Solution 4.3:9

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Using the formula for double angles on sin
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Using the formula for double angles on <math>\sin 160^{\circ}</math> gives
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<math>160^{\circ }</math>
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gives
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{{Displayed math||<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}}
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<math>\sin 160^{\circ }=2\cos 80^{\circ }\sin 80^{\circ }</math>
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On the right-hand side, we see that the factor <math>\cos 80^{\circ}</math> has appeared, and if we use the formula for double angles on the second factor (<math>\sin 80^{\circ}</math>),
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{{Displayed math||<math>2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,,</math>}}
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On the right-hand side, we see that the factor
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we obtain a further factor <math>\cos 40^{\circ}</math>. A final application of the formula for double angles on <math>\sin 40^{\circ }</math> gives us all three cosine factors,
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<math>\cos 80^{\circ }</math>
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has appeared, and if we use the formula for double angles on the second factor (
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<math>\sin 80^{\circ }</math>
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),
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<math>2\cos 80^{\circ }\sin 80^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\sin 40^{\circ }</math>
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we obtain a further factor
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<math>\cos 40^{\circ }</math>. A final application of the formula for double angles on
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<math>\sin 40^{\circ }</math>
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gives us all three cosine factors:
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<math>2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot \sin 40^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot 2\cos 20^{\circ }\sin 20^{\circ }</math>
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{{Displayed math||<math>2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·}</math>}}
We have thus succeeded in showing that
We have thus succeeded in showing that
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{{Displayed math||<math>\sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ}</math>}}
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<math>\sin 160^{\circ }=8\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }\sin 20^{\circ }</math>
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which can also be written as
which can also be written as
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{{Displayed math||<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}}
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<math>\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}</math>
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[[Image:4_3_9.gif||right]]
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If we draw the unit circle, we see that <math>160^{\circ}</math> makes an angle of
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<math>20^{\circ}</math> with the negative ''x''-axis, and therefore the angles
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<math>20^{\circ}</math> and <math>160^{\circ}</math> have the same ''y''-coordinate in the unit circle, i.e.
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<center><math>\sin 20^{\circ} = \sin 160^{\circ}\,\textrm{.}</math></center>
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If we draw the unit circle, we see that
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<math>160^{\circ }</math>
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makes an angle of
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<math>20^{\circ }</math>
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with the negative
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<math>x</math>
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-axis, and therefore the angles
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<math>20^{\circ }</math>
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and
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<math>160^{\circ }</math>
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have the same
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<math>y</math>
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-coordinate in the unit circle, i.e.
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<math>\sin 20^{\circ }=\sin 160^{\circ }</math>.
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[[Image:4_3_9.gif|center]]
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This shows that
This shows that
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<center><math>\cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}} = \frac{1}{8}\,\textrm{.}</math></center>
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<math>\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}</math>
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Current revision

Using the formula for double angles on \displaystyle \sin 160^{\circ} gives

\displaystyle \sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}

On the right-hand side, we see that the factor \displaystyle \cos 80^{\circ} has appeared, and if we use the formula for double angles on the second factor (\displaystyle \sin 80^{\circ}),

\displaystyle 2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,,

we obtain a further factor \displaystyle \cos 40^{\circ}. A final application of the formula for double angles on \displaystyle \sin 40^{\circ } gives us all three cosine factors,

\displaystyle 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·}

We have thus succeeded in showing that

\displaystyle \sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ}

which can also be written as

\displaystyle \cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}

If we draw the unit circle, we see that \displaystyle 160^{\circ} makes an angle of \displaystyle 20^{\circ} with the negative x-axis, and therefore the angles \displaystyle 20^{\circ} and \displaystyle 160^{\circ} have the same y-coordinate in the unit circle, i.e.

\displaystyle \sin 20^{\circ} = \sin 160^{\circ}\,\textrm{.}

This shows that

\displaystyle \cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}} = \frac{1}{8}\,\textrm{.}
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