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ich soll überprüfen, ob diese trigonometrische umformung gilt:

 

(sin(ax))5 = 10/16 sin(ax) - 5/16 sin(3ax) + 1/16 sin(5ax)

 

könntet ihr mir weiterhelfen?:)

 28.03.2016
bearbeitet von Gast  28.03.2016

Beste Antwort 

 #3
avatar+25990 
+5

ich soll überprüfen, ob diese trigonometrische umformung gilt:

(sin(ax))5 = 10/16 sin(ax) - 5/16 sin(3ax) + 1/16 sin(5ax)

 

Es genügt \(\sin^5(x) = \frac{10}{16} \sin(x) - \frac{5}{16} \sin(3x) + \frac{1}{16} \sin(5x)\) zu beweisen.

 


B

 

Aus der Formelsammlung: \( \boxed{~ \sin^2(x) + \cos^2(x) = 1 \qquad \cos^2(x)=1-\sin^2(x) ~}\)

 

1.

 

\(\begin{array}{rcll} \sin(2x) &=& \sin(x+x) \\ &=& \sin(x)\cos(x) + \cos(x)\sin(x)\\ &=& 2\sin(x)\cos(x) \\ \boxed{~ \sin(2x) = 2\sin(x)\cos(x) ~} \end{array}\)

 

2.

\(\begin{array}{rcll} \cos(2x) &=& \cos(x+x) \\ &=& \cos(x)\cos(x) - \sin(x)\sin(x)\\ &=& \cos^2(x) - \sin^2(x) \\ &=& [1-\sin^2(x)] - \sin^2(x) \\ &=& 1-2\sin^2(x) \\ \boxed{~ \cos(2x) = 1-2\sin^2(x) ~} \end{array}\)

 

3.

\(\begin{array}{rcll} \sin(3x) &=& \sin(x+2x) \\ &=& \sin(x)\cdot \underbrace{\cos(2x)}_{=[1-2\sin^2(x)]} + \cos(x)\cdot \underbrace{\sin(2x)}_{=[2\sin(x)\cos(x)]} \\ &=& \sin(x) \cdot [1-2\sin^2(x)] + \cos(x)\cdot [2\sin(x)\cos(x)] \\ &=& \sin(x) -2\sin^3(x) + 2\cdot \sin(x)\cos^2(x) \\ &=& \sin(x) -2\sin^3(x) + 2\cdot \sin(x)[1-\sin^2(x)] \\ &=& \sin(x) -2\sin^3(x) + 2 \sin(x) -2\sin^3(x) \\ &=& 3\sin(x) -4\sin^3(x) \\ && \boxed{~ \begin{array}{rcll} \sin(3x) &=& 3\sin(x) -4\sin^3(x)\\ \text{ oder } \quad \sin^3(x) &=& \frac14[3\sin(x)-\sin(3x)] \end{array} ~} \end{array}\)

 

4.

\(\begin{array}{rcll} \cos(3x) &=& \cos(x+2x) \\ &=& \cos(x)\cdot \underbrace{\cos(2x)}_{=[1-2\sin^2(x)]} - \sin(x)\cdot \underbrace{\sin(2x)}_{=[2\sin(x)\cos(x)]}\\ &=& \cos(x) \cdot [1-2\sin^2(x)] - \sin(x)\cdot [2\sin(x)\cos(x)] \\ &=& \cos(x) -2 \cos(x)\sin^2(x) - 2\sin^2(x) \cos(x)\\ &=& \cos(x) -4 \cos(x)\sin^2(x)\\ &=& \cos(x)[1 -4\sin^2(x)]\\ && \boxed{~ \cos(3x) = \cos(x)\cdot [1 -4\sin^2(x)] ~} \end{array}\)

 

5.

\(\begin{array}{rcll} \sin(5x) &=& \sin(2x+3x) \\ &=& \underbrace{\sin(2x)}_{=[2\sin(x)\cos(x)]} \cdot \underbrace{\cos(3x)}_{=\cos(x)\cdot [1 -4\sin^2(x)]} + \underbrace{\cos(2x)}_{=[1-2\sin^2(x)]} \cdot \underbrace{\sin(3x)}_{=[3\sin(x) -4\sin^3(x)]} \\\\ &=& [2\sin(x)\cos(x)] \cdot \cos(x)\cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)\cos^2(x) \cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)[1-\sin^2(x)] \cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)\cdot [1 -4\sin^2(x)]-2\sin^3(x)\cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)-8\sin^3(x)-2\sin^3(x)+8\sin^5(x)\\ &+&3\sin(x)-4\sin^3(x)-6\sin^3(x)+8\sin^5(x) \\\\ \sin(5x) &=& 5\sin(x) -20\cdot \underbrace{\sin^3(x)}_{=\frac14[3\sin(x)-\sin(3x)]} +16\sin^5(x) \\ &=& 5\sin(x) -20\cdot \frac14[3\sin(x)-\sin(3x)] +16\sin^5(x) \\ &=& 5\sin(x) -5\cdot [3\sin(x)-\sin(3x)] +16\sin^5(x) \\ &=& 5\sin(x) -15\sin(x)+5\sin(3x) +16\sin^5(x) \\\\ \sin(5x) &=& -10\sin(x)+5\sin(3x) +16\sin^5(x) \\ 16\sin^5(x) &=& 10\sin(x)-5\sin(3x)+\sin(5x)\\ \sin^5(x) &=& \frac{1}{16}\cdot[ 10\sin(x)-5\sin(3x)+\sin(5x) ] \\\\ \mathbf{ \sin^5(x) }& \mathbf{=} & \mathbf{\frac{10}{16}\sin(x)-\frac{5}{16}\sin(3x)+\frac{1}{16}\sin(5x) } \end{array}\)

 

laugh

 29.03.2016
 #1
avatar+14537 
+5

Hallo und guten Tag,

leider kann ich nicht beweisen, dass die Umformung stimmt. (Sie ist richtig !)

Ich hoffe, dass sich noch ein Experte  ( heureka ? ) findet.

Ostergruß radix smiley !

 

 28.03.2016
 #2
avatar+25990 
+5

ich soll überprüfen, ob diese trigonometrische umformung gilt:

(sin(ax))5 = 10/16 sin(ax) - 5/16 sin(3ax) + 1/16 sin(5ax)

 

Es genügt \(\sin^5(x) = \frac{10}{16} \sin(x) - \frac{5}{16} \sin(3x) + \frac{1}{16} \sin(5x)\) zu beweisen.

 

A Methode 1 mit komplexen Zahlen

B Methode 2 ohne komplexen Zahlen

 

 

A

Aus der Formelsammlung: \(\boxed{~ \sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix}) ~}\)

 

\(\begin{array}{rcll} \sin^5(x) &=& \left[~ \frac{1}{2i}(e^{ix}-e^{-ix}) ~\right]^5 \\\\ &=& \left[~ \frac{1}{2^5i^5}(e^{ix}-e^{-ix}) ~\right]^5 & | \quad i^5 = i\\\\ &=& \frac{1}{32i}(e^{ix}-e^{-ix})^5 \\\\ &=& \frac{1}{16\cdot 2i}(e^{ix}-e^{-ix})^5 \\\\ \text{Binom auflösen:}&&\boxed{~ \binom50=\binom55=1 \quad \binom51=\binom54=5 \quad \binom52=\binom53=10 ~}\\\\ \sin^5(x) &=& \frac{1}{16\cdot 2i} [~ \binom50 e^{i5x}-\binom51 ( e^{i4x} \cdot e^{-ix} )\\ &+&\binom52 ( e^{i3x} \cdot e^{-i2x} )-\binom53 ( e^{i2x} \cdot e^{-i3x} )\\ &+&\binom54 ( e^{ix} \cdot e^{-i4x} )-\binom55 e^{-i5x} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ e^{i5x}-5 \cdot( e^{i4x} \cdot e^{-ix} ) + 10 \cdot( e^{i3x} \cdot e^{-i2x} )-10 \cdot( e^{i2x} \cdot e^{-i3x} )\\ &+&5 \cdot( e^{ix} \cdot e^{-i4x} )- e^{-i5x} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ e^{i5x}-5 \cdot( e^{i4x-ix} ) + 10 \cdot( e^{i3x-i2x} )-10 \cdot( e^{i2x-i3x} )\\ &+&5 \cdot( e^{ix-i4x} )- e^{-i5x} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ e^{i5x}-5 \cdot( e^{ix(4-1)} ) + 10 \cdot( e^{ix(3-2)} )-10 \cdot( e^{ix(2-3)} )\\ &+&5 \cdot( e^{ix(1-4)} )- e^{-i5x} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ e^{i5x}-5 \cdot e^{i3x} + 10 \cdot e^{ix} -10 \cdot e^{-ix} \\ &+&5 \cdot e^{-i3x} - e^{-i5x} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ e^{i5x}- e^{-i5x} -5 \cdot e^{i3x} +5 \cdot e^{-i3x} + 10 \cdot e^{ix} -10 \cdot e^{-ix} ~ ] \\\\ &=& \frac{1}{16\cdot 2i} [~ ( e^{i5x}- e^{-i5x} ) - 5\cdot( e^{i3x} - e^{-i3x} ) + 10\cdot(e^{ix} - e^{-ix} ) ~ ] \\\\ &=& \frac{1}{16} [~ \underbrace{\frac{1}{2i}( e^{i5x}- e^{-i5x} )}_{=\sin(5x)} - 5\cdot \underbrace{\frac{1}{2i}( e^{i3x} - e^{-i3x} )}_{=\sin(3x)} + 10\cdot\underbrace{\frac{1}{2i}(e^{ix} - e^{-ix} )}_{=\sin(x)} ~ ] \\\\ &=& \frac{1}{16} [~ \sin(5x) - 5\cdot\sin(3x)+ 10\cdot\sin(x) ~ ] \\\\ \sin^5(x) &=& \frac{1}{16}\sin(5x) - \frac{5}{16}\cdot\sin(3x)+ \frac{10}{16}\cdot\sin(x) \end{array}\)

heureka  29.03.2016
 #3
avatar+25990 
+5
Beste Antwort

ich soll überprüfen, ob diese trigonometrische umformung gilt:

(sin(ax))5 = 10/16 sin(ax) - 5/16 sin(3ax) + 1/16 sin(5ax)

 

Es genügt \(\sin^5(x) = \frac{10}{16} \sin(x) - \frac{5}{16} \sin(3x) + \frac{1}{16} \sin(5x)\) zu beweisen.

 


B

 

Aus der Formelsammlung: \( \boxed{~ \sin^2(x) + \cos^2(x) = 1 \qquad \cos^2(x)=1-\sin^2(x) ~}\)

 

1.

 

\(\begin{array}{rcll} \sin(2x) &=& \sin(x+x) \\ &=& \sin(x)\cos(x) + \cos(x)\sin(x)\\ &=& 2\sin(x)\cos(x) \\ \boxed{~ \sin(2x) = 2\sin(x)\cos(x) ~} \end{array}\)

 

2.

\(\begin{array}{rcll} \cos(2x) &=& \cos(x+x) \\ &=& \cos(x)\cos(x) - \sin(x)\sin(x)\\ &=& \cos^2(x) - \sin^2(x) \\ &=& [1-\sin^2(x)] - \sin^2(x) \\ &=& 1-2\sin^2(x) \\ \boxed{~ \cos(2x) = 1-2\sin^2(x) ~} \end{array}\)

 

3.

\(\begin{array}{rcll} \sin(3x) &=& \sin(x+2x) \\ &=& \sin(x)\cdot \underbrace{\cos(2x)}_{=[1-2\sin^2(x)]} + \cos(x)\cdot \underbrace{\sin(2x)}_{=[2\sin(x)\cos(x)]} \\ &=& \sin(x) \cdot [1-2\sin^2(x)] + \cos(x)\cdot [2\sin(x)\cos(x)] \\ &=& \sin(x) -2\sin^3(x) + 2\cdot \sin(x)\cos^2(x) \\ &=& \sin(x) -2\sin^3(x) + 2\cdot \sin(x)[1-\sin^2(x)] \\ &=& \sin(x) -2\sin^3(x) + 2 \sin(x) -2\sin^3(x) \\ &=& 3\sin(x) -4\sin^3(x) \\ && \boxed{~ \begin{array}{rcll} \sin(3x) &=& 3\sin(x) -4\sin^3(x)\\ \text{ oder } \quad \sin^3(x) &=& \frac14[3\sin(x)-\sin(3x)] \end{array} ~} \end{array}\)

 

4.

\(\begin{array}{rcll} \cos(3x) &=& \cos(x+2x) \\ &=& \cos(x)\cdot \underbrace{\cos(2x)}_{=[1-2\sin^2(x)]} - \sin(x)\cdot \underbrace{\sin(2x)}_{=[2\sin(x)\cos(x)]}\\ &=& \cos(x) \cdot [1-2\sin^2(x)] - \sin(x)\cdot [2\sin(x)\cos(x)] \\ &=& \cos(x) -2 \cos(x)\sin^2(x) - 2\sin^2(x) \cos(x)\\ &=& \cos(x) -4 \cos(x)\sin^2(x)\\ &=& \cos(x)[1 -4\sin^2(x)]\\ && \boxed{~ \cos(3x) = \cos(x)\cdot [1 -4\sin^2(x)] ~} \end{array}\)

 

5.

\(\begin{array}{rcll} \sin(5x) &=& \sin(2x+3x) \\ &=& \underbrace{\sin(2x)}_{=[2\sin(x)\cos(x)]} \cdot \underbrace{\cos(3x)}_{=\cos(x)\cdot [1 -4\sin^2(x)]} + \underbrace{\cos(2x)}_{=[1-2\sin^2(x)]} \cdot \underbrace{\sin(3x)}_{=[3\sin(x) -4\sin^3(x)]} \\\\ &=& [2\sin(x)\cos(x)] \cdot \cos(x)\cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)\cos^2(x) \cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)[1-\sin^2(x)] \cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)\cdot [1 -4\sin^2(x)]-2\sin^3(x)\cdot [1 -4\sin^2(x)]\\ &+& [1-2\sin^2(x)] \cdot [3\sin(x) -4\sin^3(x)] \\\\ &=& 2\sin(x)-8\sin^3(x)-2\sin^3(x)+8\sin^5(x)\\ &+&3\sin(x)-4\sin^3(x)-6\sin^3(x)+8\sin^5(x) \\\\ \sin(5x) &=& 5\sin(x) -20\cdot \underbrace{\sin^3(x)}_{=\frac14[3\sin(x)-\sin(3x)]} +16\sin^5(x) \\ &=& 5\sin(x) -20\cdot \frac14[3\sin(x)-\sin(3x)] +16\sin^5(x) \\ &=& 5\sin(x) -5\cdot [3\sin(x)-\sin(3x)] +16\sin^5(x) \\ &=& 5\sin(x) -15\sin(x)+5\sin(3x) +16\sin^5(x) \\\\ \sin(5x) &=& -10\sin(x)+5\sin(3x) +16\sin^5(x) \\ 16\sin^5(x) &=& 10\sin(x)-5\sin(3x)+\sin(5x)\\ \sin^5(x) &=& \frac{1}{16}\cdot[ 10\sin(x)-5\sin(3x)+\sin(5x) ] \\\\ \mathbf{ \sin^5(x) }& \mathbf{=} & \mathbf{\frac{10}{16}\sin(x)-\frac{5}{16}\sin(3x)+\frac{1}{16}\sin(5x) } \end{array}\)

 

laugh

heureka  29.03.2016

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