Chapter 5: Problem 46
Show that $$\sin u \sin v=\frac{\cos (u-v)-\cos (u+v)}{2}$$ for all \(u, v\).
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Chapter 5: Problem 46
Show that $$\sin u \sin v=\frac{\cos (u-v)-\cos (u+v)}{2}$$ for all \(u, v\).
These are the key concepts you need to understand to accurately answer the question.
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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 \(5.5 .]\) $$\sin 82.5^{\circ}$$
Suppose \(-\frac{\pi}{2}<\theta<0\) and \(\cos \theta=0.3\). (a) Without using a double-angle formula, evaluate \(\cos (2 \theta)\) by first finding \(\theta\) using an inverse trigonometric function. (b) Without using an inverse trigonometric function, evaluate \(\cos (2 \theta)\) again by using a double-angle formula.
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 \(\left(-\frac{\pi}{2}, 0\right) .\) $$\sin (x+y)$$
Show that $$\tan \frac{x}{4}=\frac{\sqrt{2-2 \cos x}-\sin x}{1-\cos x}$$ for all \(x\) in the interval \((0,2 \pi)\). [Hint: Start with a half-angle formula for tangent to express \(\tan \frac{x}{4}\) in terms of \(\sin \frac{x}{2}\) and \(\cos \frac{x}{2}\). Then use half-angle formulas for cosine and sine, along with algebraic manipulations.]
Show that $$\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}$$ for all \(x, y\).
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