Solution 4.3:4f
From Förberedande kurs i matematik 1
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- | {{ | + | Using the addition formula for cosine, we can express <math>\cos (v-\pi/3)</math> |
- | < | + | in terms of <math>\cos v</math> and <math>\sin v</math>, |
- | {{ | + | |
+ | {{Displayed math||<math>\cos\Bigl(v-\frac{\pi}{3}\Bigr) = \cos v\cdot \cos\frac{\pi }{3} + \sin v\cdot \sin\frac{\pi}{3}\,\textrm{.}</math>}} | ||
+ | |||
+ | Since <math>\cos v = b</math> and <math>\sin v = \sqrt{1-b^2}</math> we obtain | ||
+ | |||
+ | {{Displayed math||<math>\cos\Bigl(v-\frac{\pi}{3}\Bigr) = b\cdot\frac{1}{2} + \sqrt{1-b^2}\cdot\frac{\sqrt{3}}{2}\,\textrm{.}</math>}} |
Current revision
Using the addition formula for cosine, we can express \displaystyle \cos (v-\pi/3) in terms of \displaystyle \cos v and \displaystyle \sin v,
\displaystyle \cos\Bigl(v-\frac{\pi}{3}\Bigr) = \cos v\cdot \cos\frac{\pi }{3} + \sin v\cdot \sin\frac{\pi}{3}\,\textrm{.} |
Since \displaystyle \cos v = b and \displaystyle \sin v = \sqrt{1-b^2} we obtain
\displaystyle \cos\Bigl(v-\frac{\pi}{3}\Bigr) = b\cdot\frac{1}{2} + \sqrt{1-b^2}\cdot\frac{\sqrt{3}}{2}\,\textrm{.} |