Solution 4.3:3d

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The expression for the angle
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The expression for the angle <math>\pi/2 - v</math> differs from <math>\pi/2</math> by as much as <math>-v</math> differs from <math>0</math>. This means that <math>\pi/2-v</math> makes the same angle with the positive ''y''-axis as <math>-v</math> makes with the positive ''x''-axis.
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<math>{\pi }/{2}\;-v</math>
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differs from
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<math>{\pi }/{2}\;</math>
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by as much as
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<math>-v\text{ }</math>
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differs from
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<math>0</math>. This means that
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<math>{\pi }/{2}\;</math>
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makes the same angle with the positive
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<math>y</math>
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-axis as
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<math>-v\text{ }</math>
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makes with the positive
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<math>x</math>
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-axis.
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{| align="center"
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| align="center" |[[Image:4_3_3_d-1.gif]]
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| width="20px"|&nbsp;
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| align="center" |[[Image:4_3_3_d-2.gif]]
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|-
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| align="center" |<small>Angle&nbsp;''v''</small>
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||
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| align="center" |<small>Angle&nbsp;π/2&nbsp;-&nbsp;''v''</small>
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|}
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[[Image:4_3_3_d.gif|center]]
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Therefore, the angle <math>\pi/2 - v</math> has a ''y''-coordinate which is equal to the ''x''-coordinate for the angle ''v'', i.e.
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Angle
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{{Displayed math||<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}
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<math>v</math>
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angle
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<math>\pi -v</math>
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Therefore, the angle
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and from exercise c, we know that <math>\cos v = \sqrt{1-a^2}\,</math>,
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<math>{\pi }/{2}\;-v</math>
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has a
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<math>y</math>
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-coordinate which is equal to the
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<math>x</math>
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-coordinate for the angle
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<math>v</math>, i.e.
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{{Displayed math||<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}
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<math>\sin \left( {\pi }/{2}\;-v \right)=\cos v</math>
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and from exercise c, we know that
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<math>\cos v=\sqrt{1-a^{2}}</math>
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<math>\sin \left( \frac{\pi }{2}-v \right)=\sqrt{1-a^{2}}</math>
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Current revision

The expression for the angle \displaystyle \pi/2 - v differs from \displaystyle \pi/2 by as much as \displaystyle -v differs from \displaystyle 0. This means that \displaystyle \pi/2-v makes the same angle with the positive y-axis as \displaystyle -v makes with the positive x-axis.

Image:4_3_3_d-1.gif   Image:4_3_3_d-2.gif
Angle v Angle π/2 - v

Therefore, the angle \displaystyle \pi/2 - v has a y-coordinate which is equal to the x-coordinate for the angle v, i.e.

\displaystyle \sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v

and from exercise c, we know that \displaystyle \cos v = \sqrt{1-a^2}\,,

\displaystyle \sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}
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