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受欢迎的三角函数 >

证明 cos(4B)=1-8sin^2(B)+8sin^4(B)

解答

证明

cos (4 B) = 1 - 8 sin^{2} (B) + 8 sin^{4} (B)

解答

真

求解步骤

cos (4 B) = 1 - 8 sin^{2} (B) + 8 sin^{4} (B)

调整左侧

cos (4 B)

使用三角恒等式改写

cos (4 B)

= cos (2 \cdot 2 B)

使用倍角公式:

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

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

使用倍角公式:

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

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

化简

2 (1 - 2 sin^{2} (B))^{2} - 1 : 8 sin^{4} (B) - 8 sin^{2} (B) + 1

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

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

使用完全平方公式:

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

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

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

化简

1^{2} - 2 \cdot 1 \cdot 2 sin^{2} (B) + (2 sin^{2} (B))^{2} : 1 - 4 sin^{2} (B) + 4 sin^{4} (B)

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

使用法则

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot 2 sin^{2} (B) = 4 sin^{2} (B)

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

数字相乘:

2 \cdot 1 \cdot 2 = 4

= 4 sin^{2} (B)

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

(2 sin^{2} (B))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} (sin^{2} (B))^{2}

(sin^{2} (B))^{2} : sin^{4} (B)

使用指数法则:

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

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

数字相乘:

2 \cdot 2 = 4

= sin^{4} (B)

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

2^{2} = 4

= 4 sin^{4} (B)

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

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

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

乘开

2 (1 - 4 sin^{2} (B) + 4 sin^{4} (B)) : 2 - 8 sin^{2} (B) + 8 sin^{4} (B)

2 (1 - 4 sin^{2} (B) + 4 sin^{4} (B))

打开括号

= 2 \cdot 1 + 2 (- 4 sin^{2} (B)) + 2 \cdot 4 sin^{4} (B)

使用加减运算法则

+ (- a) = - a

= 2 \cdot 1 - 2 \cdot 4 sin^{2} (B) + 2 \cdot 4 sin^{4} (B)

化简

2 \cdot 1 - 2 \cdot 4 sin^{2} (B) + 2 \cdot 4 sin^{4} (B) : 2 - 8 sin^{2} (B) + 8 sin^{4} (B)

2 \cdot 1 - 2 \cdot 4 sin^{2} (B) + 2 \cdot 4 sin^{4} (B)

数字相乘:

2 \cdot 1 = 2

= 2 - 2 \cdot 4 sin^{2} (B) + 2 \cdot 4 sin^{4} (B)

数字相乘:

2 \cdot 4 = 8

= 2 - 8 sin^{2} (B) + 8 sin^{4} (B)

= 2 - 8 sin^{2} (B) + 8 sin^{4} (B)

= 2 - 8 sin^{2} (B) + 8 sin^{4} (B) - 1

化简

2 - 8 sin^{2} (B) + 8 sin^{4} (B) - 1 : 8 sin^{4} (B) - 8 sin^{2} (B) + 1

2 - 8 sin^{2} (B) + 8 sin^{4} (B) - 1

对同类项分组

= - 8 sin^{2} (B) + 8 sin^{4} (B) + 2 - 1

数字相加/相减:

2 - 1 = 1

= 8 sin^{4} (B) - 8 sin^{2} (B) + 1

= 8 sin^{4} (B) - 8 sin^{2} (B) + 1

= 8 sin^{4} (B) - 8 sin^{2} (B) + 1

= 8 sin^{4} (B) - 8 sin^{2} (B) + 1

= 1 - 8 sin^{2} (B) + 8 sin^{4} (B)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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