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受欢迎的三角函数 >

sin(x-pi/2)tan(x)=sin(x)

解答

sin (x - \frac{π}{2}) tan (x) = sin (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

sin (x - \frac{π}{2}) tan (x) = sin (x)

使用三角恒等式改写

sin (x - \frac{π}{2}) tan (x) = sin (x)

使用三角恒等式改写

sin (x - \frac{π}{2})

使用角差恒等式:

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

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

化简

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

sin (x) cos (\frac{π}{2}) - cos (x) sin (\frac{π}{2})

sin (x) cos (\frac{π}{2}) = 0

sin (x) cos (\frac{π}{2})

化简

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

使用以下普通恒等式:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot sin (x)

使用法则

0 \cdot a = 0

= 0

cos (x) sin (\frac{π}{2}) = cos (x)

cos (x) sin (\frac{π}{2})

化简

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

使用以下普通恒等式:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot cos (x)

乘以:

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

= cos (x)

= 0 - cos (x)

0 - cos (x) = - cos (x)

= - cos (x)

= - cos (x)

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

化简

(- cos (x)) tan (x) : - cos (x) tan (x)

(- cos (x)) tan (x)

去除括号:

(- a) = - a

= - cos (x) tan (x)

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

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

两边减去

sin (x)

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

分式相乘:

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

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

约分:

cos (x)

= sin (x)

= - sin (x) - sin (x)

同类项相加:

- sin (x) - sin (x) = - 2 sin (x)

= - 2 sin (x)

= - 2 sin (x)

- 2 sin (x) = 0

两边除以

- 2

- 2 sin (x) = 0

两边除以

- 2

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

化简

sin (x) = 0

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

作图

流行的例子

8 sin (x) + 12 = 16

cos (7 x) = - 1

8sin(t)+8cos(t)=8

8 sin (t) + 8 cos (t) = 8

5 sin (\frac{π}{3} x) = 2

4cos^2(x)-5cos(x)+1=0

4 cos^{2} (x) - 5 cos (x) + 1 = 0