Модули противоположных чисел равны: поэтому |х-1|=|1-х|
И теперь воспользуемся свойством:
|а|=а <=> а≥0
Решение во вложении.Надеюсь будет понятно.
<span>Доказать тождества :
</span><span>1 .
(sinх - siny)² + (cosx-cosy)² = 4sin²(x-y)/2
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</span>(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 -<span>cos(x-y) ) = 2*2sin</span>²(x-y)/2 = <span>4sin²(x-y)/2 .
</span>мо<span>жно доказать и так</span><span> :
</span>(sinх - siny)²+(cosx-cosy)²=(2sin(x-y)/2 *cos(x+y)/2 )²+(-2sin(x-y)/2 *sin(x+y)/2 )²=<span>
4</span>sin²(x-y)/2 *(cos² (x+y)/2 +sin² (x+y)/2 ) = 4sin²(x-y)/2 *1 = 4sin<span>²(x-y)/2 .</span><span>
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2.
(sinα+sinβ)²+(cosα+cosβ)² = 4cos²(α-β)/2
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</span>(sinα+sinβ)²+(cosα+cosβ)² =
(sin²α+2sinα*sinβ+sin²β)+(cos²α+2cosα*cosβ+cos²β)= (sin²α +cos²α) +(sin²β+cos²β) +2(cosα*cosβ+sinα*sinβ) =2<span> + </span>2cos(α-β) <span>=
2</span>(1 +cos(α-β) ) = 2*2cos²(α-β)/2 = 4cos²(α-β)/2<span> .
</span><span>или по другому:
</span>(sinα+ sinβ)² + (cosα+cosβ)² =
(2sin(α+β)/2 *cos(α-β)/2 )² +(2cos(α-β)/2 *cos(α+β)/2 )² =<span>
4</span>cos²(α-β)/2 *(sin² (α+β)/2 +cos² (α+β)/2 ) = 4cos²(α-β)/2 .<span>
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3.
cos² (α+β) - cos²(α-β) = - sin2α * sin2β ;
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</span>cos² (α + β) - cos²(α - β) = (1+cos2(α+β) )/2 - (1+cos2(<span>α-β) ) /2 =
</span>( cos(2α+2β) - cos(2α-2β) )/2 = - sin2α * sin2β .
<span>* * * cosA - cosB = -2sin(A - B)/2* sin(A+B)/2 * * *
мо</span><span>жно доказать и так
</span>cos² (α+β) - cos²(α-β) = (cos (α+β) - cos(α-β) )* <span> (</span>cos (α+β) + cos(α-β) ) =<span>
= (-2sin</span>βsinα) * (2cosαcosβ)= - (2sinαcosα)*(2sinβcosβ) = - sin2α * sin2β.<span>
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</span><span>4.
sin² (x+y) - sin²(x-y)= sin2x * sin2y.
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</span>sin² (x+y) - sin²(x-y) =(1 - cos2(x+y) )/2 - (1 -cos2(<span>x - y) )/2 =
(c</span>os(2x - 2y) - cos(2<span>x +2y) ) /2 = -sin(-2y)*sin2x = sin2x*sin2y .
</span>
* * * У ДАЧИ !!! * * *