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受欢迎的三角函数 >

sec^2(θ)cot(θ)=8

解答

sec^{2} (θ) cot (θ) = 8

解答

θ = \frac{0.25268 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.25268 \dots}{2} + πn

+1

度数

θ = 7.23875 \dots^{\circ} + 18 0^{\circ} n, θ = 82.76124 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (θ) cot (θ) = 8

两边减去

8

sec^{2} (θ) cot (θ) - 8 = 0

用 sin, cos 表示

- 8 + cot (θ) sec^{2} (θ)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

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

\frac{cos ( θ )}{sin ( θ )} (\frac{1}{cos ( θ )})^{2}

分式相乘:

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

= \frac{cos ( θ ) ( \frac{1}{c o s ( θ )} ) ^{2}}{sin ( θ )}

(\frac{1}{cos ( θ )})^{2} = \frac{1}{cos ^{2} ( θ )}

(\frac{1}{cos ( θ )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= \frac{\frac{1}{c o s ^{2} ( θ )} cos ( θ )}{sin ( θ )}

乘

cos (θ) \frac{1}{cos ^{2} ( θ )} : \frac{1}{cos ( θ )}

cos (θ) \frac{1}{cos ^{2} ( θ )}

分式相乘:

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

= \frac{1 \cdot cos ( θ )}{cos ^{2} ( θ )}

乘以:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ )}{cos ^{2} ( θ )}

约分:

cos (θ)

= \frac{1}{cos ( θ )}

= \frac{\frac{1}{c o s ( θ )}}{sin ( θ )}

使用分式法则:

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

= \frac{1}{cos ( θ ) sin ( θ )}

= - 8 + \frac{1}{cos ( θ ) sin ( θ )}

将项转换为分式:

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

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

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

\frac{1 - 8 cos ( θ ) sin ( θ )}{cos ( θ ) sin ( θ )} = 0

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

1 - 8 cos (θ) sin (θ) = 0

使用三角恒等式改写

1 - 8 cos (θ) sin (θ)

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

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

= 1 - 8 \cdot \frac{sin ( 2 θ )}{2}

1 - 8 \cdot \frac{sin ( 2 θ )}{2} = 0

8 \cdot \frac{sin ( 2 θ )}{2} = 4 sin (2 θ)

8 \cdot \frac{sin ( 2 θ )}{2}

分式相乘:

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

= \frac{sin ( 2 θ ) \cdot 8}{2}

数字相除:

\frac{8}{2} = 4

= 4 sin (2 θ)

1 - 4 sin (2 θ) = 0

将

1

到右边

1 - 4 sin (2 θ) = 0

两边减去

1

1 - 4 sin (2 θ) - 1 = 0 - 1

化简

- 4 sin (2 θ) = - 1

- 4 sin (2 θ) = - 1

两边除以

- 4

- 4 sin (2 θ) = - 1

两边除以

- 4

\frac{- 4 sin ( 2 θ )}{- 4} = \frac{- 1}{- 4}

化简

sin (2 θ) = \frac{1}{4}

sin (2 θ) = \frac{1}{4}

使用反三角函数性质

sin (2 θ) = \frac{1}{4}

sin (2 θ) = \frac{1}{4}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 θ = arcsin (\frac{1}{4}) + 2 πn, 2 θ = π - arcsin (\frac{1}{4}) + 2 πn

2 θ = arcsin (\frac{1}{4}) + 2 πn, 2 θ = π - arcsin (\frac{1}{4}) + 2 πn

解

2 θ = arcsin (\frac{1}{4}) + 2 πn : θ = \frac{arcsin ( \frac{1}{4} )}{2} + πn

2 θ = arcsin (\frac{1}{4}) + 2 πn

两边除以

2

2 θ = arcsin (\frac{1}{4}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arcsin ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arcsin ( \frac{1}{4} )}{2} + πn

θ = \frac{arcsin ( \frac{1}{4} )}{2} + πn

解

2 θ = π - arcsin (\frac{1}{4}) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

2 θ = π - arcsin (\frac{1}{4}) + 2 πn

两边除以

2

2 θ = π - arcsin (\frac{1}{4}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

θ = \frac{arcsin ( \frac{1}{4} )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

以小数形式表示解

θ = \frac{0.25268 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.25268 \dots}{2} + πn

作图

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