Solution 4.3:4e
From Förberedande kurs i matematik 1
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| - | {{  | + | The addition formula for sine gives us that  | 
| - | <  | + | |
| - | {{  | + | |
| + | <math>\sin \left( v+\frac{\pi }{4} \right)=\sin v\centerdot \cos \frac{\pi }{4}+\cos v\centerdot \sin \frac{\pi }{4}.</math>  | ||
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| + | Because we know from exercise b that   | ||
| + | <math>\sin v=\sqrt{1-b^{2}}</math>  | ||
| + | we use that    | ||
| + | <math>\cos \frac{\pi }{4}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}</math>  | ||
| + | to obtain  | ||
| + | |||
| + | <math>\sin \left( v+\frac{\pi }{4} \right)=\sqrt{1-b^{2}}\centerdot \frac{1}{\sqrt{2}}+b\centerdot \frac{1}{\sqrt{2}}.</math>  | ||
Revision as of 11:54, 29 September 2008
The addition formula for sine gives us that
\displaystyle \sin \left( v+\frac{\pi }{4} \right)=\sin v\centerdot \cos \frac{\pi }{4}+\cos v\centerdot \sin \frac{\pi }{4}.
Because we know from exercise b that 
\displaystyle \sin v=\sqrt{1-b^{2}}
we use that  
\displaystyle \cos \frac{\pi }{4}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}
to obtain
\displaystyle \sin \left( v+\frac{\pi }{4} \right)=\sqrt{1-b^{2}}\centerdot \frac{1}{\sqrt{2}}+b\centerdot \frac{1}{\sqrt{2}}.
