Задача 45496 Подчеркнутое уравнение (№ 4.41)...

soursemaster

2020-03-28 13:53:20

Подчеркнутое уравнение (№ 4.41)

sova

2020-03-28 13:55:05

[m]sin(\frac{3x}{2}+\frac{\pi}{4})=sin\frac{3x}{2}\cdot cos\frac{\pi}{4}+cos\frac{3x}{2}\cdot sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}(sin\frac{3x}{2}+cos\frac{3x}{2})[/m]

[m]-2\sqrt{2} sin(\frac{3x}{2}+\frac{\pi}{4})=-2(sin\frac{3x}{2}+cos\frac{3x}{2})[/m]

[m]sin3x=2sin\frac{3x}{2}\cdot cos\frac{3x}{2}[/m]


[m]\frac{4}{\sqrt{3}+1}=\frac{4(\sqrt{3}-1)}{(\sqrt(3))^2-1)}=\frac{4(\sqrt{3}-1)}{2}=2\sqrt{3}-2[/m]



[m]tg14x=\frac{sin14x}{cos14}[/m]


[m]ctg7x=\frac{sin14x}{1-cos14x}[/m]


[m]tg14x+3ctg7x=\frac{sin14x}{cos14x}+\frac{3sin14x}{1-cos14x}=\frac{sin14x\cdot(1-cos14x+3cos14x}{cos14x\cdot (1-cos14x)}=\frac{sin14x\cdot(1+2cos14x}{cos14x\cdot (1-cos14x)}=[/m][m]=tg14x\cdot \frac{1+2cos14x}{1-cos14x}[/m]

Итак:

[m] tg14x+3ctg7x+sin3x-2\sqrt{2} sin(\frac{3x}{2}+\frac{\pi}{4})=\frac{4}{\sqrt{3}+1}[/m]

[m]tg14x\cdot \frac{1+2cos14x}{1-cos14x}-2\sqrt{3}+(2sin\frac{3x}{2}cos\frac{3x}{2}-2sin\frac{3x}{2}-2cos\frac{3x}{2}+2)=0[/m]