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受欢迎的三角函数 >

sin^2(a)+1/(sec(a))= 5/4

解答

sin^{2} (a) + \frac{1}{sec ( a )} = \frac{5}{4}

解答

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

+1

度数

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

sin^{2} (a) + \frac{1}{sec ( a )} = \frac{5}{4}

两边减去

\frac{5}{4}

sin^{2} (a) + \frac{1}{sec ( a )} - \frac{5}{4} = 0

化简

sin^{2} (a) + \frac{1}{sec ( a )} - \frac{5}{4} : \frac{4 sin ^{2} ( a ) sec ( a ) + 4 - 5 sec ( a )}{4 sec ( a )}

sin^{2} (a) + \frac{1}{sec ( a )} - \frac{5}{4}

将项转换为分式:

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

= \frac{sin ^{2} ( a )}{1} + \frac{1}{sec ( a )} - \frac{5}{4}

1, sec (a), 4

的最小公倍数:

4 sec (a)

1, sec (a), 4

最小公倍数 (LCM)

1, 4

的最小公倍数:

4

1, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

1

或

4

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

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

= 4 sec (a)

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

4 sec (a)

对于

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

将分母和分子乘以

4 sec (a)

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

对于

\frac{1}{sec ( a )} :

将分母和分子乘以

4

\frac{1}{sec ( a )} = \frac{1 \cdot 4}{sec ( a ) \cdot 4} = \frac{4}{4 sec ( a )}

对于

\frac{5}{4} :

将分母和分子乘以

sec (a)

\frac{5}{4} = \frac{5 sec ( a )}{4 sec ( a )}

= \frac{sin ^{2} ( a ) \cdot 4 sec ( a )}{4 sec ( a )} + \frac{4}{4 sec ( a )} - \frac{5 sec ( a )}{4 sec ( 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{sin ^{2} ( a ) \cdot 4 sec ( a ) + 4 - 5 sec ( a )}{4 sec ( a )}

\frac{4 sin ^{2} ( a ) sec ( a ) + 4 - 5 sec ( a )}{4 sec ( a )} = 0

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

4 sin^{2} (a) sec (a) + 4 - 5 sec (a) = 0

用 sin, cos 表示

4 sin^{2} (a) \frac{1}{cos ( a )} + 4 - 5 \cdot \frac{1}{cos ( a )} = 0

化简

4 sin^{2} (a) \frac{1}{cos ( a )} + 4 - 5 \cdot \frac{1}{cos ( a )} : \frac{4 sin ^{2} ( a ) - 5 + 4 cos ( a )}{cos ( a )}

4 sin^{2} (a) \frac{1}{cos ( a )} + 4 - 5 \cdot \frac{1}{cos ( a )}

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

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

分式相乘:

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

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

数字相乘:

1 \cdot 4 = 4

= \frac{4 sin ^{2} ( a )}{cos ( a )}

5 \cdot \frac{1}{cos ( a )} = \frac{5}{cos ( a )}

5 \cdot \frac{1}{cos ( a )}

分式相乘:

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

= \frac{1 \cdot 5}{cos ( a )}

数字相乘:

1 \cdot 5 = 5

= \frac{5}{cos ( a )}

= \frac{4 sin ^{2} ( a )}{cos ( a )} + 4 - \frac{5}{cos ( a )}

合并分式

\frac{4 sin ^{2} ( a )}{cos ( a )} - \frac{5}{cos ( a )} : \frac{4 sin ^{2} ( a ) - 5}{cos ( a )}

使用法则

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

= \frac{4 sin ^{2} ( a ) - 5}{cos ( a )}

= \frac{4 sin ^{2} ( a ) - 5}{cos ( a )} + 4

将项转换为分式:

4 = \frac{4 cos ( a )}{cos ( a )}

= \frac{4 sin ^{2} ( a ) - 5}{cos ( a )} + \frac{4 cos ( a )}{cos ( 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{4 sin ^{2} ( a ) - 5 + 4 cos ( a )}{cos ( a )}

\frac{4 sin ^{2} ( a ) - 5 + 4 cos ( a )}{cos ( a )} = 0

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

4 sin^{2} (a) - 5 + 4 cos (a) = 0

两边减去

4 cos (a)

4 sin^{2} (a) - 5 = - 4 cos (a)

两边进行平方

(4 sin^{2} (a) - 5)^{2} = (- 4 cos (a))^{2}

两边减去

(- 4 cos (a))^{2}

(4 sin^{2} (a) - 5)^{2} - 16 cos^{2} (a) = 0

分解

(4 sin^{2} (a) - 5)^{2} - 16 cos^{2} (a) : (4 sin^{2} (a) - 5 + 4 cos (a)) (4 sin^{2} (a) - 5 - 4 cos (a))

(4 sin^{2} (a) - 5)^{2} - 16 cos^{2} (a)

将

(4 sin^{2} (a) - 5)^{2} - 16 cos^{2} (a)

改写为

(4 sin^{2} (a) - 5)^{2} - (4 cos (a))^{2}

(4 sin^{2} (a) - 5)^{2} - 16 cos^{2} (a)

将

16

改写为

4^{2}

= (4 sin^{2} (a) - 5)^{2} - 4^{2} cos^{2} (a)

使用指数法则:

a^{m} b^{m} = (ab)^{m}

4^{2} cos^{2} (a) = (4 cos (a))^{2}

= (4 sin^{2} (a) - 5)^{2} - (4 cos (a))^{2}

= (4 sin^{2} (a) - 5)^{2} - (4 cos (a))^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(4 sin^{2} (a) - 5)^{2} - (4 cos (a))^{2} = ((4 sin^{2} (a) - 5) + 4 cos (a)) ((4 sin^{2} (a) - 5) - 4 cos (a))

= ((4 sin^{2} (a) - 5) + 4 cos (a)) ((4 sin^{2} (a) - 5) - 4 cos (a))

整理后得

= (4 sin^{2} (a) + 4 cos (a) - 5) (4 sin^{2} (a) - 4 cos (a) - 5)

(4 sin^{2} (a) - 5 + 4 cos (a)) (4 sin^{2} (a) - 5 - 4 cos (a)) = 0

分别求解每个部分

4 sin^{2} (a) - 5 + 4 cos (a) = 0 or 4 sin^{2} (a) - 5 - 4 cos (a) = 0

4 sin^{2} (a) - 5 + 4 cos (a) = 0 : a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

4 sin^{2} (a) - 5 + 4 cos (a) = 0

使用三角恒等式改写

- 5 + 4 cos (a) + 4 sin^{2} (a)

使用毕达哥拉斯恒等式:

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

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

= - 5 + 4 cos (a) + 4 (1 - cos^{2} (a))

化简

- 5 + 4 cos (a) + 4 (1 - cos^{2} (a)) : 4 cos (a) - 4 cos^{2} (a) - 1

- 5 + 4 cos (a) + 4 (1 - cos^{2} (a))

乘开

4 (1 - cos^{2} (a)) : 4 - 4 cos^{2} (a)

4 (1 - cos^{2} (a))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (a)

= 4 \cdot 1 - 4 cos^{2} (a)

数字相乘:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (a)

= - 5 + 4 cos (a) + 4 - 4 cos^{2} (a)

化简

- 5 + 4 cos (a) + 4 - 4 cos^{2} (a) : 4 cos (a) - 4 cos^{2} (a) - 1

- 5 + 4 cos (a) + 4 - 4 cos^{2} (a)

对同类项分组

= 4 cos (a) - 4 cos^{2} (a) - 5 + 4

数字相加/相减:

- 5 + 4 = - 1

= 4 cos (a) - 4 cos^{2} (a) - 1

= 4 cos (a) - 4 cos^{2} (a) - 1

= 4 cos (a) - 4 cos^{2} (a) - 1

- 1 + 4 cos (a) - 4 cos^{2} (a) = 0

用替代法求解

- 1 + 4 cos (a) - 4 cos^{2} (a) = 0

令

: cos (a) = u

- 1 + 4 u - 4 u^{2} = 0

- 1 + 4 u - 4 u^{2} = 0 : u = \frac{1}{2}

- 1 + 4 u - 4 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

- 4 u^{2} + 4 u - 1 = 0

使用求根公式求解

- 4 u^{2} + 4 u - 1 = 0

二次方程求根公式:

若

a = - 4, b = 4, c = - 1

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 4 ) ( - 1 )}{2 ( - 4 )}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 4 ) ( - 1 )}{2 ( - 4 )}

4^{2} - 4 (- 4) (- 1) = 0

4^{2} - 4 (- 4) (- 1)

使用法则

- (- a) = a

= 4^{2} - 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- 4 \pm 0}{2 ( - 4 )}

u = \frac{- 4}{2 ( - 4 )}

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

\frac{- 4}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 4}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4}{8}

约分:

4

= \frac{1}{2}

u = \frac{1}{2}

二次方程组的解是:

u = \frac{1}{2}

u = cos (a)

代回

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

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

cos (a) = \frac{1}{2} : a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

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

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

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}

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

合并所有解

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

4 sin^{2} (a) - 5 - 4 cos (a) = 0 : a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn

4 sin^{2} (a) - 5 - 4 cos (a) = 0

使用三角恒等式改写

- 5 - 4 cos (a) + 4 sin^{2} (a)

使用毕达哥拉斯恒等式:

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

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

= - 5 - 4 cos (a) + 4 (1 - cos^{2} (a))

化简

- 5 - 4 cos (a) + 4 (1 - cos^{2} (a)) : - 4 cos^{2} (a) - 4 cos (a) - 1

- 5 - 4 cos (a) + 4 (1 - cos^{2} (a))

乘开

4 (1 - cos^{2} (a)) : 4 - 4 cos^{2} (a)

4 (1 - cos^{2} (a))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (a)

= 4 \cdot 1 - 4 cos^{2} (a)

数字相乘:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (a)

= - 5 - 4 cos (a) + 4 - 4 cos^{2} (a)

化简

- 5 - 4 cos (a) + 4 - 4 cos^{2} (a) : - 4 cos^{2} (a) - 4 cos (a) - 1

- 5 - 4 cos (a) + 4 - 4 cos^{2} (a)

对同类项分组

= - 4 cos (a) - 4 cos^{2} (a) - 5 + 4

数字相加/相减:

- 5 + 4 = - 1

= - 4 cos^{2} (a) - 4 cos (a) - 1

= - 4 cos^{2} (a) - 4 cos (a) - 1

= - 4 cos^{2} (a) - 4 cos (a) - 1

- 1 - 4 cos (a) - 4 cos^{2} (a) = 0

用替代法求解

- 1 - 4 cos (a) - 4 cos^{2} (a) = 0

令

: cos (a) = u

- 1 - 4 u - 4 u^{2} = 0

- 1 - 4 u - 4 u^{2} = 0 : u = - \frac{1}{2}

- 1 - 4 u - 4 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

- 4 u^{2} - 4 u - 1 = 0

使用求根公式求解

- 4 u^{2} - 4 u - 1 = 0

二次方程求根公式:

若

a = - 4, b = - 4, c = - 1

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 4 ) ( - 1 )}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 4 ) ( - 1 )}{2 ( - 4 )}

(- 4)^{2} - 4 (- 4) (- 1) = 0

(- 4)^{2} - 4 (- 4) (- 1)

使用法则

- (- a) = a

= (- 4)^{2} - 4 \cdot 4 \cdot 1

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 ( - 4 )}

u = \frac{- ( - 4 )}{2 ( - 4 )}

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

\frac{- ( - 4 )}{2 ( - 4 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{4}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{4}{- 8}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{4}{8}

约分:

4

= - \frac{1}{2}

u = - \frac{1}{2}

二次方程组的解是:

u = - \frac{1}{2}

u = cos (a)

代回

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

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

cos (a) = - \frac{1}{2} : a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn

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

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

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}

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

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

合并所有解

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

合并所有解

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn, a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn

将解代入原方程进行验证

将它们代入

sin^{2} (a) + \frac{1}{sec ( a )} = \frac{5}{4}

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{3} + 2 πn

的解:

真

\frac{π}{3} + 2 πn

代入

n = 1

\frac{π}{3} + 2 π 1

对于

sin^{2} (a) + \frac{1}{sec ( a )} = \frac{5}{4}

代入

a = \frac{π}{3} + 2 π 1

sin^{2} (\frac{π}{3} + 2 π 1) + \frac{1}{sec ( \frac{π}{3} + 2 π 1 )} = \frac{5}{4}

整理后得

1.25 = 1.25

\Rightarrow 真

检验

\frac{5 π}{3} + 2 πn

的解:

真

\frac{5 π}{3} + 2 πn

代入

n = 1

\frac{5 π}{3} + 2 π 1

对于

sin^{2} (a) + \frac{1}{sec ( a )} = \frac{5}{4}

代入

a = \frac{5 π}{3} + 2 π 1

sin^{2} (\frac{5 π}{3} + 2 π 1) + \frac{1}{sec ( \frac{5 π}{3} + 2 π 1 )} = \frac{5}{4}

整理后得

1.25 = 1.25

\Rightarrow 真

检验

\frac{2 π}{3} + 2 πn

的解:

假

\frac{2 π}{3} + 2 πn

代入

n = 1

\frac{2 π}{3} + 2 π 1

对于

sin^{2} (a) + \frac{1}{sec ( a )} = \frac{5}{4}

代入

a = \frac{2 π}{3} + 2 π 1

sin^{2} (\frac{2 π}{3} + 2 π 1) + \frac{1}{sec ( \frac{2 π}{3} + 2 π 1 )} = \frac{5}{4}

整理后得

0.25 = 1.25

\Rightarrow 假

检验

\frac{4 π}{3} + 2 πn

的解:

假

\frac{4 π}{3} + 2 πn

代入

n = 1

\frac{4 π}{3} + 2 π 1

对于

sin^{2} (a) + \frac{1}{sec ( a )} = \frac{5}{4}

代入

a = \frac{4 π}{3} + 2 π 1

sin^{2} (\frac{4 π}{3} + 2 π 1) + \frac{1}{sec ( \frac{4 π}{3} + 2 π 1 )} = \frac{5}{4}

整理后得

0.25 = 1.25

\Rightarrow 假

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

作图

流行的例子

(csc(x))/(cot(x))=sqrt(2)

\frac{csc ( x )}{cot ( x )} = 2

sin(4x)+6cos(2x)=0

sin (4 x) + 6 cos (2 x) = 0

sec(a)= 41/12 ,2pi<a<(5pi)/2 ,sin(a/2)

sec (a) = \frac{41}{12}, 2 π < a < \frac{5 π}{2}, sin (\frac{a}{2})

sin (2 x) = 0.3

sin (2 x) = 0.6