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Solution 4.3:6b

From Förberedande kurs i matematik 1

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m (Lösning 4.3:6b moved to Solution 4.3:6b: Robot: moved page)
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{{NAVCONTENT_START}}
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We draw an angle
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<center> [[Image:4_3_6b-1(2).gif]] </center>
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<math>v</math>
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{{NAVCONTENT_STOP}}
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in the unit circle, and the fact that
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{{NAVCONTENT_START}}
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<math>\text{sin }v\text{ }=\frac{3}{10}</math>
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<center> [[Image:4_3_6b-2(2).gif]] </center>
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means that its
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{{NAVCONTENT_STOP}}
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<math>y</math>
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-coordinate equals
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<math>\frac{3}{10}</math>.
[[Image:4_3_6_b1.gif|center]]
[[Image:4_3_6_b1.gif|center]]
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With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of
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<math>\text{1}</math>
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and a vertical side of length
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<math>\frac{3}{10}</math>.
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[[Image:4_3_6_b2.gif|center]]
[[Image:4_3_6_b2.gif|center]]
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We can determine the triangle's remaining side by using Pythagoras' theorem,
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<math>a^{2}+\left( \frac{3}{10} \right)^{2}=1^{2}</math>
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which gives that
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<math>a=\sqrt{1-\left( \frac{3}{10} \right)^{2}}=\sqrt{1-\frac{9}{100}}=\sqrt{\frac{91}{100}}=\frac{\sqrt{91}}{10}</math>
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This means that the angle's
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<math>x</math>
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-coordinate is
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<math>-a</math>, i.e. we have
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 +
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<math>\cos v=-\frac{\sqrt{91}}{10}</math>
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and thus
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<math>\tan v=\frac{\sin v}{\cos v}=\frac{\frac{3}{10}}{-\frac{\sqrt{91}}{10}}=-\frac{3}{\sqrt{91}}</math>

Revision as of 09:15, 10 October 2008

We draw an angle v in the unit circle, and the fact that sin v =310 means that its y -coordinate equals 310.

With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of 1 and a vertical side of length 310.


We can determine the triangle's remaining side by using Pythagoras' theorem,


a2+3102=12 


which gives that


a=1−3102=1−9100=91100=1091 

This means that the angle's x -coordinate is −a, i.e. we have


cosv=−1091 

and thus


tanv=sinvcosv=310−1091=−391

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