Übung
$\frac{4\cot3x}{\cot\left(60-x\right)+\cot\left(60+x\right)}$
Schritt-für-Schritt-Lösung
Learn how to solve problems step by step online. (4cot(3x))/(cot(60-x)+cot(60+x)). Applying the trigonometric identity: \cot\left(\theta \right) = \frac{\cos\left(\theta \right)}{\sin\left(\theta \right)}. Applying the trigonometric identity: \cot\left(\theta \right) = \frac{\cos\left(\theta \right)}{\sin\left(\theta \right)}. Wenden Sie die Formel an: a+\frac{b}{c}=\frac{b+ac}{c}, wobei a=\cot\left(60+x\right), b=\cos\left(60-x\right), c=\sin\left(60-x\right), a+b/c=\frac{\cos\left(60-x\right)}{\sin\left(60-x\right)}+\cot\left(60+x\right) und b/c=\frac{\cos\left(60-x\right)}{\sin\left(60-x\right)}. Wenden Sie die Formel an: \frac{\frac{a}{b}}{\frac{c}{f}}=\frac{af}{bc}, wobei a=4\cos\left(3x\right), b=\sin\left(3x\right), a/b/c/f=\frac{\frac{4\cos\left(3x\right)}{\sin\left(3x\right)}}{\frac{\cos\left(60-x\right)+\cot\left(60+x\right)\sin\left(60-x\right)}{\sin\left(60-x\right)}}, c=\cos\left(60-x\right)+\cot\left(60+x\right)\sin\left(60-x\right), a/b=\frac{4\cos\left(3x\right)}{\sin\left(3x\right)}, f=\sin\left(60-x\right) und c/f=\frac{\cos\left(60-x\right)+\cot\left(60+x\right)\sin\left(60-x\right)}{\sin\left(60-x\right)}.
(4cot(3x))/(cot(60-x)+cot(60+x))
Endgültige Antwort auf das Problem
$\frac{4\cos\left(3x\right)\sin\left(60-x\right)\sin\left(60+x\right)}{\cos\left(60+x\right)\sin\left(60-x\right)\sin\left(3x\right)+\cos\left(60-x\right)\sin\left(3x\right)\sin\left(60+x\right)}$