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受欢迎的三角函数 >

tan(x)=sin(x)cos(x)

解答

tan (x) = sin (x) cos (x)

解答

x = 2 πn, x = π + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

tan (x) = sin (x) cos (x)

两边减去

sin (x) cos (x)

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

用 sin, cos 表示

tan (x) - cos (x) sin (x)

使用基本三角恒等式:

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

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

化简

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

\frac{sin ( x )}{cos ( x )} - cos (x) sin (x)

将项转换为分式:

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

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

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 ( x ) - cos ( x ) sin ( x ) cos ( x )}{cos ( x )}

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

sin (x) - cos (x) sin (x) cos (x)

cos (x) sin (x) cos (x) = cos^{2} (x) sin (x)

cos (x) sin (x) cos (x)

使用指数法则:

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

cos (x) cos (x) = cos^{1 + 1} (x)

= sin (x) cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin (x) cos^{2} (x)

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

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

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

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

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

sin (x) - cos^{2} (x) sin (x) = 0

分解

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

sin (x) - cos^{2} (x) sin (x)

因式分解出通项

- sin (x)

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

分解

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

cos^{2} (x) - 1

将

1

改写为

1^{2}

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

使用平方差公式:

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

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

= (cos (x) + 1) (cos (x) - 1)

= - sin (x) (cos (x) + 1) (cos (x) - 1)

- sin (x) (cos (x) + 1) (cos (x) - 1) = 0

分别求解每个部分

sin (x) = 0 or cos (x) + 1 = 0 or cos (x) - 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

将

1

到右边

cos (x) + 1 = 0

两边减去

1

cos (x) + 1 - 1 = 0 - 1

化简

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1

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}

x = π + 2 πn

x = π + 2 πn

cos (x) - 1 = 0 : x = 2 πn

cos (x) - 1 = 0

将

1

到右边

cos (x) - 1 = 0

两边加上

1

cos (x) - 1 + 1 = 0 + 1

化简

cos (x) = 1

cos (x) = 1

cos (x) = 1

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}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

合并所有解

x = 2 πn, x = π + 2 πn

作图

流行的例子

sqrt(3)sin(θ)-cos(θ)=1

3 sin (θ) - cos (θ) = 1

csc (y) = 2

16sin^2(θ)-1=0

16 sin^{2} (θ) - 1 = 0

8sin^2(x)+16sin(x)+8=0

8 sin^{2} (x) + 16 sin (x) + 8 = 0

tan(θ)=(2sqrt(3))/3 sin(θ),0<= θ<2pi

tan (θ) = \frac{2 3}{3} sin (θ), 0 \leq θ < 2 π