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受欢迎的三角函数 >

tan(2arccos(1/4))

解答

tan (2 arccos (\frac{1}{4}))

解答

- \frac{15}{7}

+1

十进制

- 0.55328 \dots

求解步骤

tan (2 arccos (\frac{1}{4}))

使用三角恒等式改写:

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

tan (2 arccos (\frac{1}{4}))

使用倍角公式:

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

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

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

使用三角恒等式改写:

tan (arccos (\frac{1}{4})) = 15

tan (arccos (\frac{1}{4}))

使用三角恒等式改写:

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

利用以下特性:

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

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

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

化简

= 15

= \frac{2 15}{1 - ( 15 ) ^{2}}

化简

\frac{2 15}{1 - ( 15 ) ^{2}} : - \frac{15}{7}

\frac{2 15}{1 - ( 15 ) ^{2}}

(15)^{2} = 15

(15)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (1 5^{\frac{1}{2}})^{2}

使用指数法则:

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

= 1 5^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 15

= \frac{2 15}{1 - 15}

数字相减:

1 - 15 = - 14

= \frac{2 15}{- 14}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2 15}{14}

约分:

2

= - \frac{15}{7}

= - \frac{15}{7}

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