Chapter 5: Problem 75
Show that $$|\sin (2 \theta)| \leq 2|\sin \theta|$$ for every angle \(\theta\).
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Chapter 5: Problem 75
Show that $$|\sin (2 \theta)| \leq 2|\sin \theta|$$ for every angle \(\theta\).
These are the key concepts you need to understand to accurately answer the question.
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Evaluate the indicated expressions assuming that $$ \begin{array}{ll} \cos x=\frac{1}{3} & \text { and } \sin y=\frac{1}{4} \\ \sin u=\frac{2}{3} & \text { and } \cos v=\frac{1}{5} \end{array} $$ Assume also that \(x\) and \(u\) are in the interval \(\left(0, \frac{\pi}{2}\right),\) that \(y\) is in the interval \(\left(\frac{\pi}{2}, \pi\right),\) and that \(v\) is in the interval \(\left(-\frac{\pi}{2}, 0\right) .\) $$\tan (u-v)$$
Find a formula for \(\tan \left(\theta-\frac{\pi}{4}\right)\).
Find angles \(u\) and \(v\) such that \(\sin (2 u)=\sin (2 v)\) but \(|\sin u| \neq|\sin v|\).
Find exact expressions for the indicated quantities. The following information will be useful: $$ \begin{array}{l} \cos 22.5^{\circ}=\frac{\sqrt{2+\sqrt{2}}}{2} \text { and } \sin 22.5^{\circ}=\frac{\sqrt{2-\sqrt{2}}}{2} \\ \cos 18^{\circ}=\sqrt{\frac{\sqrt{5}+5}{8}} \text { and } \sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \end{array} $$ [The value for \(\sin 22.5^{\circ}\) used here was derived in Example 4 in Section \(5.5 ;\) the other values were derived in Exercise 64 and Problems 102 and 103 in Section \(5.5 .]\) $$\sin 37.5^{\circ}$$
Evaluate \(\cos \left(\cos ^{-1} \frac{2}{3}+\tan ^{-1} 3\right)\).
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