Solution 4.3:6a

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If we think of the angle v as an angle in the unit circle, then
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If we think of the angle v as an angle in the unit circle, then ''v'' lies in the fourth quadrant and has ''x''-coordinate 3/4.
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<math>v</math>
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lies in the fourth quadrant and has
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<math>x</math>
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-coordinate
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<math>\frac{3}{4}</math>.
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[[Image:4_3_6_a1.gif|center]]
[[Image:4_3_6_a1.gif|center]]
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If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to
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If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to 1 and an opposite side equal to 3/4.
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<math>\text{1}</math>
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and an opposite side equal to
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<math>\frac{3}{4}</math>.
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[[Image:4_3_6_a2.gif|center]]
[[Image:4_3_6_a2.gif|center]]
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Using Pythagoras' theorem, it is possible to determine the remaining side from
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Using the Pythagorean theorem, it is possible to determine the remaining side from
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<math>b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2}</math>
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{{Displayed math||<math>b^2 + \Bigl(\frac{3}{4}\Bigr)^2 = 1^2</math>}}
which gives that
which gives that
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{{Displayed math||<math>b = \sqrt{1-\Bigl(\frac{3}{4}\Bigr)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\,\textrm{.}</math>}}
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<math>\begin{align}
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Because the angle ''v'' belongs to the fourth quadrant, its ''y''-coordinate is negative and is therefore equal to <math>-b</math>, i.e.
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& b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2} \\
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& b=\sqrt{1-\left( \frac{3}{4} \right)^{2}}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4} \\
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\end{align}</math>
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Because the angle
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<math>v</math>
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belongs to the fourth quadrant, its
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<math>y</math>
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-coordinate is negative and is therefore equal to
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<math>-b</math>, i.e.
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<math>\sin v=-\frac{\sqrt{7}}{4}</math>
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{{Displayed math||<math>\sin v=-\frac{\sqrt{7}}{4}\,\textrm{.}</math>}}
Thus, we have directly that
Thus, we have directly that
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{{Displayed math||<math>\tan v = \frac{\sin v}{\cos v} = \frac{-\sqrt{7}/4}{3/4} = -\frac{\sqrt{7}}{3}\,\textrm{.}</math>}}
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<math>\tan v=\frac{\sin v}{\cos v}=\frac{-\frac{\sqrt{7}}{4}}{\frac{3}{4}}=-\frac{\sqrt{7}}{3}</math>
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Current revision

If we think of the angle v as an angle in the unit circle, then v lies in the fourth quadrant and has x-coordinate 3/4.

If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to 1 and an opposite side equal to 3/4.

Using the Pythagorean theorem, it is possible to determine the remaining side from

\displaystyle b^2 + \Bigl(\frac{3}{4}\Bigr)^2 = 1^2

which gives that

\displaystyle b = \sqrt{1-\Bigl(\frac{3}{4}\Bigr)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\,\textrm{.}

Because the angle v belongs to the fourth quadrant, its y-coordinate is negative and is therefore equal to \displaystyle -b, i.e.

\displaystyle \sin v=-\frac{\sqrt{7}}{4}\,\textrm{.}

Thus, we have directly that

\displaystyle \tan v = \frac{\sin v}{\cos v} = \frac{-\sqrt{7}/4}{3/4} = -\frac{\sqrt{7}}{3}\,\textrm{.}
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