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受欢迎的三角函数 >

证明 tan(8x)-tan(8x)tan^2(4x)=2tan(4x)

解答

证明

tan (8 x) - tan (8 x) tan^{2} (4 x) = 2 tan (4 x)

解答

真

求解步骤

tan (8 x) - tan (8 x) tan^{2} (4 x) = 2 tan (4 x)

调整左侧

tan (8 x) - tan (8 x) tan^{2} (4 x)

使用三角恒等式改写

tan (8 x) - tan (8 x) tan^{2} (4 x)

= tan (2 \cdot 4 x) - tan (2 \cdot 4 x) tan^{2} (4 x)

使用倍角公式:

tan (2 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

= \frac{2 tan ( 4 x )}{1 - tan ^{2} ( 4 x )} - \frac{2 tan ( 4 x )}{1 - tan ^{2} ( 4 x )} tan^{2} (4 x)

化简

\frac{2 tan ( 4 x )}{1 - tan ^{2} ( 4 x )} - \frac{2 tan ( 4 x )}{1 - tan ^{2} ( 4 x )} tan^{2} (4 x) : 2 tan (4 x)

\frac{2 tan ( 4 x )}{1 - tan ^{2} ( 4 x )} - \frac{2 tan ( 4 x )}{1 - tan ^{2} ( 4 x )} tan^{2} (4 x)

\frac{2 tan ( 4 x )}{1 - tan ^{2} ( 4 x )} tan^{2} (4 x) = \frac{2 tan ^{3} ( 4 x )}{1 - tan ^{2} ( 4 x )}

\frac{2 tan ( 4 x )}{1 - tan ^{2} ( 4 x )} tan^{2} (4 x)

分式相乘:

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

= \frac{2 tan ( 4 x ) tan ^{2} ( 4 x )}{1 - tan ^{2} ( 4 x )}

2 tan (4 x) tan^{2} (4 x) = 2 tan^{3} (4 x)

2 tan (4 x) tan^{2} (4 x)

使用指数法则:

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

tan (4 x) tan^{2} (4 x) = tan^{1 + 2} (4 x)

= 2 tan^{1 + 2} (4 x)

数字相加:

1 + 2 = 3

= 2 tan^{3} (4 x)

= \frac{2 tan ^{3} ( 4 x )}{1 - tan ^{2} ( 4 x )}

= \frac{2 tan ( 4 x )}{- tan ^{2} ( 4 x ) + 1} - \frac{2 tan ^{3} ( 4 x )}{- tan ^{2} ( 4 x ) + 1}

使用法则

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

= \frac{2 tan ( 4 x ) - 2 tan ^{3} ( 4 x )}{1 - tan ^{2} ( 4 x )}

分解

2 tan (4 x) - 2 tan^{3} (4 x) : 2 tan (4 x) (1 - tan^{2} (4 x))

2 tan (4 x) - 2 tan^{3} (4 x)

使用指数法则:

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

tan^{3} (4 x) = tan (4 x) tan^{2} (4 x)

= 2 tan (4 x) - 2 tan (4 x) tan^{2} (4 x)

改写为

= 1 \cdot 2 tan (4 x) - 2 tan (4 x) tan^{2} (4 x)

因式分解出通项

2 tan (4 x)

= 2 tan (4 x) (1 - tan^{2} (4 x))

= \frac{2 tan ( 4 x ) ( 1 - tan ^{2} ( 4 x ) )}{1 - tan ^{2} ( 4 x )}

约分:

1 - tan^{2} (4 x)

= 2 tan (4 x)

= 2 tan (4 x)

= 2 tan (4 x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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