Answer 5.1:2

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
(New page: {| width="100%" cellspacing="10px" |a) |width="50%" | \cos{v} = \cos{\displaystyle \frac{3\pi}{2}} |b) |width="50%" | \tan\displaystyle\frac{u}{2}=\frac{\sin u}{1+\cos u} |- |c) || \left\{...)
Line 1: Line 1:
{| width="100%" cellspacing="10px"
{| width="100%" cellspacing="10px"
|a)
|a)
-
|width="50%" | \cos{v} = \cos{\displaystyle \frac{3\pi}{2}}
+
|width="50%" | <math>\cos{v} = \cos{\displaystyle \frac{3\pi}{2}}</math>
|b)
|b)
-
|width="50%" | \tan\displaystyle\frac{u}{2}=\frac{\sin u}{1+\cos u}
+
|width="50%" | <math>\tan\displaystyle\frac{u}{2}=\frac{\sin u}{1+\cos u}</math>
|-
|-
|c)
|c)
-
|| \left\{\eqalign{
+
|| &lt;math&gt;\left\{\eqalign{ <br/>
-
x&=n\pi\cr
+
x&=n\pi\cr <br/>
-
x&=\displaystyle \frac{\pi}{4}+\displaystyle \frac{n\pi}{2}
+
x&=\displaystyle \frac{\pi}{4}+\displaystyle \frac{n\pi}{2} <br/>
-
}\right.
+
}\right.&lt;/math&gt;
|d)
|d)
|| &lt;math&gt; \bigl(\sqrt[\scriptstyle4]3\,\bigr)^3\ \sqrt{2 + \sqrt{4}}&lt;/math&gt;
|| &lt;math&gt; \bigl(\sqrt[\scriptstyle4]3\,\bigr)^3\ \sqrt{2 + \sqrt{4}}&lt;/math&gt;
|}
|}

Revision as of 13:12, 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>
Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Answer_5.1:2"