11. Упростите выражение: a) 2 sin a cos β-sin (α-β); cos (a-β)-2 sin a sin β б) ; 1-cos a+ cos 2a sin 2a-sin a В) ; √2 cos a-2 cos(+a) 2 sin(+a)-√2 sin a Г) ctg² a (1-cos 2a)+cos² a.

11. Упростите выражение:

a) $$\frac{2 sin α cos β-sin (α-β)}{cos (α-β)-2 sin α sin β}$$;

$$\frac{2 sin α cos β-sin (α-β)}{cos (α-β)-2 sin α sin β}$$ = $$\frac{2 sin α cos β-(sin α cos β - cos α sin β)}{cos α cos β + sin α sin β -2 sin α sin β}$$ = $$\frac{2 sin α cos β-sin α cos β + cos α sin β}{cos α cos β - sin α sin β}$$ = $$\frac{sin α cos β + cos α sin β}{cos α cos β - sin α sin β}$$ = $$\frac{sin (α + β)}{cos (α + β)}$$ = tg (α + β)

Ответ: tg (α + β)

б) $$\frac{1-cos α+ cos 2α}{sin 2α-sin α}$$;

$$\frac{1-cos α+ cos 2α}{sin 2α-sin α}$$ = $$\frac{1 - cos α + 2 cos² α - 1}{2 sin α cos α - sin α}$$ = $$\frac{-cos α + 2 cos² α}{2 sin α cos α - sin α}$$ = $$\frac{cos α (2 cos α - 1)}{sin α (2 cos α - 1)}$$ = $$\frac{cos α}{sin α}$$ = ctg α

Ответ: ctg α

В) $$\frac{\sqrt{2} cos α-2 cos(\frac{π}{4}+α)}}{2 sin(\frac{π}{4}+α)-\sqrt{2} sin α}$$;

$$\frac{\sqrt{2} cos α-2 cos(\frac{π}{4}+α)}}{2 sin(\frac{π}{4}+α)-\sqrt{2} sin α}$$ = $$\frac{\sqrt{2} cos α - 2 (cos \frac{π}{4} cos α - sin \frac{π}{4} sin α)}{2(sin \frac{π}{4} cos α + cos \frac{π}{4} sin α) - \sqrt{2} sin α}$$ = $$\frac{\sqrt{2} cos α - 2 (\frac{\sqrt{2}}{2} cos α - \frac{\sqrt{2}}{2} sin α)}{2(\frac{\sqrt{2}}{2} cos α + \frac{\sqrt{2}}{2} sin α) - \sqrt{2} sin α}$$ = $$\frac{\sqrt{2} cos α - \sqrt{2} cos α + \sqrt{2} sin α}{\sqrt{2} cos α + \sqrt{2} sin α - \sqrt{2} sin α}$$ = $$\frac{\sqrt{2} sin α}{\sqrt{2} cos α}$$ = tg α

Ответ: tg α

Г) ctg² a (1-cos 2a)+cos² a.

ctg² a (1-cos 2a)+cos² a = $$\frac{cos² α}{sin² α}$$ (1 - (cos² α - sin² α)) + cos² α = $$\frac{cos² α}{sin² α}$$ (1 - cos² α + sin² α) + cos² α = $$\frac{cos² α}{sin² α}$$ (sin² α + sin² α) + cos² α = $$\frac{cos² α}{sin² α}$$ * 2sin² α + cos² α = 2cos² α + cos² α = 3cos² α

Ответ: 3cos² α