Solution 4.4:1f

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Current revision (13:41, 10 October 2008) (edit) (undo)
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We can translate the equation
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We can translate the equation <math>\sin v=-1/2</math> to the problem of finding those angles in the unit circle which have a ''y''-coordinate of <math>-1/2</math>. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied <math>\sin v = +1/2</math>, then the situation is the same, except that the angles now lie under, rather than above, the ''x''-axis, due to reflectional symmetry.
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<math>\sin v=-\frac{1}{2}</math>
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to the problem of finding those angles in the unit circle which have a y-coordinate of
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<math>-\frac{1}{2}</math>. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied
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<math>\sin v=+\frac{1}{2}</math>, then the situation is the same, except that the angles now lie under, rather than above, the
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<math>x</math>
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-axis, due to reflectional symmetry.
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{| align="center"
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| align="center" |[[Image:4_4_1_f-1.gif|center]]
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| width="20px" |&nbsp;
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| align="center" |[[Image:4_4_1_f-2.gif|center]]
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|-
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| align="center" |<small>Angle 2π&nbsp;-&nbsp;π/6&nbsp;=&nbsp;11π/6</small>
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| align="center" |<small>Angle π&nbsp;+&nbsp;π/6&nbsp;=&nbsp;7π/6</small>
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|}
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[[Image:4_4_1_f.gif|center]]
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The two angles which satisfy <math>\sin v=-1/2</math> lie in the third and fourth quadrants and are <math>v=2\pi - \pi/6 = 11\pi/6</math> and <math>v = \pi + \pi/6 = 7\pi/6</math>.
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Angle
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<math>2\pi -\frac{\pi }{6}=\frac{11\pi }{6}</math>
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Angle
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<math>\pi +\frac{\pi }{6}=\frac{7\pi }{6}</math>
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The two angles which satisfy
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<math>\sin v=-\frac{1}{2}</math>
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lie in the third and fourth quadrants and are
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<math>v=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}</math>
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and
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<math>v=\pi +\frac{\pi }{6}=\frac{7\pi }{6}</math>
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Current revision

We can translate the equation \displaystyle \sin v=-1/2 to the problem of finding those angles in the unit circle which have a y-coordinate of \displaystyle -1/2. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied \displaystyle \sin v = +1/2, then the situation is the same, except that the angles now lie under, rather than above, the x-axis, due to reflectional symmetry.

 
Angle 2π - π/6 = 11π/6 Angle π + π/6 = 7π/6

The two angles which satisfy \displaystyle \sin v=-1/2 lie in the third and fourth quadrants and are \displaystyle v=2\pi - \pi/6 = 11\pi/6 and \displaystyle v = \pi + \pi/6 = 7\pi/6.

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