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受欢迎的三角函数 >

cos(2arctan(3))

解答

cos (2 arctan (3))

解答

- \frac{4}{5}

+1

十进制

- 0.8

求解步骤

cos (2 arctan (3))

使用三角恒等式改写:

2 cos^{2} (arctan (3)) - 1

cos (2 arctan (3))

使用倍角公式:

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

= 2 cos^{2} (arctan (3)) - 1

= 2 cos^{2} (arctan (3)) - 1

使用三角恒等式改写:

cos (arctan (3)) = \frac{10}{10}

cos (arctan (3))

使用三角恒等式改写:

cos (arctan (3)) = \frac{1 + 3 ^{2}}{1 + 3 ^{2}}

利用以下特性:

cos (arctan (x)) = \frac{1 + x ^{2}}{1 + x ^{2}}

= \frac{1 + 3 ^{2}}{1 + 3 ^{2}}

= \frac{1 + 3 ^{2}}{1 + 3 ^{2}}

化简

= \frac{10}{10}

= 2 (\frac{10}{10})^{2} - 1

化简

2 (\frac{10}{10})^{2} - 1 : - \frac{4}{5}

2 (\frac{10}{10})^{2} - 1

2 (\frac{10}{10})^{2} = \frac{1}{5}

2 (\frac{10}{10})^{2}

(\frac{10}{10})^{2} = \frac{1}{10}

(\frac{10}{10})^{2}

使用指数法则:

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

= \frac{( 10 ) ^{2}}{1 0 ^{2}}

(10)^{2} : 10

使用根式运算法则:

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

= (1 0^{\frac{1}{2}})^{2}

使用指数法则:

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

= 1 0^{\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

= 10

= \frac{10}{1 0 ^{2}}

约分:

10

= \frac{1}{10}

= 2 \cdot \frac{1}{10}

分式相乘:

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

= \frac{1 \cdot 2}{10}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{10}

约分:

2

= \frac{1}{5}

= \frac{1}{5} - 1

将项转换为分式:

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

= - \frac{1 \cdot 5}{5} + \frac{1}{5}

因为分母相等，所以合并分式:

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

= \frac{- 1 \cdot 5 + 1}{5}

- 1 \cdot 5 + 1 = - 4

- 1 \cdot 5 + 1

数字相乘:

1 \cdot 5 = 5

= - 5 + 1

数字相加/相减:

- 5 + 1 = - 4

= - 4

= \frac{- 4}{5}

使用分式法则:

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

= - \frac{4}{5}

= - \frac{4}{5}

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sin (43. 5^{\circ})

cos((2pi)/3)+sin((2pi)/3)

cos (\frac{2 π}{3}) + sin (\frac{2 π}{3})

tan (\frac{25 π}{4})

e^{- 1} sin (1)

e^{-pi/2}sin(pi/2)

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