From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	By subtracting 360° from 495°, we do not change the value of the tangent,
-	<center> [[Bild:4_2_5d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\tan 495^{\circ} = \tan (495^{\circ} - 360^{\circ}) = \tan 135^{\circ}\,\textrm{.}</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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		+	{{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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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{.}