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Solution 4.3:3f

From Förberedande kurs i matematik 1

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Current revision (14:03, 9 October 2008) (edit) (undo)
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In this case, it is perhaps simplest to use the addition formula for sine,
In this case, it is perhaps simplest to use the addition formula for sine,
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{{Displayed math||<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \sin\frac{\pi }{3}\cdot \cos v + \cos\frac{\pi}{3}\cdot\sin v\,\textrm{.}</math>}}
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<math>\sin \left( \frac{\pi }{3}+v \right)=\sin \frac{\pi }{3}\centerdot \cos v+\cos \frac{\pi }{3}\centerdot \sin v.</math>
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Since <math>\sin (\pi/3) = \sqrt{3}/\!2</math>, <math>\cos (\pi/3) = 1/2</math>, <math>\sin v = a</math>, and <math>\cos v=\sqrt{1-a^2}</math> this can be written as
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Since
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{{Displayed math||<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \frac{\sqrt{3}}{2}\sqrt{1-a^2} + \frac{1}{2}a\,\textrm{.}</math>}}
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<math>\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2},\ \ \cos \frac{\pi }{3}=\frac{1}{2},\ \ \sin v=a</math>, and
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<math>\cos v=\sqrt{1-a^{2}}</math>
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this can be written as
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<math>\sin \left( \frac{\pi }{3}+v \right)=\frac{\sqrt{3}}{2}\sqrt{1-a^{2}}+\frac{1}{2}a.</math>
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Current revision

In this case, it is perhaps simplest to use the addition formula for sine,

sin3+v=sin3cosv+cos3sinv. 

Since sin(3)=32 , cos(3)=12, sinv=a, and cosv=1−a2  this can be written as

sin3+v=231−a2+21a. 
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