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Solution 4.3:6c

From Förberedande kurs i matematik 1

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m (Lösning 4.3:6c moved to Solution 4.3:6c: Robot: moved page)
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{{NAVCONTENT_START}}
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Because the angle
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<center> [[Image:4_3_6c-1(2).gif]] </center>
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<math>v</math>
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{{NAVCONTENT_STOP}}
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satisfies
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{{NAVCONTENT_START}}
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<math>\pi \le v\le \frac{3\pi }{2}</math>,
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<center> [[Image:4_3_6c-2(2).gif]] </center>
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<math>v</math>
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{{NAVCONTENT_STOP}}
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belongs to the third quadrant in the unit circle. Furthermore,
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<math>\text{tan }v=\text{3 }</math>
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gives that the line which corresponds to the angle
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<math>v</math>
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 +
<math>v</math>
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has a gradient of
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<math>\text{3}</math>.
 +
 
[[Image:4_3_6_c1.gif|center]]
[[Image:4_3_6_c1.gif|center]]
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 +
slope 3
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 +
 +
In the third quadrant, we can introduce a right-angled triangle in which the hypotenuse is
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<math>\text{1}</math>
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and the sides have a
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<math>\text{3}:\text{1 }</math>
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ratio.
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[[Image:4_3_6_c2.gif|center]]
[[Image:4_3_6_c2.gif|center]]
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If we now use Pythagoras' theorem on the triangle, we see that the horizontal side
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<math>\text{a}</math>
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satisfies
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 +
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<math>a^{2}+\left( 3a \right)^{2}=1^{2}</math>
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 +
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which gives us that
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 +
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<math>10a^{2}=1</math>
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i.e.
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<math>a=\frac{1}{\sqrt{10}}</math>
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 +
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Thus, the angle
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<math>v</math>'s
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<math>x</math>
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-coordinate is
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<math>-\frac{1}{\sqrt{10}}</math>
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and
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<math>y</math>
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-coordinate is
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<math>-\frac{3}{\sqrt{10}}</math>, i.e.
 +
 +
<math>\cos v=--\frac{1}{\sqrt{10}}</math>
 +
 +
 +
<math>\sin v=-\frac{3}{\sqrt{10}}</math>

Revision as of 09:38, 30 September 2008

Because the angle v satisfies v23, v belongs to the third quadrant in the unit circle. Furthermore, tan v=3 gives that the line which corresponds to the angle v

v has a gradient of 3.


slope 3


In the third quadrant, we can introduce a right-angled triangle in which the hypotenuse is 1 and the sides have a 3:1 ratio.

If we now use Pythagoras' theorem on the triangle, we see that the horizontal side a satisfies


a2+3a2=12 


which gives us that


10a2=1 i.e. a=110


Thus, the angle v's x -coordinate is −110 and y -coordinate is −310, i.e.

cosv=−−110


sinv=−310

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