Solution 4.2:4c
From Förberedande kurs i matematik 1
(Difference between revisions)
m (Lösning 4.2:4c moved to Solution 4.2:4c: Robot: moved page) |
m |
||
(One intermediate revision not shown.) | |||
Line 1: | Line 1: | ||
- | {{ | + | In exercise 4.2:3e, we studied the angle <math>3\pi/4</math> and found that |
- | < | + | |
- | {{ | + | {{Displayed math||<math>\cos\frac{3\pi }{4} = -\frac{1}{\sqrt{2}}\qquad\text{and}\qquad\sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}\,\textrm{.}</math>}} |
+ | |||
+ | Because <math>\tan x</math> is defined as <math>\frac{\sin x}{\cos x}</math>, we get immediately that | ||
+ | |||
+ | {{Displayed math||<math>\tan\frac{3\pi}{4} = \frac{\sin\dfrac{3\pi}{4}}{\cos \dfrac{3\pi}{4}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}</math>}} |
Current revision
In exercise 4.2:3e, we studied the angle \displaystyle 3\pi/4 and found that
\displaystyle \cos\frac{3\pi }{4} = -\frac{1}{\sqrt{2}}\qquad\text{and}\qquad\sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}\,\textrm{.} |
Because \displaystyle \tan x is defined as \displaystyle \frac{\sin x}{\cos x}, we get immediately that
\displaystyle \tan\frac{3\pi}{4} = \frac{\sin\dfrac{3\pi}{4}}{\cos \dfrac{3\pi}{4}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.} |