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受欢迎的三角函数 >

sec^2(a)cot(a)=8

解答

sec^{2} (a) cot (a) = 8

解答

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

+1

度数

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

求解步骤

sec^{2} (a) cot (a) = 8

两边减去

8

sec^{2} (a) cot (a) - 8 = 0

用 sin, cos 表示

- 8 + cot (a) sec^{2} (a)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

分式相乘:

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

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

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

(\frac{1}{cos ( a )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

乘

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

cos (a) \frac{1}{cos ^{2} ( a )}

分式相乘:

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

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

乘以:

1 \cdot cos (a) = cos (a)

= \frac{cos ( a )}{cos ^{2} ( a )}

约分:

cos (a)

= \frac{1}{cos ( a )}

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

使用分式法则:

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

= \frac{1}{cos ( a ) sin ( a )}

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

将项转换为分式:

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

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

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 ( a ) sin ( a ) + 1}{cos ( a ) sin ( a )}

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

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

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

1 - 8 cos (a) sin (a) = 0

使用三角恒等式改写

1 - 8 cos (a) sin (a)

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 a )}{2}

1 - 8 \cdot \frac{sin ( 2 a )}{2} = 0

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

8 \cdot \frac{sin ( 2 a )}{2}

分式相乘:

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

= \frac{sin ( 2 a ) \cdot 8}{2}

数字相除:

\frac{8}{2} = 4

= 4 sin (2 a)

1 - 4 sin (2 a) = 0

将

1

到右边

1 - 4 sin (2 a) = 0

两边减去

1

1 - 4 sin (2 a) - 1 = 0 - 1

化简

- 4 sin (2 a) = - 1

- 4 sin (2 a) = - 1

两边除以

- 4

- 4 sin (2 a) = - 1

两边除以

- 4

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

化简

sin (2 a) = \frac{1}{4}

sin (2 a) = \frac{1}{4}

使用反三角函数性质

sin (2 a) = \frac{1}{4}

sin (2 a) = \frac{1}{4}

的通解

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

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

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

解

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

2 a = arcsin (\frac{1}{4}) + 2 πn

两边除以

2

2 a = arcsin (\frac{1}{4}) + 2 πn

两边除以

2

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

化简

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

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

解

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

2 a = π - arcsin (\frac{1}{4}) + 2 πn

两边除以

2

2 a = π - arcsin (\frac{1}{4}) + 2 πn

两边除以

2

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

化简

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

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

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

以小数形式表示解

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

作图

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