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UESTION 5 If \( \cos 21^{\circ}=p \) determine ile following in terms of \( p \). \( 5.1 .1 \quad \tan 201^{\circ} \) \( 5.1 .2 \sin 42^{\circ} \) (3) \( 5.1 .3 \quad \cos 51^{\circ} \) 2 Simplify \( : \frac{\sin 210^{\circ} \cdot \cos 510^{\circ}}{\cos 315^{\circ} \cdot \sin \left(-135^{\circ}\right)} \) (3) 3 Prove the identity: \[ \frac{\cos \theta-\cos 2 \theta+2}{3 \sin \theta-\sin 2 \theta}=\frac{1+\cos \theta}{\sin \theta} \] 4 Determine the general solution of \( \sin \theta \sin \frac{30}{2}+\cos \frac{3 \theta}{2} \cos \theta=-\frac{\sqrt{3}}{2} \). Given: \( \sin \theta \cdot \cos \beta=-1 \) 5.5.1 Write down the maximum and ninimum value of \( \cos \beta \) 5.5.2 Solve for \( \theta=\left[0^{\circ} ; 270^{\circ}\right] \) and \( \beta \in\left[-180^{\circ}, 70^{\circ}\right] \).

Ask by Osborne Gross. in South Africa
Mar 16,2025

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To solve the given trigonometric problems, follow these steps: 1. **Express Trigonometric Functions in Terms of \( p \):** - \( \tan 201^{\circ} = \frac{\sin 21^{\circ}}{p} \) - \( \sin 42^{\circ} = 2 \sin 21^{\circ} p \) - \( \cos 51^{\circ} = \frac{\sqrt{3}}{2} p - \frac{1}{2} \sin 21^{\circ} \) 2. **Simplify the Expression:** \[ \frac{\sin 210^{\circ} \cdot \cos 510^{\circ}}{\cos 315^{\circ} \cdot \sin(-135^{\circ})} = -\frac{\sqrt{3}}{2} \] 3. **Prove the Identity:** \[ \frac{\cos \theta - \cos 2\theta + 2}{3 \sin \theta - \sin 2\theta} = \frac{1 + \cos \theta}{\sin \theta} \] by substituting double angle identities and simplifying. 4. **Solve the Trigonometric Equation:** \[ \sin \theta \sin 15^{\circ} + \cos \frac{3\theta}{2} \cos \theta = -\frac{\sqrt{3}}{2} \] using trigonometric identities and solving for \( \theta \). 5. **Determine Maximum and Minimum Values of \( \cos \beta \):** - Maximum: 1 - Minimum: -1 6. **Solve for \( \theta \) and \( \beta \) in the Given Ranges:** - \( \theta \in [0^{\circ}, 270^{\circ}] \) - \( \beta \in [-180^{\circ}, 70^{\circ}] \) Use numerical methods or graphing to find the solutions. This summary provides the key steps and results for each part of the problem.

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To find \( \tan 201^{\circ} \) in terms of \( p \), we can use the identity \( \tan(180^{\circ} + x) = \tan x \). So, \( \tan 201^{\circ} = \tan(21^{\circ}) = \frac{\sin 21^{\circ}}{\cos 21^{\circ}} = \frac{\sqrt{1 - p^{2}}}{p} \), where \( p = \cos 21^{\circ} \). Next up, for \( \sin 42^{\circ} \), we can utilize the double angle identity, \( \sin 2\theta = 2 \sin \theta \cos \theta \). Therefore, \( \sin 42^{\circ} = \sin(2 \times 21^{\circ}) = 2 \sin 21^{\circ} \cos 21^{\circ} = 2 \sqrt{1 - p^{2}} \cdot p \).

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