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Solution 4.3:6b

From Förberedande kurs i matematik 1

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We draw an angle <math>v</math> in the unit circle, and the fact that <math>\sin v = 3/10</math> means that its ''y''-coordinate equals <math>3/10</math>.
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[[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 1 and a vertical side of length 3/10.
[[Image:4_3_6_b2.gif|center]]
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We can determine the triangle's remaining side by using the Pythagorean theorem,
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{{Displayed math||<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}
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which gives that
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{{Displayed math||<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}
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This means that the angle's ''x''-coordinate is <math>-a</math>, i.e. we have
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{{Displayed math||<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}
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and thus
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{{Displayed math||<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}

Current revision

We draw an angle v in the unit circle, and the fact that sinv=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 3/10.

We can determine the triangle's remaining side by using the Pythagorean 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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