[tex]\mathrm{tg}\,8\alpha - \mathrm{ctg}\,8\alpha = -2\,\mathrm{ctg}\,16\alpha[/tex]
Преобразуем левую часть к правой:
[tex]\mathrm{tg}\,8\alpha - \mathrm{ctg}\,8\alpha =\dfrac{\sin8\alpha }{\cos8\alpha } -\dfrac{\cos8\alpha }{\sin8\alpha } =\dfrac{\sin^28\alpha-\cos^28\alpha }{\sin8\alpha\cos8\alpha } =[/tex]
[tex]=2\cdot\dfrac{-(\cos^28\alpha -\sin^28\alpha)}{2\sin8\alpha\cos8\alpha } =-2\cdot\dfrac{\cos^28\alpha -\sin^28\alpha}{2\sin8\alpha\cos8\alpha } =[/tex]
[tex]=-2\cdot\dfrac{\cos16\alpha }{\sin16\alpha } =-2\,\mathrm{ctg}\,16\alpha[/tex]
Copyright © 2025 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
[tex]\mathrm{tg}\,8\alpha - \mathrm{ctg}\,8\alpha = -2\,\mathrm{ctg}\,16\alpha[/tex]
Преобразуем левую часть к правой:
[tex]\mathrm{tg}\,8\alpha - \mathrm{ctg}\,8\alpha =\dfrac{\sin8\alpha }{\cos8\alpha } -\dfrac{\cos8\alpha }{\sin8\alpha } =\dfrac{\sin^28\alpha-\cos^28\alpha }{\sin8\alpha\cos8\alpha } =[/tex]
[tex]=2\cdot\dfrac{-(\cos^28\alpha -\sin^28\alpha)}{2\sin8\alpha\cos8\alpha } =-2\cdot\dfrac{\cos^28\alpha -\sin^28\alpha}{2\sin8\alpha\cos8\alpha } =[/tex]
[tex]=-2\cdot\dfrac{\cos16\alpha }{\sin16\alpha } =-2\,\mathrm{ctg}\,16\alpha[/tex]