Given \(\sin \theta = \frac{\sqrt{3}}{2}, 0° \leq \theta \leq 90°\), find \(\tan 2\theta\) in surd...
FURTHER MATHEMATICS
WAEC 2009
Given \(\sin \theta = \frac{\sqrt{3}}{2}, 0° \leq \theta \leq 90°\), find \(\tan 2\theta\) in surd form.
- A. \(- \sqrt{3}\)
- B. \(-\frac{\sqrt{3}}{2}\)
- C. \(\frac{\sqrt{3}}{2}\)
- D. \(\sqrt{3}\)
Correct Answer: A. \(- \sqrt{3}\)
Explanation
\(\sin \theta = \frac{\sqrt{3}}{2} \implies opp = \sqrt{3}; hyp = 2\)
\(adj^{2} = 2^{2} - (\sqrt{3})^{2} = 1 \implies adj = 1\)
\(\cos \theta = \frac{1}{2}\)
\(\sin 2\theta = \sin (180 - \theta) = \sin \theta = \frac{\sqrt{3}}{2}\)
\(\cos 2\theta = \cos (180 - \theta) = -\cos \theta = -\frac{1}{2}\)
\(\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\)
= \(- \sqrt{3}\)
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