Solution 4.3:4e

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (14:20, 9 October 2008) (edit) (undo)
m
 
Line 1: Line 1:
The addition formula for sine gives us that
The addition formula for sine gives us that
 +
{{Displayed math||<math>\sin\Bigl(v+\frac{\pi}{4}\Bigr) = \sin v\cdot\cos\frac{\pi }{4} + \cos v\cdot\sin\frac{\pi}{4}\,\textrm{.}</math>}}
-
<math>\sin \left( v+\frac{\pi }{4} \right)=\sin v\centerdot \cos \frac{\pi }{4}+\cos v\centerdot \sin \frac{\pi }{4}.</math>
+
Because we know from exercise b that <math>\sin v = \sqrt{1-b^2}</math> we use that
 +
<math>\cos (\pi/4) = \sin (\pi/4) = 1/\!\sqrt{2}</math> to obtain
-
 
+
{{Displayed math||<math>\sin\Bigl(v+\frac{\pi }{4}\Bigr) = \sqrt{1-b^2}\cdot\frac{1}{\sqrt{2}} + b\cdot\frac{1}{\sqrt{2}}\,\textrm{.}</math>}}
-
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>
+

Current revision

The addition formula for sine gives us that

\displaystyle \sin\Bigl(v+\frac{\pi}{4}\Bigr) = \sin v\cdot\cos\frac{\pi }{4} + \cos v\cdot\sin\frac{\pi}{4}\,\textrm{.}

Because we know from exercise b that \displaystyle \sin v = \sqrt{1-b^2} we use that \displaystyle \cos (\pi/4) = \sin (\pi/4) = 1/\!\sqrt{2} to obtain

\displaystyle \sin\Bigl(v+\frac{\pi }{4}\Bigr) = \sqrt{1-b^2}\cdot\frac{1}{\sqrt{2}} + b\cdot\frac{1}{\sqrt{2}}\,\textrm{.}
Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_4.3:4e"