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受欢迎的三角函数 >

8sin(θ)cos(θ)tan(θ)=csc(θ)

解答

8 sin (θ) cos (θ) tan (θ) = csc (θ)

解答

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

+1

度数

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

8 sin (θ) cos (θ) tan (θ) = csc (θ)

两边减去

csc (θ)

8 sin (θ) cos (θ) tan (θ) - csc (θ) = 0

用 sin, cos 表示

- csc (θ) + 8 cos (θ) sin (θ) tan (θ)

使用基本三角恒等式:

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

= - \frac{1}{sin ( θ )} + 8 cos (θ) sin (θ) tan (θ)

使用基本三角恒等式:

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

= - \frac{1}{sin ( θ )} + 8 cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

化简

- \frac{1}{sin ( θ )} + 8 cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )} : \frac{- 1 + 8 sin ^{3} ( θ )}{sin ( θ )}

- \frac{1}{sin ( θ )} + 8 cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

8 cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )} = 8 sin^{2} (θ)

8 cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

分式相乘:

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

= \frac{sin ( θ ) \cdot 8 cos ( θ ) sin ( θ )}{cos ( θ )}

约分:

cos (θ)

= sin (θ) \cdot 8 sin (θ)

使用指数法则:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= 8 sin^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= 8 sin^{2} (θ)

= - \frac{1}{sin ( θ )} + 8 sin^{2} (θ)

将项转换为分式:

8 sin^{2} (θ) = \frac{8 sin ^{2} ( θ ) sin ( θ )}{sin ( θ )}

= - \frac{1}{sin ( θ )} + \frac{8 sin ^{2} ( θ ) sin ( θ )}{sin ( θ )}

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 + 8 sin ^{2} ( θ ) sin ( θ )}{sin ( θ )}

- 1 + 8 sin^{2} (θ) sin (θ) = - 1 + 8 sin^{3} (θ)

- 1 + 8 sin^{2} (θ) sin (θ)

8 sin^{2} (θ) sin (θ) = 8 sin^{3} (θ)

8 sin^{2} (θ) sin (θ)

使用指数法则:

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

sin^{2} (θ) sin (θ) = sin^{2 + 1} (θ)

= 8 sin^{2 + 1} (θ)

数字相加:

2 + 1 = 3

= 8 sin^{3} (θ)

= - 1 + 8 sin^{3} (θ)

= \frac{- 1 + 8 sin ^{3} ( θ )}{sin ( θ )}

= \frac{- 1 + 8 sin ^{3} ( θ )}{sin ( θ )}

\frac{- 1 + 8 sin ^{3} ( θ )}{sin ( θ )} = 0

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

- 1 + 8 sin^{3} (θ) = 0

用替代法求解

- 1 + 8 sin^{3} (θ) = 0

令

: sin (θ) = u

- 1 + 8 u^{3} = 0

- 1 + 8 u^{3} = 0 : u = \frac{1}{2}, u = - \frac{1}{4} + i \frac{3}{4}, u = - \frac{1}{4} - i \frac{3}{4}

- 1 + 8 u^{3} = 0

将

1

到右边

- 1 + 8 u^{3} = 0

两边加上

1

- 1 + 8 u^{3} + 1 = 0 + 1

化简

8 u^{3} = 1

8 u^{3} = 1

两边除以

8

8 u^{3} = 1

两边除以

8

\frac{8 u ^{3}}{8} = \frac{1}{8}

化简

u^{3} = \frac{1}{8}

u^{3} = \frac{1}{8}

对于

x^{3} = f (a)

解为

x = 3 f (a), 3 f (a) \frac{- 1 - 3 i}{2}, 3 f (a) \frac{- 1 + 3 i}{2}

u = 3 \frac{1}{8}, u = 3 \frac{1}{8} \frac{- 1 + 3 i}{2}, u = 3 \frac{1}{8} \frac{- 1 - 3 i}{2}

3 \frac{1}{8} = \frac{1}{2}

3 \frac{1}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{3 1}{3 8}

38 = 2

38

因式分解数字:

8 = 2^{3}

= 3 2^{3}

使用根式运算法则:

n a^{n} = a

3 2^{3} = 2

= 2

= \frac{3 1}{2}

使用法则

n 1 = 1

31 = 1

= \frac{1}{2}

化简

3 \frac{1}{8} \frac{- 1 + 3 i}{2} : - \frac{1}{4} + i \frac{3}{4}

3 \frac{1}{8} \frac{- 1 + 3 i}{2}

3 \frac{1}{8} = \frac{1}{2}

3 \frac{1}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{3 1}{3 8}

38 = 2

38

因式分解数字:

8 = 2^{3}

= 3 2^{3}

使用根式运算法则:

n a^{n} = a

3 2^{3} = 2

= 2

= \frac{3 1}{2}

使用法则

n 1 = 1

31 = 1

= \frac{1}{2}

= \frac{1}{2} \cdot \frac{- 1 + 3 i}{2}

分式相乘:

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

= \frac{1 \cdot ( - 1 + 3 i )}{2 \cdot 2}

1 \cdot (- 1 + 3 i) = - 1 + 3 i

1 \cdot (- 1 + 3 i)

乘以:

1 \cdot (- 1 + 3 i) = (- 1 + 3 i)

= (- 1 + 3 i)

去除括号:

(- a) = - a

= - 1 + 3 i

= \frac{- 1 + 3 i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 1 + 3 i}{4}

将

\frac{- 1 + 3 i}{4}

改写成标准复数形式:

- \frac{1}{4} + \frac{3}{4} i

\frac{- 1 + 3 i}{4}

使用分式法则:

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

\frac{- 1 + 3 i}{4} = - \frac{1}{4} + \frac{3 i}{4}

= - \frac{1}{4} + \frac{3 i}{4}

= - \frac{1}{4} + \frac{3}{4} i

化简

3 \frac{1}{8} \frac{- 1 - 3 i}{2} : - \frac{1}{4} - i \frac{3}{4}

3 \frac{1}{8} \frac{- 1 - 3 i}{2}

3 \frac{1}{8} = \frac{1}{2}

3 \frac{1}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{3 1}{3 8}

38 = 2

38

因式分解数字:

8 = 2^{3}

= 3 2^{3}

使用根式运算法则:

n a^{n} = a

3 2^{3} = 2

= 2

= \frac{3 1}{2}

使用法则

n 1 = 1

31 = 1

= \frac{1}{2}

= \frac{1}{2} \cdot \frac{- 1 - 3 i}{2}

分式相乘:

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

= \frac{1 \cdot ( - 1 - 3 i )}{2 \cdot 2}

1 \cdot (- 1 - 3 i) = - 1 - 3 i

1 \cdot (- 1 - 3 i)

乘以:

1 \cdot (- 1 - 3 i) = (- 1 - 3 i)

= (- 1 - 3 i)

去除括号:

(- a) = - a

= - 1 - 3 i

= \frac{- 1 - 3 i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 1 - 3 i}{4}

将

\frac{- 1 - 3 i}{4}

改写成标准复数形式:

- \frac{1}{4} - \frac{3}{4} i

\frac{- 1 - 3 i}{4}

使用分式法则:

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

\frac{- 1 - 3 i}{4} = - \frac{1}{4} - \frac{3 i}{4}

= - \frac{1}{4} - \frac{3 i}{4}

= - \frac{1}{4} - \frac{3}{4} i

u = \frac{1}{2}, u = - \frac{1}{4} + i \frac{3}{4}, u = - \frac{1}{4} - i \frac{3}{4}

u = sin (θ)

代回

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{4} + i \frac{3}{4}, sin (θ) = - \frac{1}{4} - i \frac{3}{4}

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{4} + i \frac{3}{4}, sin (θ) = - \frac{1}{4} - i \frac{3}{4}

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

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}

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = - \frac{1}{4} + i \frac{3}{4} :

无解

sin (θ) = - \frac{1}{4} + i \frac{3}{4}

无解

sin (θ) = - \frac{1}{4} - i \frac{3}{4} :

无解

sin (θ) = - \frac{1}{4} - i \frac{3}{4}

无解

合并所有解

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

作图

流行的例子

(cot(θ))/(sec(θ))=sin(θ)

\frac{cot ( θ )}{sec ( θ )} = sin (θ)

16 sin (2 x) = 0

cos(2α)+sin(α)=0

cos (2 α) + sin (α) = 0

sin (9 0^{\circ} - x^{\circ}) = \frac{4}{5}

sin (x) = sin (8 0^{\circ})