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受欢迎的三角函数 >

证明 2sin(a+b)sin(a-b)=cos(2b)-cos(2a)

解答

证明

2 sin (a + b) sin (a - b) = cos (2 b) - cos (2 a)

解答

真

求解步骤

2 sin (a + b) sin (a - b) = cos (2 b) - cos (2 a)

调整左侧

2 sin (a + b) sin (a - b)

使用三角恒等式改写

2 sin (a + b) sin (a - b)

使用积化和差恒等式:

sin (s) sin (t) = \frac{1}{2} (cos (s - t) - cos (s + t))

= 2 \cdot \frac{1}{2} (cos (a + b - (a - b)) - cos (a + b + a - b))

化简

2 \cdot \frac{1}{2} (cos (a + b - (a - b)) - cos (a + b + a - b)) : cos (2 b) - cos (2 a)

2 \cdot \frac{1}{2} (cos (a + b - (a - b)) - cos (a + b + a - b))

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2 ( cos ( a + b - ( a - b ) ) - cos ( a + b + a - b ) )}{2}

约分:

2

= 1 \cdot (cos (a + b - (a - b)) - cos (a + b + a - b))

乘开

a + b - (a - b) : 2 b

a + b - (a - b)

- (a - b) : - a + b

- (a - b)

打开括号

= - (a) - (- b)

使用加减运算法则

- (- a) = a, - (a) = - a

= - a + b

= a + b - a + b

化简

a + b - a + b : 2 b

a + b - a + b

对同类项分组

= a - a + b + b

同类项相加:

a - a = 0

= b + b

同类项相加:

b + b = 2 b

= 2 b

= 2 b

= 1 \cdot (cos (2 b) - cos (a + a + b - b))

化简

a + b + a - b : 2 a

a + b + a - b

对同类项分组

= a + a + b - b

同类项相加:

a + a = 2 a

= 2 a + b - b

同类项相加:

b - b = 0

= 2 a

= 1 \cdot (cos (2 b) - cos (2 a))

整理后得

= cos (2 b) - cos (2 a)

= cos (2 b) - cos (2 a)

= cos (2 b) - cos (2 a)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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