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Algebra Difficulty 6.0 National olympiad Prove it

3.045. ctg⁡22α−12ctg⁡2α−cos⁡8αctg⁡4α=sin⁡8α\frac{\operatorname{ctg}^{2} 2 \alpha-1}{2 \operatorname{ctg} 2 \alpha}-\cos 8 \alpha \operatorname{ctg} 4 \alpha=\sin 8 \alpha.

3.045. cot⁡22α−12cot⁡2α−cos⁡8αcot⁡4α=sin⁡8α\frac{\cot^{2} 2 \alpha-1}{2 \cot 2 \alpha}-\cos 8 \alpha \cot 4 \alpha=\sin 8 \alpha.

Solution

Solution.

ctg⁡22α−12ctg⁡2α−cos⁡8αctg⁡4α=1tg⁡22α−12tg⁡2α−cos⁡8αctg⁡4α= \frac{\operatorname{ctg}^{2} 2 \alpha-1}{2 \operatorname{ctg} 2 \alpha}-\cos 8 \alpha \operatorname{ctg} 4 \alpha=\frac{\frac{1}{\operatorname{tg}^{2} 2 \alpha}-1}{\frac{2}{\operatorname{tg} 2 \alpha}}-\cos 8 \alpha \operatorname{ctg} 4 \alpha=

=1−tg⁡22αtg⁡22α⋅tg⁡2α2−cos⁡8αctg⁡4α=1−tg⁡22α2tg⁡2α−cos⁡8αctg⁡4α==1tg⁡4α−cos⁡8αctg⁡4α=1tg⁡4α−cos⁡8α1tg⁡4α=1tg⁡4α(1−cos⁡8α)==11−cos⁡8αcos⁡8α⋅(1−cos⁡8α)=sin⁡8α(1−cos⁡8α)1−cos⁡8α=sin⁡8α \begin{aligned} & =\frac{1-\operatorname{tg}^{2} 2 \alpha}{\operatorname{tg}^{2} 2 \alpha} \cdot \frac{\operatorname{tg} 2 \alpha}{2}-\cos 8 \alpha \operatorname{ctg} 4 \alpha=\frac{1-\operatorname{tg}^{2} 2 \alpha}{2 \operatorname{tg} 2 \alpha}-\cos 8 \alpha \operatorname{ctg} 4 \alpha= \\ & =\frac{1}{\operatorname{tg} 4 \alpha}-\cos 8 \alpha \operatorname{ctg} 4 \alpha=\frac{1}{\operatorname{tg} 4 \alpha}-\cos 8 \alpha \frac{1}{\operatorname{tg} 4 \alpha}=\frac{1}{\operatorname{tg} 4 \alpha}(1-\cos 8 \alpha)= \\ & =\frac{1}{\frac{1-\cos 8 \alpha}{\cos 8 \alpha}} \cdot(1-\cos 8 \alpha)=\frac{\sin 8 \alpha(1-\cos 8 \alpha)}{1-\cos 8 \alpha}=\sin 8 \alpha \end{aligned}

The identity is proven.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.