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受欢迎的三角函数 >

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

解答

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

解答

x = 2 πn

+1

度数

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

求解步骤

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

两边减去

1

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

分式相乘:

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

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

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

sin (x) sin (x)

使用指数法则:

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

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

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

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

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

将项转换为分式:

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

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

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

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

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

1 \cdot cos (x)

乘以:

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

= cos (x)

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

cos (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)

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= - cos (x) + 1

- 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

两边除以

- 1

- cos (x) = - 1

两边除以

- 1

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

作图

流行的例子

6 sin (t) = 0

(500*9.8*l(1-cos(θ)))=0.018

(500 \cdot 9.8 \cdot l (1 - cos (θ))) = 0.018

sec(x)=(sin(x))/(cos(x))

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

-cos(20.02t)=sin(20t)

- cos (20.02 t) = sin (20 t)

sin^3(x)+3sin^2(x)+2sin(x)=0,0<= x<2pi

sin^{3} (x) + 3 sin^{2} (x) + 2 sin (x) = 0, 0 \leq x < 2 π