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受欢迎的三角函数 >

证明 csc^4(x)-cot^4(x)=1+2cot^2(x)

解答

证明

csc^{4} (x) - cot^{4} (x) = 1 + 2 cot^{2} (x)

解答

真

求解步骤

csc^{4} (x) - cot^{4} (x) = 1 + 2 cot^{2} (x)

调整左侧

csc^{4} (x) - cot^{4} (x)

分解

- cot^{4} (x) + csc^{4} (x) : (csc^{2} (x) + cot^{2} (x)) (csc (x) + cot (x)) (csc (x) - cot (x))

- cot^{4} (x) + csc^{4} (x)

将

csc^{4} (x) - cot^{4} (x)

改写为

(csc^{2} (x))^{2} - (cot^{2} (x))^{2}

csc^{4} (x) - cot^{4} (x)

使用指数法则:

a^{b c} = (a^{b})^{c}

cot^{4} (x) = (cot^{2} (x))^{2}

= csc^{4} (x) - (cot^{2} (x))^{2}

使用指数法则:

a^{b c} = (a^{b})^{c}

csc^{4} (x) = (csc^{2} (x))^{2}

= (csc^{2} (x))^{2} - (cot^{2} (x))^{2}

= (csc^{2} (x))^{2} - (cot^{2} (x))^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(csc^{2} (x))^{2} - (cot^{2} (x))^{2} = (csc^{2} (x) + cot^{2} (x)) (csc^{2} (x) - cot^{2} (x))

= (csc^{2} (x) + cot^{2} (x)) (csc^{2} (x) - cot^{2} (x))

分解

csc^{2} (x) - cot^{2} (x) : (csc (x) + cot (x)) (csc (x) - cot (x))

csc^{2} (x) - cot^{2} (x)

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

csc^{2} (x) - cot^{2} (x) = (csc (x) + cot (x)) (csc (x) - cot (x))

= (csc (x) + cot (x)) (csc (x) - cot (x))

= (csc^{2} (x) + cot^{2} (x)) (csc (x) + cot (x)) (csc (x) - cot (x))

= (- cot (x) + csc (x)) (cot (x) + csc (x)) (cot^{2} (x) + csc^{2} (x))

使用三角恒等式改写

(- cot (x) + csc (x)) (cot (x) + csc (x)) (cot^{2} (x) + csc^{2} (x))

乘开

(csc (x) + cot (x)) (csc (x) - cot (x)) : csc^{2} (x) - cot^{2} (x)

(csc (x) + cot (x)) (csc (x) - cot (x))

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = csc (x), b = cot (x)

= csc^{2} (x) - cot^{2} (x)

= (csc^{2} (x) - cot^{2} (x)) (cot^{2} (x) + csc^{2} (x))

使用毕达哥拉斯恒等式:

csc^{2} (x) = 1 + cot^{2} (x)

csc^{2} (x) - cot^{2} (x) = 1

= 1 \cdot (cot^{2} (x) + csc^{2} (x))

化简

= cot^{2} (x) + csc^{2} (x)

使用毕达哥拉斯恒等式:

csc^{2} (x) = 1 + cot^{2} (x)

= cot^{2} (x) + 1 + cot^{2} (x)

化简

cot^{2} (x) + 1 + cot^{2} (x) : 2 cot^{2} (x) + 1

cot^{2} (x) + 1 + cot^{2} (x)

对同类项分组

= cot^{2} (x) + cot^{2} (x) + 1

同类项相加:

cot^{2} (x) + cot^{2} (x) = 2 cot^{2} (x)

= 2 cot^{2} (x) + 1

= 2 cot^{2} (x) + 1

= 2 cot^{2} (x) + 1

= 1 + 2 cot^{2} (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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