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8. sin 2a = Zsn AcosA 24 25 12 25 9. cos 2a =

8. sin 2a = Zsn AcosA 24 25 12 25 9. cos 2a =

Trigonometry (MindTap Course List)
Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter2: Right Triangle Trigonometry
Section: Chapter Questions
Problem 6GP
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Did I do 8 and 9 correctly?

## Trigonometric Identities and Calculations ### Given: - \( \sin \alpha = -\frac{3}{5}, \quad \frac{3\pi}{2} \leq \alpha \leq 2\pi \) - \( \cos \beta = -\frac{8}{17}, \quad \pi \leq \beta \leq \frac{3\pi}{2} \) ### 1. Finding \( \cos \alpha \) Using the identity: \[ \cos^2 \alpha = 1 - \sin^2 \alpha \] \[ \sin^2 \alpha = \left(-\frac{3}{5}\right)^2 = \frac{9}{25} \] \[ \cos^2 \alpha = 1 - \frac{9}{25} = \frac{16}{25} \] Therefore, \( \cos \alpha = \sqrt{\frac{16}{25}} = \frac{4}{5} \) (noting the quadrant) ### 2. Finding \( \sin \beta \) Using the identity: \[ \sin^2 \beta = 1 - \cos^2 \beta \] \[ \cos^2 \beta = \left(-\frac{8}{17}\right)^2 = \frac{64}{289} \] \[ \sin^2 \beta = 1 - \frac{64}{289} = \frac{225}{289} \] Therefore, \( \sin \beta = \sqrt{\frac{225}{289}} = \frac{15}{17} \) (noting the quadrant) ### 3. Finding \( \tan \alpha \) Using the identity: \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \] \[ \tan \alpha = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4} \] ### 4. Finding \( \tan \beta \) Using the identity: \[ \tan \beta = \frac{\sin \beta}{\cos \beta} \] \[ \tan \beta = \frac{\frac{15}{17}}{-\frac{8}{17}} = -\frac{15}{8} \] ### 5. Finding \( \sin(\alpha + \beta) \
Transcribed Image Text:## Trigonometric Identities and Calculations ### Given: - \( \sin \alpha = -\frac{3}{5}, \quad \frac{3\pi}{2} \leq \alpha \leq 2\pi \) - \( \cos \beta = -\frac{8}{17}, \quad \pi \leq \beta \leq \frac{3\pi}{2} \) ### 1. Finding \( \cos \alpha \) Using the identity: \[ \cos^2 \alpha = 1 - \sin^2 \alpha \] \[ \sin^2 \alpha = \left(-\frac{3}{5}\right)^2 = \frac{9}{25} \] \[ \cos^2 \alpha = 1 - \frac{9}{25} = \frac{16}{25} \] Therefore, \( \cos \alpha = \sqrt{\frac{16}{25}} = \frac{4}{5} \) (noting the quadrant) ### 2. Finding \( \sin \beta \) Using the identity: \[ \sin^2 \beta = 1 - \cos^2 \beta \] \[ \cos^2 \beta = \left(-\frac{8}{17}\right)^2 = \frac{64}{289} \] \[ \sin^2 \beta = 1 - \frac{64}{289} = \frac{225}{289} \] Therefore, \( \sin \beta = \sqrt{\frac{225}{289}} = \frac{15}{17} \) (noting the quadrant) ### 3. Finding \( \tan \alpha \) Using the identity: \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \] \[ \tan \alpha = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4} \] ### 4. Finding \( \tan \beta \) Using the identity: \[ \tan \beta = \frac{\sin \beta}{\cos \beta} \] \[ \tan \beta = \frac{\frac{15}{17}}{-\frac{8}{17}} = -\frac{15}{8} \] ### 5. Finding \( \sin(\alpha + \beta) \
Certainly! Below is the transcription and explanation of the image content for an educational website: --- ### Trigonometric Identities and Examples **Example 8: Calculation of \(\sin 2\alpha\)** The identity used here is: \[ \sin 2\alpha = 2 \sin A \cos A \] Given: \[ \sin A = -\frac{3}{5}, \quad \cos A = -\frac{4}{5} \] Calculation: \[ 2 \left( -\frac{3}{5} \right) \left( -\frac{4}{5} \right) = 2 \left( \frac{12}{25} \right) = \frac{24}{25} \] **Example 9: Calculation of \(\cos 2\alpha\)** The identity used here is: \[ \cos 2\alpha = 1 - 2\sin^2 A \] Given: \(\sin A = -\frac{3}{5}\) Calculation: \[ 1 - 2 \left( \left( -\frac{3}{5} \right)^2 \right) \] First, calculate \(\sin^2 A\): \[ \left( -\frac{3}{5} \right)^2 = \frac{9}{25} \] Then: \[ 1 - 2 \left( \frac{9}{25} \right) = 1 - \frac{18}{25} = \frac{7}{25} \] **Example 10: Calculation of \(\sin 3\alpha\)** The identity noted here is: \[ \sin 3\alpha = 3 \sin A \cos A \] **Hint:** Why did I ask #8 and #9? --- These examples illustrate the application of trigonometric identities to calculate specific trigonometric values, given certain initial sine and cosine values. The work shown uses algebraic manipulation to solve and verify these identities.
Transcribed Image Text:Certainly! Below is the transcription and explanation of the image content for an educational website: --- ### Trigonometric Identities and Examples **Example 8: Calculation of \(\sin 2\alpha\)** The identity used here is: \[ \sin 2\alpha = 2 \sin A \cos A \] Given: \[ \sin A = -\frac{3}{5}, \quad \cos A = -\frac{4}{5} \] Calculation: \[ 2 \left( -\frac{3}{5} \right) \left( -\frac{4}{5} \right) = 2 \left( \frac{12}{25} \right) = \frac{24}{25} \] **Example 9: Calculation of \(\cos 2\alpha\)** The identity used here is: \[ \cos 2\alpha = 1 - 2\sin^2 A \] Given: \(\sin A = -\frac{3}{5}\) Calculation: \[ 1 - 2 \left( \left( -\frac{3}{5} \right)^2 \right) \] First, calculate \(\sin^2 A\): \[ \left( -\frac{3}{5} \right)^2 = \frac{9}{25} \] Then: \[ 1 - 2 \left( \frac{9}{25} \right) = 1 - \frac{18}{25} = \frac{7}{25} \] **Example 10: Calculation of \(\sin 3\alpha\)** The identity noted here is: \[ \sin 3\alpha = 3 \sin A \cos A \] **Hint:** Why did I ask #8 and #9? --- These examples illustrate the application of trigonometric identities to calculate specific trigonometric values, given certain initial sine and cosine values. The work shown uses algebraic manipulation to solve and verify these identities.
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