From Förberedande kurs i matematik 1

Revision as of 09:49, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_4e.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (14:20, 9 October 2008) (edit) (undo) Tek (Talk | contribs) m
(3 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	The addition formula for sine gives us that
-	<center> [[Bild:4_3_4e.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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>}}
		+
		+	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>}}

---

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{.}