From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	Using the addition formula for cosine, we can express <math>\cos (v-\pi/3)</math>
-	<center> [[Image:4_3_4f.gif]] </center>	+	in terms of <math>\cos v</math> and <math>\sin v</math>,
-	{{NAVCONTENT_STOP}}	+
		+	{{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>}}
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		+	Since <math>\cos v = b</math> and <math>\sin v = \sqrt{1-b^2}</math> we obtain
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		+	{{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>}}

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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{.}