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受欢迎的三角函数 >

1-tan^4(a)cos^4(a)=1-2sin^2(a)

解答

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

解答

a = 2 πn, a = π + 2 πn

+1

度数

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

1 - 2 sin^{2} (a)

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

分解

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

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

将

tan^{4} (a) cos^{4} (a)

改写为

(tan (a) cos (a))^{4}

tan^{4} (a) cos^{4} (a)

使用指数法则:

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

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

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

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

将

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

改写为

(2 sin (a))^{2} - ((tan (a) cos (a))^{2})^{2}

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

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

使用指数法则:

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

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

= (2)^{2} sin^{2} (a) - ((tan (a) cos (a))^{2})^{2}

使用指数法则:

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

(2)^{2} sin^{2} (a) = (2 sin (a))^{2}

= (2 sin (a))^{2} - ((tan (a) cos (a))^{2})^{2}

= (2 sin (a))^{2} - ((tan (a) cos (a))^{2})^{2}

使用平方差公式:

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

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

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

化简

(2 sin (a) + (tan (a) cos (a))^{2}) (2 sin (a) - (tan (a) cos (a))^{2}) : (2 sin (a) + tan^{2} (a) cos^{2} (a)) (2 sin (a) - tan^{2} (a) cos^{2} (a))

(2 sin (a) + (tan (a) cos (a))^{2}) (2 sin (a) - (tan (a) cos (a))^{2})

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

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

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

分别求解每个部分

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

2 sin (a) + tan^{2} (a) cos^{2} (a) = 0 : a = 2 πn, a = π + 2 πn

2 sin (a) + tan^{2} (a) cos^{2} (a) = 0

使用三角恒等式改写

cos^{2} (a) tan^{2} (a) + sin (a) 2

使用基本三角恒等式:

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

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

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

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

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

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

使用指数法则:

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

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

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

分式相乘:

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

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

约分:

cos^{2} (a)

= sin^{2} (a)

= sin^{2} (a) + 2 sin (a)

sin^{2} (a) + sin (a) 2 = 0

用替代法求解

sin^{2} (a) + sin (a) 2 = 0

令

: sin (a) = u

u^{2} + u 2 = 0

u^{2} + u 2 = 0 : u = 0, u = - 2

u^{2} + u 2 = 0

改写成标准形式

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

u^{2} + 2 u = 0

使用求根公式求解

u^{2} + 2 u = 0

二次方程求根公式:

若

a = 1, b = 2, c = 0

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

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

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

(2)^{2} - 4 \cdot 1 \cdot 0

(2)^{2} = 2

(2)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

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

使用指数法则:

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

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 2 - 0

数字相减:

2 - 0 = 2

= 2

u_{1, 2} = \frac{- 2 \pm 2}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 2 + 2}{2 \cdot 1}, u_{2} = \frac{- 2 - 2}{2 \cdot 1}

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

\frac{- 2 + 2}{2 \cdot 1}

同类项相加:

- 2 + 2 = 0

= \frac{0}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 2 - 2}{2 \cdot 1} : - 2

\frac{- 2 - 2}{2 \cdot 1}

同类项相加:

- 2 - 2 = - 22

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2 2}{2}

使用分式法则:

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

= - \frac{2 2}{2}

数字相除:

\frac{2}{2} = 1

= - 2

二次方程组的解是:

u = 0, u = - 2

u = sin (a)

代回

sin (a) = 0, sin (a) = - 2

sin (a) = 0, sin (a) = - 2

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

sin (a) = 0

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}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

解

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

sin (a) = - 2 :

无解

sin (a) = - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

a = 2 πn, a = π + 2 πn

2 sin (a) - tan^{2} (a) cos^{2} (a) = 0 : a = 2 πn, a = π + 2 πn

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

使用三角恒等式改写

- cos^{2} (a) tan^{2} (a) + sin (a) 2

使用基本三角恒等式:

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

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

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

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

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

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

使用指数法则:

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

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

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

分式相乘:

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

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

约分:

cos^{2} (a)

= sin^{2} (a)

= - sin^{2} (a) + 2 sin (a)

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

用替代法求解

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

令

: sin (a) = u

- u^{2} + u 2 = 0

- u^{2} + u 2 = 0 : u = 0, u = 2

- u^{2} + u 2 = 0

改写成标准形式

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

- u^{2} + 2 u = 0

使用求根公式求解

- u^{2} + 2 u = 0

二次方程求根公式:

若

a = - 1, b = 2, c = 0

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

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

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

(2)^{2} - 4 (- 1) \cdot 0

使用法则

- (- a) = a

= (2)^{2} + 4 \cdot 1 \cdot 0

(2)^{2} = 2

(2)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

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

使用指数法则:

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

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 2 + 0

数字相加:

2 + 0 = 2

= 2

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

将解分隔开

u_{1} = \frac{- 2 + 2}{2 ( - 1 )}, u_{2} = \frac{- 2 - 2}{2 ( - 1 )}

u = \frac{- 2 + 2}{2 ( - 1 )} : 0

\frac{- 2 + 2}{2 ( - 1 )}

去除括号:

(- a) = - a

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

同类项相加:

- 2 + 2 = 0

= \frac{0}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{- 2}

使用分式法则:

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

= - \frac{0}{2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 2 - 2}{2 ( - 1 )} : 2

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

去除括号:

(- a) = - a

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

同类项相加:

- 2 - 2 = - 22

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2 2}{- 2}

使用分式法则:

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

= \frac{2 2}{2}

数字相除:

\frac{2}{2} = 1

= 2

二次方程组的解是:

u = 0, u = 2

u = sin (a)

代回

sin (a) = 0, sin (a) = 2

sin (a) = 0, sin (a) = 2

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

sin (a) = 0

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}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

解

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

sin (a) = 2 :

无解

sin (a) = 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

a = 2 πn, a = π + 2 πn

合并所有解

a = 2 πn, a = π + 2 πn

作图

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