trigonometrie , umformung

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.

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

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 !