Solution 4.2:5d

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Current revision (11:15, 9 October 2008) (edit) (undo)
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By subtracting
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By subtracting 360° from 495°, we do not change the value of the tangent,
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<math>360^{\circ }</math>
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from
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<math>\text{495}^{\circ }</math>, we do not change the value of the tangent:
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{{Displayed math||<math>\tan 495^{\circ} = \tan (495^{\circ} - 360^{\circ}) = \tan 135^{\circ}\,\textrm{.}</math>}}
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<math>\tan \text{495}^{\circ }=\tan \left( \text{495}^{\circ }-360^{\circ } \right)=\tan \text{135}^{\circ }</math>
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We know from exercise a that <math>\cos 135^{\circ} = -1/\!\sqrt{2}</math> and <math>\sin 135^{\circ} = 1/\!\sqrt{2}\,</math>, which gives
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We know from exercise a that
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{{Displayed math||<math>\tan 135^{\circ} = \frac{\sin 135^{\circ}}{\cos 135^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}</math>}}
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<math>\cos 135^{\circ }=-\frac{1}{\sqrt{2}}</math>
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and
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<math>\sin 135^{\circ }=\frac{1}{\sqrt{2}}</math>, which gives
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<math>\tan 135^{\circ }=\frac{\sin 135^{\circ }}{\cos 135^{\circ }}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1</math>
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Current revision

By subtracting 360° from 495°, we do not change the value of the tangent,

\displaystyle \tan 495^{\circ} = \tan (495^{\circ} - 360^{\circ}) = \tan 135^{\circ}\,\textrm{.}

We know from exercise a that \displaystyle \cos 135^{\circ} = -1/\!\sqrt{2} and \displaystyle \sin 135^{\circ} = 1/\!\sqrt{2}\,, which gives

\displaystyle \tan 135^{\circ} = \frac{\sin 135^{\circ}}{\cos 135^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}
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