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Solution 4.2:5d

From Förberedande kurs i matematik 1

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m (Lösning 4.2:5d moved to Solution 4.2:5d: Robot: moved page)
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{{NAVCONTENT_START}}
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By subtracting
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<center> [[Image:4_2_5d.gif]] </center>
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<math>360^{\circ }</math>
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{{NAVCONTENT_STOP}}
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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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<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
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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>

Revision as of 08:16, 29 September 2008

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


tan495=tan495−360=tan135 

We know from exercise a that cos135=−12 and sin135=12, which gives


tan135=sin135cos135=12−12=−1

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