Solution 4.2:4f

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Current revision (10:52, 9 October 2008) (edit) (undo)
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If we add <math>2\pi</math> to <math>-5\pi/3\,</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle <math>-5\pi/3</math> and consequently has the same tangent value,
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<center> [[Image:4_2_4f.gif]] </center>
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{{Displayed math||<math>\begin{align}
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\tan\Bigl(-\frac{5\pi}{3}\Bigr)
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= \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr)
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= \tan\frac{\pi}{3}
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= \frac{\sin\dfrac{\pi}{3}}{\cos\dfrac{\pi}{3}}
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= \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}
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= \sqrt{3}\,\textrm{.}
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\end{align}</math>}}

Current revision

If we add \displaystyle 2\pi to \displaystyle -5\pi/3\,, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle \displaystyle -5\pi/3 and consequently has the same tangent value,

\displaystyle \begin{align}

\tan\Bigl(-\frac{5\pi}{3}\Bigr) = \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr) = \tan\frac{\pi}{3} = \frac{\sin\dfrac{\pi}{3}}{\cos\dfrac{\pi}{3}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3}\,\textrm{.} \end{align}

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