Chapter 7: Problem 9
Find the following exactly in radians and degrees. $$\cos ^{-1} \frac{\sqrt{3}}{2}$$
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Chapter 7: Problem 9
Find the following exactly in radians and degrees. $$\cos ^{-1} \frac{\sqrt{3}}{2}$$
These are the key concepts you need to understand to accurately answer the question.
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The acceleration due to gravity is often denoted by \(g\) in a formula such as \(S=\frac{1}{2} g t^{2},\) where \(S\) is the distance that an object falls in time \(t .\) The number \(g\) relates to motion near the earth's surface and is generally considered constant. In fact, however, \(g\) is not constant, but varies slightly with latitude. Latitude is used to measure north-south location on the earth between the equator and the poles. If \(\phi\) stands for latitude, in degrees, \(g\) is given with good approximation by the formula \(g=9.78049\left(1+0.005288 \sin ^{2} \phi-0.000006 \sin ^{2} 2 \phi\right)\), where \(g\) is measured in meters per second per second at sea level. a) Chicago has latitude \(42^{\circ} \mathrm{N}\). Find \(g .\) b) Philadelphia has latitude \(40^{\circ} \mathrm{N}\). Find \(g\). c) Express \(g\) in terms of \(\sin \phi\) only. That is, eliminate the double angle.
Solve in \([0,2 \pi)\),. $$e^{\sin x}=1$$
Simplify. Check your results using a graphing calculator. $$\sin \left(\frac{\pi}{2}-x\right)[\sec x-\cos x]$$
Use the given substitution to express the given radical expression as a trigonometric function without radicals. Assume that \(a>0\) and \(0<\theta<\pi / 2 .\) Then find expressions for the indicated trigonometric functions. Let \(x=a \sec \theta\) in \(\sqrt{x^{2}-a^{2}} .\) Then find \(\sin \theta\) and \(\cos \theta\)
Prove the identity. \(\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}},\) for \(x \geq 0\)
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