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Solution 4.3:6c

From Förberedande kurs i matematik 1

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Current revision (07:48, 10 October 2008) (edit) (undo)
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Because the angle
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Because the angle <math>v</math> satisfies <math>\pi \le v\le 3\pi/2\,</math>, <math>v</math> belongs to the third quadrant in the unit circle. Furthermore, <math>\tan v = 3</math> gives that the line which corresponds to the angle
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<math>v</math>
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<math>v</math> has slope 3.
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satisfies
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<math>\pi \le v\le \frac{3\pi }{2}</math>,
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<math>v</math>
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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>.
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[[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 1 and the sides have a 3:1 ratio.
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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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If we now use the Pythagorean theorem on the triangle, we see that the horizontal side ''a'' satisfies
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<math>\text{a}</math>
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satisfies
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<math>a^{2}+\left( 3a \right)^{2}=1^{2}</math>
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which gives us that
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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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Thus, the angle
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{{Displayed math||<math>a^2 + (3a)^2 = 1^2</math>}}
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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.
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<math>\cos v=--\frac{1}{\sqrt{10}}</math>
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which gives us that <math>10a^{2}=1</math> i.e. <math>a = 1/\!\sqrt{10}\,\textrm{.}</math>
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Thus, the angle ''v'''s ''x''-coordinate is <math>-1/\!\sqrt{10}</math> and ''y''-coordinate is <math>-3/\!\sqrt{10}</math>, i.e.
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<math>\sin v=-\frac{3}{\sqrt{10}}</math>
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{{Displayed math||<math>\begin{align}
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\cos v &= -\frac{1}{\sqrt{10}}\,,\\[5pt]
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\sin v &= -\frac{3}{\sqrt{10}}\,\textrm{.}
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\end{align}</math>}}

Current revision

Because the angle v satisfies v32, v belongs to the third quadrant in the unit circle. Furthermore, tanv=3 gives that the line which corresponds to the angle v has 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 the Pythagorean theorem on the triangle, we see that the horizontal side a satisfies

a2+(3a)2=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.

cosvsinv=−110=−310.
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