Помогите пожалуйста !!!
Доказать тождества:
1 . (sinх - siny)²+ (cosx-cosy)²= 4sin²(x-y)/2;
2. (sinα+sinβ)²+(cosα+cosβ)²=4cos²(α-β)/2;
3. cos² (α+β) - cos²(α-β)= - sin2α * sin2β;
4. sin² (x+y) - sin²(x-y)= sin2x * sin2y.
Доказать тождества:
1 . (sinх - siny)²+ (cosx-cosy)²= 4sin²(x-y)/2;
2. (sinα+sinβ)²+(cosα+cosβ)²=4cos²(α-β)/2;
3. cos² (α+β) - cos²(α-β)= - sin2α * sin2β;
4. sin² (x+y) - sin²(x-y)= sin2x * sin2y.
More Questions From This User See All
Answers & Comments
Verified answer
Доказать тождества :1 .
(sinх - siny)² + (cosx-cosy)² = 4sin²(x-y)/2
---
(sinх - siny)²+(cosx - cosy)² =
(sin²х -2sinx*siny +sin²y) + (cos²x -2cosx*cosy+ cos²y) =
(sin²х +cos²x) +(sin²y+cos²y) -2(cosx*cosy+sinx*siny) =2 - 2cos(x-y) =
2(1 -cos(x-y) ) = 2*2sin²(x-y)/2 = 4sin²(x-y)/2 .
можно доказать и так :
(sinх - siny)²+(cosx-cosy)²=(2sin(x-y)/2 *cos(x+y)/2 )²+(-2sin(x-y)/2 *sin(x+y)/2 )²=
4sin²(x-y)/2 *(cos² (x+y)/2 +sin² (x+y)/2 ) = 4sin²(x-y)/2 *1 = 4sin²(x-y)/2 .
=======
2.
(sinα+sinβ)²+(cosα+cosβ)² = 4cos²(α-β)/2
---
(sinα+sinβ)²+(cosα+cosβ)² =
(sin²α+2sinα*sinβ+sin²β)+(cos²α+2cosα*cosβ+cos²β)= (sin²α +cos²α) +(sin²β+cos²β) +2(cosα*cosβ+sinα*sinβ) =2 + 2cos(α-β) =
2(1 +cos(α-β) ) = 2*2cos²(α-β)/2 = 4cos²(α-β)/2 .
или по другому:
(sinα+ sinβ)² + (cosα+cosβ)² =
(2sin(α+β)/2 *cos(α-β)/2 )² +(2cos(α-β)/2 *cos(α+β)/2 )² =
4cos²(α-β)/2 *(sin² (α+β)/2 +cos² (α+β)/2 ) = 4cos²(α-β)/2 .
=======
3.
cos² (α+β) - cos²(α-β) = - sin2α * sin2β ;
---
cos² (α + β) - cos²(α - β) = (1+cos2(α+β) )/2 - (1+cos2(α-β) ) /2 =
( cos(2α+2β) - cos(2α-2β) )/2 = - sin2α * sin2β .
* * * cosA - cosB = -2sin(A - B)/2* sin(A+B)/2 * * *
можно доказать и так
cos² (α+β) - cos²(α-β) = (cos (α+β) - cos(α-β) )* (cos (α+β) + cos(α-β) ) =
= (-2sinβsinα) * (2cosαcosβ)= - (2sinαcosα)*(2sinβcosβ) = - sin2α * sin2β.
=======
4.
sin² (x+y) - sin²(x-y)= sin2x * sin2y.
---
sin² (x+y) - sin²(x-y) =(1 - cos2(x+y) )/2 - (1 -cos2(x - y) )/2 =
(cos(2x - 2y) - cos(2x +2y) ) /2 = -sin(-2y)*sin2x = sin2x*sin2y .
* * * У ДАЧИ !!! * * *