Solution 4.2:4c
From Förberedande kurs i matematik 1
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| - | {{  | + | In exercise e, we studied the angle   | 
| - | <  | + | <math>\frac{3\pi }{4}</math>  | 
| - | {{  | + | and found that  | 
| + | |||
| + | <math>\cos \frac{3\pi }{4}=-\frac{1}{\sqrt{2}}</math>  | ||
| + | and    | ||
| + | <math>\sin \frac{3\pi }{4}=\frac{1}{\sqrt{2}}</math>  | ||
| + | |||
| + | |||
| + | Because   | ||
| + | <math>\text{tan }x</math>  | ||
| + | is defined as   | ||
| + | <math>\frac{\sin x}{\cos x}</math>, we get immediately that  | ||
| + | |||
| + | |||
| + | <math>\tan \frac{3\pi }{4}=\frac{\sin \frac{3\pi }{4}}{\cos \frac{3\pi }{4}}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1</math>  | ||
Revision as of 13:11, 28 September 2008
In exercise e, we studied the angle \displaystyle \frac{3\pi }{4} and found that
\displaystyle \cos \frac{3\pi }{4}=-\frac{1}{\sqrt{2}} and \displaystyle \sin \frac{3\pi }{4}=\frac{1}{\sqrt{2}}
Because 
\displaystyle \text{tan }x
is defined as 
\displaystyle \frac{\sin x}{\cos x}, we get immediately that
\displaystyle \tan \frac{3\pi }{4}=\frac{\sin \frac{3\pi }{4}}{\cos \frac{3\pi }{4}}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1
