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受欢迎的三角函数 >

证明 cot^4(x)+cot^2(x)=csc^4(x)-csc^2(x)

解答

证明

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

解答

真

求解步骤

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

调整右侧

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

分解

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

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

使用指数法则:

a^{b + c} = a^{b} a^{c}

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

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

因式分解出通项

csc^{2} (x)

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

化简

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

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

乘开

(1 + cot^{2} (x)) cot^{2} (x) : cot^{2} (x) + cot^{4} (x)

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

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

使用分配律:

a (b + c) = ab + a c

a = cot^{2} (x), b = 1, c = cot^{2} (x)

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

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

化简

1 \cdot cot^{2} (x) + cot^{2} (x) cot^{2} (x) : cot^{2} (x) + cot^{4} (x)

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

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

1 \cdot cot^{2} (x)

乘以:

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

= cot^{2} (x)

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

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

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

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

= cot^{2 + 2} (x)

数字相加:

2 + 2 = 4

= cot^{4} (x)

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

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

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

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

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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