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受欢迎的三角函数 >

证明 (sin^2(x)-1)^2=cos(2x)+sin^4(x)

解答

证明

(sin^{2} (x) - 1)^{2} = cos (2 x) + sin^{4} (x)

解答

真

求解步骤

(sin^{2} (x) - 1)^{2} = cos (2 x) + sin^{4} (x)

调整左侧

(sin^{2} (x) - 1)^{2}

乘开

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

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

使用完全平方公式:

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

a = - 1, b = sin^{2} (x)

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

化简

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

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

去除括号:

(- a) = - a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

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

2 \cdot 1 \cdot sin^{2} (x)

数字相乘:

2 \cdot 1 = 2

= 2 sin^{2} (x)

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

(sin^{2} (x))^{2}

使用指数法则:

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

= sin^{2 \cdot 2} (x)

数字相乘:

2 \cdot 2 = 4

= sin^{4} (x)

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

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

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

调整右侧

cos (2 x) + sin^{4} (x)

使用三角恒等式改写

cos (2 x) + sin^{4} (x)

使用倍角公式:

cos (2 x) = 1 - 2 sin^{2} (x)

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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