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]] | [[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 \displaystyle v in the unit circle, and the fact that \displaystyle \sin v = 3/10 means that its y-coordinate equals \displaystyle 3/10.
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,
\displaystyle a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2 |
which gives that
\displaystyle a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.} |
This means that the angle's x-coordinate is \displaystyle -a, i.e. we have
\displaystyle \cos v=-\frac{\sqrt{91}}{10} |
and thus
\displaystyle \tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.} |