Solution 4.4:5b

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Let's first investigate when the equality
Let's first investigate when the equality
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{{Displayed math||<math>\tan u=\tan v</math>}}
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<math>\tan u=\tan v</math>
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is satisfied. Because <math>\tan u</math> can be interpreted as the slope (gradient) of the line which makes an angle ''u'' with the positive ''x''-axis, we see that for a fixed value of <math>\tan u</math>, there are two angles ''v'' in the unit circle with this slope,
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is satisfied. Because
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<math>u</math>
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can be interpreted as the slope (gradient) of the line which makes an angle
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<math>u</math>
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with the positive
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<math>x</math>
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-axis, we see that for a fixed value of tan u, there are two angles
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<math>v</math>
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in the unit circle with this slope:
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<math>v=u</math>
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and
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<math>v=u+\pi </math>
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{{Displayed math||<math>v=u\qquad\text{and}\qquad v=u+\pi\,\textrm{.}</math>}}
[[Image:4_4_5_b.gif|center]]
[[Image:4_4_5_b.gif|center]]
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slope
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The angle ''v'' has the same slope after every half turn, so if we add multiples of
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<math>=\text{ tan }u</math>
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<math>\pi</math> to ''u'', we will obtain all the angles ''v'' which satisfy the equality
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slope
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<math>=\text{ tan }u</math>
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{{Displayed math||<math>v=u+n\pi\,,</math>}}
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The angle
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where ''n'' is an arbitrary integer.
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<math>v</math>
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has the same slope after every half turn, so if we add multiples of
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<math>\pi \text{ }</math>
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to
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<math>u</math>, we will obtain all the angles
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<math>v</math>
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which satisfy the equality
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<math>v=u+n\pi </math>
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where
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<math>n</math>
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is an arbitrary integer.
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If we apply this result to the equation
If we apply this result to the equation
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{{Displayed math||<math>\tan x=\tan 4x</math>}}
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<math>\tan x=\tan 4x</math>
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we see that the solutions are given by
we see that the solutions are given by
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{{Displayed math||<math>4x = x+n\pi\qquad\text{(n is an arbitrary integer),}</math>}}
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<math>4x=x+n\pi </math>
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and solving for ''x'' gives
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(
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<math>n</math>
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an arbitrary integer),
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and solving for
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<math>x</math>
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gives
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<math>x=\frac{1}{3}n\pi </math>
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{{Displayed math||<math>x = \tfrac{1}{3}n\pi\qquad\text{(n is an arbitrary integer).}</math>}}
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(
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<math>n</math>
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an arbitrary integer).
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Current revision

Let's first investigate when the equality

\displaystyle \tan u=\tan v

is satisfied. Because \displaystyle \tan u can be interpreted as the slope (gradient) of the line which makes an angle u with the positive x-axis, we see that for a fixed value of \displaystyle \tan u, there are two angles v in the unit circle with this slope,

\displaystyle v=u\qquad\text{and}\qquad v=u+\pi\,\textrm{.}

The angle v has the same slope after every half turn, so if we add multiples of \displaystyle \pi to u, we will obtain all the angles v which satisfy the equality

\displaystyle v=u+n\pi\,,

where n is an arbitrary integer.

If we apply this result to the equation

\displaystyle \tan x=\tan 4x

we see that the solutions are given by

\displaystyle 4x = x+n\pi\qquad\text{(n is an arbitrary integer),}

and solving for x gives

\displaystyle x = \tfrac{1}{3}n\pi\qquad\text{(n is an arbitrary integer).}
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