Solution 4.2:4c

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Current revision (08:41, 9 October 2008) (edit) (undo)
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In exercise e, we studied the angle
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In exercise 4.2:3e, we studied the angle <math>3\pi/4</math> and found that
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<math>\frac{3\pi }{4}</math>
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and found that
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<math>\cos \frac{3\pi }{4}=-\frac{1}{\sqrt{2}}</math>
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{{Displayed math||<math>\cos\frac{3\pi }{4} = -\frac{1}{\sqrt{2}}\qquad\text{and}\qquad\sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}\,\textrm{.}</math>}}
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and
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<math>\sin \frac{3\pi }{4}=\frac{1}{\sqrt{2}}</math>
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Because <math>\tan x</math> is defined as <math>\frac{\sin x}{\cos x}</math>, we get immediately that
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Because
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{{Displayed math||<math>\tan\frac{3\pi}{4} = \frac{\sin\dfrac{3\pi}{4}}{\cos \dfrac{3\pi}{4}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}</math>}}
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<math>\text{tan }x</math>
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is defined as
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<math>\frac{\sin x}{\cos x}</math>, we get immediately that
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<math>\tan \frac{3\pi }{4}=\frac{\sin \frac{3\pi }{4}}{\cos \frac{3\pi }{4}}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1</math>
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Current revision

In exercise 4.2:3e, we studied the angle \displaystyle 3\pi/4 and found that

\displaystyle \cos\frac{3\pi }{4} = -\frac{1}{\sqrt{2}}\qquad\text{and}\qquad\sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}\,\textrm{.}

Because \displaystyle \tan x is defined as \displaystyle \frac{\sin x}{\cos x}, we get immediately that

\displaystyle \tan\frac{3\pi}{4} = \frac{\sin\dfrac{3\pi}{4}}{\cos \dfrac{3\pi}{4}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}
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