Answer 5.1:2
From Förberedande kurs i matematik 1
(Difference between revisions)
| Line 6: | Line 6: | ||
|- | |- | ||
|c) | |c) | ||
| - | || <math>\left\{\eqalign{ | + | || <math>\left\{\eqalign{ |
| - | x&=n\pi\cr | + | x&=n\pi\cr |
| - | x&=\displaystyle \frac{\pi}{4}+\displaystyle \frac{n\pi}{2} | + | x&=\displaystyle \frac{\pi}{4}+\displaystyle \frac{n\pi}{2} |
}\right.</math> | }\right.</math> | ||
|d) | |d) | ||
|| <math> \bigl(\sqrt[\scriptstyle4]3\,\bigr)^3\ \sqrt{2 + \sqrt{4}}</math> | || <math> \bigl(\sqrt[\scriptstyle4]3\,\bigr)^3\ \sqrt{2 + \sqrt{4}}</math> | ||
|} | |} | ||
Revision as of 13:16, 22 January 2009
| a) | <math>\cos{v} = \cos{\displaystyle \frac{3\pi}{2}}</math> | b) | <math>\tan\displaystyle\frac{u}{2}=\frac{\sin u}{1+\cos u}</math> |
| c) | <math>\left\{\eqalign{
x&=n\pi\cr x&=\displaystyle \frac{\pi}{4}+\displaystyle \frac{n\pi}{2} }\right.</math> | d) | <math> \bigl(\sqrt[\scriptstyle4]3\,\bigr)^3\ \sqrt{2 + \sqrt{4}}</math> |
