From Förberedande kurs i matematik 1

Revision as of 10:07, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_7b.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (08:11, 10 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	Using the addition formula, we rewrite <math>\sin (x+y)</math> as
-	<center> [[Bild:4_3_7b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\sin (x+y) = \sin x\cdot\cos y + \cos x\cdot\sin y\,\textrm{.}</math>}}
		+
		+	If we use the same solution procedure as in exercise a, we use the Pythagorean identity <math>\cos^2\!v + \sin^2\!v = 1</math> to express the unknown factors <math>\sin x</math> and <math>\sin y</math> in terms of <math>\cos x</math> and <math>\cos y</math>,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\sin x &= \pm\sqrt{1-\cos^2\!x} = \pm\sqrt{1-\bigl(\tfrac{2}{5}\bigr)^2} = \pm\sqrt{1-\tfrac{4}{25}} = \pm\frac{\sqrt{21}}{5}\,,\\[5pt]
		+	\sin y &= \pm\sqrt{1-\cos^2\!y} = \pm\sqrt{1-\bigl(\tfrac{3}{5}\bigr)^2} = \pm\sqrt{1-\tfrac{9}{25}} = \pm\frac{4}{5}\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	The angles ''x'' and ''y'' lie in the first quadrant and both <math>\sin x</math> and <math>\sin y</math> are therefore positive, i.e.
		+
		+	{{Displayed math||<math>\sin x = \frac{\sqrt{21}}{5}\qquad\text{and}\qquad\sin y = \frac{4}{5}\,\textrm{.}</math>}}
		+
		+	Thus, the answer is
		+
		+	{{Displayed math||<math>\sin (x+y) = \frac{\sqrt{21}}{5}\cdot\frac{3}{5} + \frac{2}{5}\cdot\frac{4}{5} = \frac{3\sqrt{21}+8}{25}\,\textrm{.}</math>}}

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Current revision

Using the addition formula, we rewrite \displaystyle \sin (x+y) as

\displaystyle \sin (x+y) = \sin x\cdot\cos y + \cos x\cdot\sin y\,\textrm{.}

If we use the same solution procedure as in exercise a, we use the Pythagorean identity \displaystyle \cos^2\!v + \sin^2\!v = 1 to express the unknown factors \displaystyle \sin x and \displaystyle \sin y in terms of \displaystyle \cos x and \displaystyle \cos y,

\displaystyle \begin{align} \sin x &= \pm\sqrt{1-\cos^2\!x} = \pm\sqrt{1-\bigl(\tfrac{2}{5}\bigr)^2} = \pm\sqrt{1-\tfrac{4}{25}} = \pm\frac{\sqrt{21}}{5}\,,\\[5pt] \sin y &= \pm\sqrt{1-\cos^2\!y} = \pm\sqrt{1-\bigl(\tfrac{3}{5}\bigr)^2} = \pm\sqrt{1-\tfrac{9}{25}} = \pm\frac{4}{5}\,\textrm{.} \end{align}

The angles x and y lie in the first quadrant and both \displaystyle \sin x and \displaystyle \sin y are therefore positive, i.e.

\displaystyle \sin x = \frac{\sqrt{21}}{5}\qquad\text{and}\qquad\sin y = \frac{4}{5}\,\textrm{.}

Thus, the answer is

\displaystyle \sin (x+y) = \frac{\sqrt{21}}{5}\cdot\frac{3}{5} + \frac{2}{5}\cdot\frac{4}{5} = \frac{3\sqrt{21}+8}{25}\,\textrm{.}