Chapter 5: Problem 42
Explain why
$$\cos ^{-1} t=\tan ^{-1} \frac{\sqrt{1-t^{2}}}{t}$$
whenever \(0
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Chapter 5: Problem 42
Explain why
$$\cos ^{-1} t=\tan ^{-1} \frac{\sqrt{1-t^{2}}}{t}$$
whenever \(0
These are the key concepts you need to understand to accurately answer the question.
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Find a formula for \(\cos \left(\theta+\frac{\pi}{4}\right)\).
Suppose \(\theta\) is an angle such that \(\cos \theta\) is rational. Explain why \(\cos (2 \theta)\) is rational.
Evaluate the given quantities assuming that \(u\) and \(v\) are both in the interval \(\left(-\frac{\pi}{2}, 0\right)\) and \(\tan u=-\frac{1}{7} \quad\) and \(\quad \tan v=-\frac{1}{8}\) $$\tan (2 v)$$
Show that $$\cos (2 \theta) \leq \cos ^{2} \theta$$ for every angle \(\theta\).
Show that $$\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{\sqrt{2+\sqrt{3}}}{2}$$ Do this without using a calculator and without using the knowledge that both expressions above are equal to \(\cos 15^{\circ}\) (see Example 2 in this section and Example 3 in Section 5.5\()\)
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