine is positive, cosine is negative. - **Quadrant III:** Both sine and cosine are negative. - **Quadrant IV:** Sine is negative, cosine is positive. ### Conclusion: Since both \( \sin (\alpha + \beta) \) and \( \cos (\alpha + \beta) \) are positive, \( (\alpha + \beta) \) lies in **Quadrant I**. **Answer for Part c):** \[ (\alpha + \beta) \text{ is in Quadrant I.} \] --- ## Summary of Answers: a) \( \cos (\alpha + \beta) = \frac{56}{65} \) b) \( \sin (\alpha + \beta) = \frac{33}{65} \) c) \( (\alpha + \beta) \) is in **Quadrant I** --- ### Visual Representation (Right Triangles) While I can't provide actual drawings here, I'll describe the right triangles for both angles in their respective quadrants. #### Triangle for \( \alpha \) in Quadrant III: - **Adjacent side (x):** -3 - **Opposite side (y):** -4 - **Hypotenuse (r):** 5 #### Triangle for \( \beta \) in Quadrant II: - **Adjacent side (x):** -12 - **Opposite side (y):** 5 - **Hypotenuse (r):** 13 These triangles help in determining the sine and cosine values based on the given trigonometric ratios and their signs in the respective quadrants. --- I hope this clarifies the problem and its solution! Let me know if you have any further questions.