Aus Online Mathematik Brückenkurs 1

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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. {| align="center" | align="center" |[[Image:4_4_1_f-1.gif|center]] | width="20px" |&nbsp; | align="center" |[[Image:4_4_1_f-2.gif|center]] |- | align="center" |<small>Angle 2π&nbsp;-&nbsp;π/6&nbsp;=&nbsp;11π/6</small> || | align="center" |<small>Angle π&nbsp;+&nbsp;π/6&nbsp;=&nbsp;7π/6</small> |} 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>.

• 4.4.1f - Solution - The unit circle with angle 2π - π/6
• 4.4.1f - Solution - The unit circle with angle π + π/6

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