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受欢迎的三角函数 >

sin(x)=sin(x+30)

解答

sin (x) = sin (x + 3 0^{\circ})

解答

x = 1.30899 \dots + 18 0^{\circ} n

+1

弧度

x = 1.30899 \dots + πn

求解步骤

sin (x) = sin (x + 3 0^{\circ})

使用三角恒等式改写

sin (x) = sin (x + 3 0^{\circ})

使用三角恒等式改写

sin (x + 3 0^{\circ})

使用角和恒等式:

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

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

化简

sin (x) cos (3 0^{\circ}) + cos (x) sin (3 0^{\circ}) : \frac{3}{2} sin (x) + \frac{1}{2} cos (x)

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

化简

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

cos (3 0^{\circ})

使用以下普通恒等式:

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

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}

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

化简

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

sin (3 0^{\circ})

使用以下普通恒等式:

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

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{3}{2} sin (x) + \frac{1}{2} cos (x)

= \frac{3}{2} sin (x) + \frac{1}{2} cos (x)

sin (x) = \frac{3}{2} sin (x) + \frac{1}{2} cos (x)

sin (x) = \frac{3}{2} sin (x) + \frac{1}{2} cos (x)

两边减去

\frac{3}{2} sin (x) + \frac{1}{2} cos (x)

\frac{2 - 3}{2} sin (x) - \frac{1}{2} cos (x) = 0

化简

\frac{2 - 3}{2} sin (x) - \frac{1}{2} cos (x) : \frac{( 2 - 3 ) sin ( x ) - cos ( x )}{2}

\frac{2 - 3}{2} sin (x) - \frac{1}{2} cos (x)

\frac{2 - 3}{2} sin (x) = \frac{( 2 - 3 ) sin ( x )}{2}

\frac{2 - 3}{2} sin (x)

分式相乘:

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

= \frac{( 2 - 3 ) sin ( x )}{2}

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

\frac{1}{2} cos (x)

分式相乘:

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

= \frac{1 \cdot cos ( x )}{2}

乘以:

1 \cdot cos (x) = cos (x)

= \frac{cos ( x )}{2}

= \frac{( 2 - 3 ) sin ( x )}{2} - \frac{cos ( x )}{2}

使用法则

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

= \frac{( 2 - 3 ) sin ( x ) - cos ( x )}{2}

\frac{( 2 - 3 ) sin ( x ) - cos ( x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

(2 - 3) sin (x) - cos (x) = 0

使用三角恒等式改写

(2 - 3) sin (x) - cos (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

\frac{( 2 - 3 ) sin ( x ) - cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

\frac{2 sin ( x )}{cos ( x )} - \frac{3 sin ( x )}{cos ( x )} - 1 = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

(2 - 3) tan (x) - 1 = 0

(2 - 3) tan (x) - 1 = 0

将

1

到右边

(2 - 3) tan (x) - 1 = 0

两边加上

1

(2 - 3) tan (x) - 1 + 1 = 0 + 1

化简

(2 - 3) tan (x) = 1

(2 - 3) tan (x) = 1

两边除以

2 - 3

(2 - 3) tan (x) = 1

两边除以

2 - 3

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

化简

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

化简

\frac{( 2 - 3 ) tan ( x )}{2 - 3} : tan (x)

\frac{( 2 - 3 ) tan ( x )}{2 - 3}

约分:

2 - 3

= tan (x)

化简

\frac{1}{2 - 3} : 2 + 3

\frac{1}{2 - 3}

乘以共轭根式

\frac{2 + 3}{2 + 3}

= \frac{1 \cdot ( 2 + 3 )}{( 2 - 3 ) ( 2 + 3 )}

1 \cdot (2 + 3) = 2 + 3

(2 - 3) (2 + 3) = 1

(2 - 3) (2 + 3)

使用平方差公式:

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

a = 2, b = 3

= 2^{2} - (3)^{2}

化简

2^{2} - (3)^{2} : 1

2^{2} - (3)^{2}

2^{2} = 4

2^{2}

2^{2} = 4

= 4

(3)^{2} = 3

(3)^{2}

使用根式运算法则:

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

= (3^{\frac{1}{2}})^{2}

使用指数法则:

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

= 3^{\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

= 3

= 4 - 3

数字相减:

4 - 3 = 1

= 1

= 1

= \frac{2 + 3}{1}

使用法则

\frac{a}{1} = a

= 2 + 3

tan (x) = 2 + 3

tan (x) = 2 + 3

tan (x) = 2 + 3

使用反三角函数性质

tan (x) = 2 + 3

tan (x) = 2 + 3

的通解

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

x = arctan (2 + 3) + 18 0^{\circ} n

x = arctan (2 + 3) + 18 0^{\circ} n

以小数形式表示解

x = 1.30899 \dots + 18 0^{\circ} n

作图

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