Solution 4.2:4f
From Förberedande kurs i matematik 1
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- | If we add | + | If we add <math>2\pi</math> to <math>-5\pi/3\,</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle <math>-5\pi/3</math> and consequently has the same tangent value, |
- | <math>2\pi </math> | + | |
- | to | + | |
- | <math>- | + | |
- | <math>- | + | |
- | and consequently has the same tangent value | + | |
- | + | {{Displayed math||<math>\begin{align} | |
- | <math>\begin{align} | + | \tan\Bigl(-\frac{5\pi}{3}\Bigr) |
- | + | = \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr) | |
- | + | = \tan\frac{\pi}{3} | |
- | \end{align}</math> | + | = \frac{\sin\dfrac{\pi}{3}}{\cos\dfrac{\pi}{3}} |
+ | = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} | ||
+ | = \sqrt{3}\,\textrm{.} | ||
+ | \end{align}</math>}} |
Current revision
If we add \displaystyle 2\pi to \displaystyle -5\pi/3\,, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle \displaystyle -5\pi/3 and consequently has the same tangent value,
\displaystyle \begin{align}
\tan\Bigl(-\frac{5\pi}{3}\Bigr) = \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr) = \tan\frac{\pi}{3} = \frac{\sin\dfrac{\pi}{3}}{\cos\dfrac{\pi}{3}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3}\,\textrm{.} \end{align} |