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Solution 4.4:1f

From Förberedande kurs i matematik 1

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m (Lösning 4.4:1f moved to Solution 4.4:1f: Robot: moved page)
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{{NAVCONTENT_START}}
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We can translate the equation
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<center> [[Image:4_4_1f.gif]] </center>
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<math>\sin v=-\frac{1}{2}</math>
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{{NAVCONTENT_STOP}}
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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.
[[Image:4_4_1_f.gif|center]]
[[Image:4_4_1_f.gif|center]]
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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>

Revision as of 12:47, 30 September 2008

We can translate the equation sinv=−21 to the problem of finding those angles in the unit circle which have a y-coordinate of −21. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied sinv=+21, then the situation is the same, except that the angles now lie under, rather than above, the x -axis, due to reflectional symmetry.


Angle 2−6=611 Angle +6=67


The two angles which satisfy sinv=−21 lie in the third and fourth quadrants and are v=2−6=611 and v=+6=67

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