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受欢迎的三角函数 >

(1+csc(γ))/(cot(γ)+cos(γ))=csc(γ)

解答

\frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} = csc (γ)

解答

γ = \frac{π}{4} + πn

+1

度数

γ = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

\frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} = csc (γ)

两边减去

csc (γ)

\frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} - csc (γ) = 0

化简

\frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} - csc (γ) : \frac{1 + csc ( γ ) - csc ( γ ) ( cot ( γ ) + cos ( γ ) )}{cot ( γ ) + cos ( γ )}

\frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} - csc (γ)

将项转换为分式:

csc (γ) = \frac{csc ( γ ) ( cot ( γ ) + cos ( γ ) )}{cot ( γ ) + cos ( γ )}

= \frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} - \frac{csc ( γ ) ( cot ( γ ) + cos ( γ ) )}{cot ( γ ) + cos ( γ )}

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 + csc ( γ ) - csc ( γ ) ( cot ( γ ) + cos ( γ ) )}{cot ( γ ) + cos ( γ )}

\frac{1 + csc ( γ ) - csc ( γ ) ( cot ( γ ) + cos ( γ ) )}{cot ( γ ) + cos ( γ )} = 0

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

1 + csc (γ) - csc (γ) (cot (γ) + cos (γ)) = 0

用 sin, cos 表示

1 + csc (γ) - (cos (γ) + cot (γ)) csc (γ)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

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

使用基本三角恒等式:

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

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

化简

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

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

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

(cos (γ) + \frac{cos ( γ )}{sin ( γ )}) \frac{1}{sin ( γ )}

分式相乘:

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

= \frac{1 \cdot ( cos ( γ ) + \frac{c o s ( γ )}{s i n ( γ )} )}{sin ( γ )}

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

1 \cdot (cos (γ) + \frac{cos ( γ )}{sin ( γ )})

乘以:

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

= (cos (γ) + \frac{cos ( γ )}{sin ( γ )})

去除括号:

(a) = a

= cos (γ) + \frac{cos ( γ )}{sin ( γ )}

= \frac{cos ( γ ) + \frac{c o s ( γ )}{s i n ( γ )}}{sin ( γ )}

化简

cos (γ) + \frac{cos ( γ )}{sin ( γ )} : \frac{cos ( γ ) sin ( γ ) + cos ( γ )}{sin ( γ )}

cos (γ) + \frac{cos ( γ )}{sin ( γ )}

将项转换为分式:

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

= \frac{cos ( γ ) sin ( γ )}{sin ( γ )} + \frac{cos ( γ )}{sin ( γ )}

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{cos ( γ ) sin ( γ ) + cos ( γ )}{sin ( γ )}

= \frac{\frac{c o s ( γ ) s i n ( γ ) + c o s ( γ )}{s i n ( γ )}}{sin ( γ )}

使用分式法则:

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

= \frac{cos ( γ ) sin ( γ ) + cos ( γ )}{sin ( γ ) sin ( γ )}

sin (γ) sin (γ) = sin^{2} (γ)

sin (γ) sin (γ)

使用指数法则:

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

sin (γ) sin (γ) = sin^{1 + 1} (γ)

= sin^{1 + 1} (γ)

数字相加:

1 + 1 = 2

= sin^{2} (γ)

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

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

将项转换为分式:

1 = \frac{1}{1}

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

1, sin (γ), sin^{2} (γ)

的最小公倍数:

sin^{2} (γ)

1, sin (γ), sin^{2} (γ)

最小公倍数 (LCM)

计算出由至少在以下一个因式表达式中出现的因子组成的表达式

= sin^{2} (γ)

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

sin^{2} (γ)

对于

\frac{1}{1} :

将分母和分子乘以

sin^{2} (γ)

\frac{1}{1} = \frac{1 \cdot sin ^{2} ( γ )}{1 \cdot sin ^{2} ( γ )} = \frac{sin ^{2} ( γ )}{sin ^{2} ( γ )}

对于

\frac{1}{sin ( γ )} :

将分母和分子乘以

sin (γ)

\frac{1}{sin ( γ )} = \frac{1 \cdot sin ( γ )}{sin ( γ ) sin ( γ )} = \frac{sin ( γ )}{sin ^{2} ( γ )}

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

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

- (cos (γ) sin (γ) + cos (γ)) : - cos (γ) sin (γ) - cos (γ)

- (cos (γ) sin (γ) + cos (γ))

打开括号

= - (cos (γ) sin (γ)) - (cos (γ))

使用加减运算法则

+ (- a) = - a

= - cos (γ) sin (γ) - cos (γ)

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

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

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

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

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

分解

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

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

因式分解出通项

cos (γ)

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

使用指数法则:

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

sin^{2} (γ) = sin (γ) sin (γ)

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

因式分解出通项

sin (γ)

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

因式分解出通项

(1 + sin (γ))

= (1 + sin (γ)) (- cos (γ) + sin (γ))

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

分别求解每个部分

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

1 + sin (γ) = 0 : γ = \frac{3 π}{2} + 2 πn

1 + sin (γ) = 0

将

1

到右边

1 + sin (γ) = 0

两边减去

1

1 + sin (γ) - 1 = 0 - 1

化简

sin (γ) = - 1

sin (γ) = - 1

sin (γ) = - 1

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}

γ = \frac{3 π}{2} + 2 πn

γ = \frac{3 π}{2} + 2 πn

- cos (γ) + sin (γ) = 0 : γ = \frac{π}{4} + πn

- cos (γ) + sin (γ) = 0

使用三角恒等式改写

- cos (γ) + sin (γ) = 0

在两边除以

cos (γ), cos (γ) \neq = 0

\frac{- cos ( γ ) + sin ( γ )}{cos ( γ )} = \frac{0}{cos ( γ )}

化简

- 1 + \frac{sin ( γ )}{cos ( γ )} = 0

使用基本三角恒等式:

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

- 1 + tan (γ) = 0

- 1 + tan (γ) = 0

将

1

到右边

- 1 + tan (γ) = 0

两边加上

1

- 1 + tan (γ) + 1 = 0 + 1

化简

tan (γ) = 1

tan (γ) = 1

tan (γ) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

γ = \frac{π}{4} + πn

γ = \frac{π}{4} + πn

合并所有解

γ = \frac{3 π}{2} + 2 πn, γ = \frac{π}{4} + πn

因为方程对以下值无定义:

\frac{3 π}{2} + 2 πn

γ = \frac{π}{4} + πn

作图

流行的例子

sin (θ) = \frac{2}{6}

3 sin (2 x) = - 2

solvefor x,arctan(y)=2arctan(x)

so l v e f or x, arctan (y) = 2 arctan (x)

sin(2x)=-sqrt(3/2)

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

tan(x)=(7.3)/(6.8)

tan (x) = \frac{7.3}{6.8}