Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 10:07, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.4:6c moved to Solution 4.4:6c: Robot: moved page) ← Previous diff		Current revision (06:55, 14 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we use the trigonometric relation <math>\sin (-x) = -\sin x</math>, the equation can be rewritten as
-	<center> [[Image:4_4_6c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\sin 2x = \sin (-x)\,\textrm{.}</math>}}
		+
		+	In exercise 4.4:5a, we saw that an equality of the type
		+
		+	{{Displayed math||<math>\sin u = \sin v</math>}}
		+
		+	is satisfied if
		+
		+	{{Displayed math||<math>u = v+2n\pi\qquad\text{or}\qquad u = \pi-v+2n\pi\,,</math>}}
		+
		+	where ''n'' is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy
		+
		+	{{Displayed math||<math>2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,,</math>}}
		+
		+	i.e.
		+
		+	{{Displayed math||<math>3x = 2n\pi\qquad\text{or}\qquad x = \pi +2n\pi\,\textrm{.}</math>}}
		+
		+	The solutions to the equation are thus
		+
		+	{{Displayed math||<math>\left\{\begin{align}
		+	x &= \frac{2n\pi}{3}\,,\\[5pt]
		+	x &= \pi + 2n\pi\,,
		+	\end{align}\right.</math>}}
		+
		+	where ''n'' is an arbitrary integer.

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Current revision

If we use the trigonometric relation sin(−x)=−sinx, the equation can be rewritten as

sin2x=sin(−x).

In exercise 4.4:5a, we saw that an equality of the type

sinu=sinv

is satisfied if

u=v+2noru=−v+2n

where n is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy

2x=−x+2nor2x=−(−x)+2n

i.e.

3x=2norx=+2n.

The solutions to the equation are thus

xx=32n=+2n

where n is an arbitrary integer.