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受欢迎的三角函数 >

2sin(15)cos(15)=tan(x)

解答

2 sin (1 5^{\circ}) cos (1 5^{\circ}) = tan (x)

解答

x = 0.46364 \dots + 18 0^{\circ} n

+1

弧度

x = 0.46364 \dots + πn

求解步骤

2 sin (1 5^{\circ}) cos (1 5^{\circ}) = tan (x)

sin (1 5^{\circ}) = \frac{6 - 2}{4}

sin (1 5^{\circ})

使用三角恒等式改写:

sin (4 5^{\circ}) cos (3 0^{\circ}) - cos (4 5^{\circ}) sin (3 0^{\circ})

sin (1 5^{\circ})

将

sin (1 5^{\circ})

写为

sin (4 5^{\circ} - 3 0^{\circ})

= sin (4 5^{\circ} - 3 0^{\circ})

使用角差恒等式:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= sin (4 5^{\circ}) cos (3 0^{\circ}) - cos (4 5^{\circ}) sin (3 0^{\circ})

= sin (4 5^{\circ}) cos (3 0^{\circ}) - cos (4 5^{\circ}) sin (3 0^{\circ})

使用以下普通恒等式:

sin (4 5^{\circ}) = \frac{2}{2}

sin (4 5^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

使用以下普通恒等式:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

使用以下普通恒等式:

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= \frac{2}{2} \cdot \frac{3}{2} - \frac{2}{2} \cdot \frac{1}{2}

化简

\frac{2}{2} \cdot \frac{3}{2} - \frac{2}{2} \cdot \frac{1}{2} : \frac{6 - 2}{4}

\frac{2}{2} \cdot \frac{3}{2} - \frac{2}{2} \cdot \frac{1}{2}

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

\frac{2}{2} \cdot \frac{3}{2}

分式相乘:

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

= \frac{2 3}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 3}{4}

化简

23 : 6

23

使用根式运算法则:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

数字相乘:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

\frac{2}{2} \cdot \frac{1}{2} = \frac{2}{4}

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

分式相乘:

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

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

乘以:

2 \cdot 1 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{6}{4} - \frac{2}{4}

使用法则

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

= \frac{6 - 2}{4}

= \frac{6 - 2}{4}

cos (1 5^{\circ}) = \frac{6 + 2}{4}

cos (1 5^{\circ})

使用三角恒等式改写:

cos (4 5^{\circ}) cos (3 0^{\circ}) + sin (4 5^{\circ}) sin (3 0^{\circ})

cos (1 5^{\circ})

将

cos (1 5^{\circ})

写为

cos (4 5^{\circ} - 3 0^{\circ})

= cos (4 5^{\circ} - 3 0^{\circ})

使用角差恒等式:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= cos (4 5^{\circ}) cos (3 0^{\circ}) + sin (4 5^{\circ}) sin (3 0^{\circ})

= cos (4 5^{\circ}) cos (3 0^{\circ}) + sin (4 5^{\circ}) sin (3 0^{\circ})

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

使用以下普通恒等式:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

使用以下普通恒等式:

sin (4 5^{\circ}) = \frac{2}{2}

sin (4 5^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

使用以下普通恒等式:

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

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

化简

\frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2} : \frac{6 + 2}{4}

\frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2}

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

\frac{2}{2} \cdot \frac{3}{2}

分式相乘:

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

= \frac{2 3}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 3}{4}

化简

23 : 6

23

使用根式运算法则:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

数字相乘:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

\frac{2}{2} \cdot \frac{1}{2} = \frac{2}{4}

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

分式相乘:

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

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

乘以:

2 \cdot 1 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{6}{4} + \frac{2}{4}

使用法则

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

= \frac{6 + 2}{4}

= \frac{6 + 2}{4}

2 \cdot \frac{6 - 2}{4} \cdot \frac{6 + 2}{4} = tan (x)

交换两边

tan (x) = 2 \cdot \frac{6 - 2}{4} \cdot \frac{6 + 2}{4}

化简

2 \cdot \frac{6 - 2}{4} \cdot \frac{6 + 2}{4} : \frac{1}{2}

2 \cdot \frac{6 - 2}{4} \cdot \frac{6 + 2}{4}

分式相乘:

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

= \frac{( 6 - 2 ) ( 6 + 2 ) \cdot 2}{4 \cdot 4}

数字相乘:

4 \cdot 4 = 16

= \frac{2 ( 6 - 2 ) ( 6 + 2 )}{16}

约分:

2

= \frac{( 6 - 2 ) ( 6 + 2 )}{8}

乘开

(6 - 2) (6 + 2) : 4

(6 - 2) (6 + 2)

使用平方差公式:

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

a = 6, b = 2

= (6)^{2} - (2)^{2}

化简

(6)^{2} - (2)^{2} : 4

(6)^{2} - (2)^{2}

(6)^{2} = 6

(6)^{2}

使用根式运算法则:

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

= (6^{\frac{1}{2}})^{2}

使用指数法则:

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

= 6^{\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

= 6

(2)^{2} = 2

(2)^{2}

使用根式运算法则:

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

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

使用指数法则:

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

= 2^{\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

= 2

= 6 - 2

数字相减:

6 - 2 = 4

= 4

= 4

= \frac{4}{8}

约分:

4

= \frac{1}{2}

使用反三角函数性质

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

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

的通解

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

x = arctan (\frac{1}{2}) + 18 0^{\circ} n

x = arctan (\frac{1}{2}) + 18 0^{\circ} n

以小数形式表示解

x = 0.46364 \dots + 18 0^{\circ} n

作图

流行的例子

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6 sin^{2} (x) + 5 cos (x) = 7

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- 3 = tan^{2} (θ) - 2 - 2 tan (θ)

tan(6x)-3tan(3x)=0

tan (6 x) - 3 tan (3 x) = 0

sin (a) = \frac{2}{5}

cos^2(θ)-cos(θ)-1=0

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