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受欢迎的三角函数 >

证明 sin(a+b)+sin(a-b)=2sin(a)cos(b)

解答

证明

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

解答

真

求解步骤

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

调整右侧

2 sin (a) cos (b)

使用三角恒等式改写

2 sin (a) cos (b)

使用积化和差恒等式:

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

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

化简

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

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

分式相乘:

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

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

约分:

2

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

整理后得

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

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p ro v e (cos (x)) (1 + tan (x))^{2} = sec (x) + 2 sin (x)

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p ro v e \frac{1 + tan ^{2} ( x )}{2 tan ( x )} = csc (2 x)

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