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受欢迎的三角函数 >

tan(θ)=sin(2θ)

解答

tan (θ) = sin (2 θ)

解答

θ = 2 πn, θ = π + 2 πn, θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n, θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

tan (θ) = sin (2 θ)

两边减去

sin (2 θ)

tan (θ) - sin (2 θ) = 0

用 sin, cos 表示

- sin (2 θ) + tan (θ)

使用基本三角恒等式:

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

= - sin (2 θ) + \frac{sin ( θ )}{cos ( θ )}

化简

- sin (2 θ) + \frac{sin ( θ )}{cos ( θ )} : \frac{- sin ( 2 θ ) cos ( θ ) + sin ( θ )}{cos ( θ )}

- sin (2 θ) + \frac{sin ( θ )}{cos ( θ )}

将项转换为分式:

sin (2 θ) = \frac{sin ( 2 θ ) cos ( θ )}{cos ( θ )}

= - \frac{sin ( 2 θ ) cos ( θ )}{cos ( θ )} + \frac{sin ( θ )}{cos ( θ )}

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{- sin ( 2 θ ) cos ( θ ) + sin ( θ )}{cos ( θ )}

= \frac{- sin ( 2 θ ) cos ( θ ) + sin ( θ )}{cos ( θ )}

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

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

sin (θ) - cos (θ) sin (2 θ) = 0

使用三角恒等式改写

sin (θ) - cos (θ) sin (2 θ)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= sin (θ) - cos (θ) \cdot 2 sin (θ) cos (θ)

cos (θ) \cdot 2 sin (θ) cos (θ) = 2 cos^{2} (θ) sin (θ)

cos (θ) \cdot 2 sin (θ) cos (θ)

使用指数法则:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= 2 sin (θ) cos^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= 2 sin (θ) cos^{2} (θ)

= sin (θ) - 2 cos^{2} (θ) sin (θ)

sin (θ) - 2 cos^{2} (θ) sin (θ) = 0

分解

sin (θ) - 2 cos^{2} (θ) sin (θ) : - sin (θ) (2 cos (θ) + 1) (2 cos (θ) - 1)

sin (θ) - 2 cos^{2} (θ) sin (θ)

因式分解出通项

- sin (θ)

= - sin (θ) (- 1 + 2 cos^{2} (θ))

分解

2 cos^{2} (θ) - 1 : (2 cos (θ) + 1) (2 cos (θ) - 1)

2 cos^{2} (θ) - 1

将

2 cos^{2} (θ) - 1

改写为

(2 cos (θ))^{2} - 1^{2}

2 cos^{2} (θ) - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} cos^{2} (θ) - 1

将

1

改写为

1^{2}

= (2)^{2} cos^{2} (θ) - 1^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(2)^{2} cos^{2} (θ) = (2 cos (θ))^{2}

= (2 cos (θ))^{2} - 1^{2}

= (2 cos (θ))^{2} - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 cos (θ))^{2} - 1^{2} = (2 cos (θ) + 1) (2 cos (θ) - 1)

= (2 cos (θ) + 1) (2 cos (θ) - 1)

= - sin (θ) (2 cos (θ) + 1) (2 cos (θ) - 1)

- sin (θ) (2 cos (θ) + 1) (2 cos (θ) - 1) = 0

分别求解每个部分

sin (θ) = 0 or 2 cos (θ) + 1 = 0 or 2 cos (θ) - 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

sin (θ) = 0

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

2 cos (θ) + 1 = 0 : θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

2 cos (θ) + 1 = 0

将

1

到右边

2 cos (θ) + 1 = 0

两边减去

1

2 cos (θ) + 1 - 1 = 0 - 1

化简

2 cos (θ) = - 1

2 cos (θ) = - 1

两边除以

2

2 cos (θ) = - 1

两边除以

2

\frac{2 cos ( θ )}{2} = \frac{- 1}{2}

化简

\frac{2 cos ( θ )}{2} = \frac{- 1}{2}

化简

\frac{2 cos ( θ )}{2} : cos (θ)

\frac{2 cos ( θ )}{2}

约分:

2

= cos (θ)

化简

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

\frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

2 cos (θ) - 1 = 0 : θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

2 cos (θ) - 1 = 0

将

1

到右边

2 cos (θ) - 1 = 0

两边加上

1

2 cos (θ) - 1 + 1 = 0 + 1

化简

2 cos (θ) = 1

2 cos (θ) = 1

两边除以

2

2 cos (θ) = 1

两边除以

2

\frac{2 cos ( θ )}{2} = \frac{1}{2}

化简

\frac{2 cos ( θ )}{2} = \frac{1}{2}

化简

\frac{2 cos ( θ )}{2} : cos (θ)

\frac{2 cos ( θ )}{2}

约分:

2

= cos (θ)

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2}

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

合并所有解

θ = 2 πn, θ = π + 2 πn, θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

作图

流行的例子

sin(θ)+cos(θ)=0.46

sin (θ) + cos (θ) = 0.46

3cos(x)-4sin(x)=5

3 cos (x) - 4 sin (x) = 5

sqrt(2)cos(x-pi/4)-1=0

2 cos (x - \frac{π}{4}) - 1 = 0

0 = 3 cos (x)

tan(pi/2-x)+tan(x)-1=0

tan (\frac{π}{2} - x) + tan (x) - 1 = 0