Solution 4.2:5b
From Förberedande kurs i matematik 1
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| - | {{ | + | If we draw the angle <math>225^{\circ} = 180^{\circ} + 45^{\circ}</math> on a unit circle, we see that it makes an angle of <math>45^{\circ}</math> with the negative ''x''-axis. |
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| + | This means that <math>\tan 225^{\circ}</math>, which is the slope of the line that makes an angle of <math>45^{\circ}</math> with the positive ''x''-axis, equals <math>\tan 45^{\circ}</math>, because the line which makes an angle of <math>45^{\circ}</math> has the same slope, | ||
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| + | {{Displayed math||<math>\tan 225^{\circ} = \tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1\,\textrm{.}</math>}} | ||
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Current revision
If we draw the angle \displaystyle 225^{\circ} = 180^{\circ} + 45^{\circ} on a unit circle, we see that it makes an angle of \displaystyle 45^{\circ} with the negative x-axis.
This means that \displaystyle \tan 225^{\circ}, which is the slope of the line that makes an angle of \displaystyle 45^{\circ} with the positive x-axis, equals \displaystyle \tan 45^{\circ}, because the line which makes an angle of \displaystyle 45^{\circ} has the same slope,
| \displaystyle \tan 225^{\circ} = \tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1\,\textrm{.} |
