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受欢迎的三角函数 >

tan(θ)=sqrt(2)sin(θ)

解答

tan (θ) = 2 sin (θ)

解答

θ = 2 πn, θ = π + 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 = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

tan (θ) = 2 sin (θ)

两边减去

2 sin (θ)

tan (θ) - 2 sin (θ) = 0

用 sin, cos 表示

tan (θ) - sin (θ) 2

使用基本三角恒等式:

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

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

化简

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

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

将项转换为分式:

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

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

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

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

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

sin (θ) - cos (θ) sin (θ) 2 = 0

分解

sin (θ) - cos (θ) sin (θ) 2 : sin (θ) (1 - 2 cos (θ))

sin (θ) - cos (θ) sin (θ) 2

因式分解出通项

sin (θ)

= sin (θ) (1 - cos (θ) 2)

sin (θ) (1 - 2 cos (θ)) = 0

分别求解每个部分

sin (θ) = 0 or 1 - 2 cos (θ) = 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

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

1 - 2 cos (θ) = 0

将

1

到右边

1 - 2 cos (θ) = 0

两边减去

1

1 - 2 cos (θ) - 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}

使用分式法则:

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

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

作图

流行的例子

0=-sin(x)+cos(x)

0 = - sin (x) + cos (x)

sin (3 θ) = \frac{1}{2}

-sin^2(θ)+cos^2(θ)=1-sin(θ)

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

1-sin(θ)=2cos^2(θ)

1 - sin (θ) = 2 cos^{2} (θ)

2sin^2(θ)-sqrt(3)sin(θ)=0

2 sin^{2} (θ) - 3 sin (θ) = 0