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受欢迎的三角函数 >

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

解答

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

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

两边减去

\frac{1}{cos ( x )}

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} - \frac{1}{cos ( x )} = 0

化简

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

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

1 + sin (x), cos (x)

的最小公倍数:

cos (x) (sin (x) + 1)

1 + sin (x), cos (x)

最小公倍数 (LCM)

计算出由出现在

1 + sin (x)

或

cos (x)

中的因子组成的表达式

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

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos (x) (sin (x) + 1)

对于

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} :

将分母和分子乘以

cos (x)

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

对于

\frac{1}{cos ( x )} :

将分母和分子乘以

sin (x) + 1

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

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

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

乘开

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

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

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

乘开

cos (x) (tan (x) + cos (x)) : cos (x) tan (x) + cos^{2} (x)

cos (x) (tan (x) + cos (x))

使用分配律:

a (b + c) = ab + a c

a = cos (x), b = tan (x), c = cos (x)

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

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

- (sin (x) + 1) : - sin (x) - 1

- (sin (x) + 1)

打开括号

= - (sin (x)) - (1)

使用加减运算法则

+ (- a) = - a

= - sin (x) - 1

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

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

\frac{cos ( x ) tan ( x ) + cos ^{2} ( x ) - sin ( x ) - 1}{cos ( x ) ( sin ( x ) + 1 )} = 0

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

cos (x) tan (x) + cos^{2} (x) - sin (x) - 1 = 0

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

化简

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

- 1 + cos^{2} (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)

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

同类项相加:

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

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

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

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

用替代法求解

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

令

: cos (x) = u

- 1 + u^{2} = 0

- 1 + u^{2} = 0 : u = 1, u = - 1

- 1 + u^{2} = 0

将

1

到右边

- 1 + u^{2} = 0

两边加上

1

- 1 + u^{2} + 1 = 0 + 1

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = cos (x)

代回

cos (x) = 1, cos (x) = - 1

cos (x) = 1, cos (x) = - 1

cos (x) = 1 : x = 2 πn

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

cos (x) = - 1 : x = π + 2 πn

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

合并所有解

x = 2 πn, x = π + 2 πn

作图

流行的例子

sin(θ)= 4/5 ,0<= θ<= pi/2

sin (θ) = \frac{4}{5}, 0 \leq θ \leq \frac{π}{2}

((tan(x)+2sin(x)))/((tan(x)-2sin(x)))=3

\frac{( tan ( x ) + 2 sin ( x ) )}{( tan ( x ) - 2 sin ( x ) )} = 3

1 - sin (2 x) = 0

2cos(3x)+cos(2x)+1=0

2 cos (3 x) + cos (2 x) + 1 = 0

sin(x)=sin(pi/5)

sin (x) = sin (\frac{π}{5})