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受欢迎的三角函数 >

cot^2(t)csc(t)=sin(t)

解答

cot^{2} (t) csc (t) = sin (t)

解答

t = 2.23703 \dots + 2 πn, t = - 2.23703 \dots + 2 πn, t = 0.90455 \dots + 2 πn, t = 2 π - 0.90455 \dots + 2 πn

+1

度数

t = 128.17270 \dots^{\circ} + 36 0^{\circ} n, t = - 128.17270 \dots^{\circ} + 36 0^{\circ} n, t = 51.82729 \dots^{\circ} + 36 0^{\circ} n, t = 308.17270 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cot^{2} (t) csc (t) = sin (t)

两边减去

sin (t)

cot^{2} (t) csc (t) - sin (t) = 0

用 sin, cos 表示

- sin (t) + cot^{2} (t) csc (t)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

- sin (t) + (\frac{cos ( t )}{sin ( t )})^{2} \frac{1}{sin ( t )} : \frac{- sin ^{4} ( t ) + cos ^{2} ( t )}{sin ^{3} ( t )}

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

(\frac{cos ( t )}{sin ( t )})^{2} \frac{1}{sin ( t )} = \frac{cos ^{2} ( t )}{sin ^{3} ( t )}

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

分式相乘:

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

= \frac{1 \cdot ( \frac{c o s ( t )}{s i n ( t )} ) ^{2}}{sin ( t )}

乘以:

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

= \frac{( \frac{c o s ( t )}{s i n ( t )} ) ^{2}}{sin ( t )}

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

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

使用指数法则:

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

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

= \frac{\frac{c o s ^{2} ( t )}{s i n ^{2} ( t )}}{sin ( t )}

使用分式法则:

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

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

sin^{2} (t) sin (t) = sin^{3} (t)

sin^{2} (t) sin (t)

使用指数法则:

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

sin^{2} (t) sin (t) = sin^{2 + 1} (t)

= sin^{2 + 1} (t)

数字相加:

2 + 1 = 3

= sin^{3} (t)

= \frac{cos ^{2} ( t )}{sin ^{3} ( t )}

= - sin (t) + \frac{cos ^{2} ( t )}{sin ^{3} ( t )}

将项转换为分式:

sin (t) = \frac{sin ( t ) sin ^{3} ( t )}{sin ^{3} ( t )}

= - \frac{sin ( t ) sin ^{3} ( t )}{sin ^{3} ( t )} + \frac{cos ^{2} ( t )}{sin ^{3} ( t )}

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 ( t ) sin ^{3} ( t ) + cos ^{2} ( t )}{sin ^{3} ( t )}

- sin (t) sin^{3} (t) + cos^{2} (t) = - sin^{4} (t) + cos^{2} (t)

- sin (t) sin^{3} (t) + cos^{2} (t)

sin (t) sin^{3} (t) = sin^{4} (t)

sin (t) sin^{3} (t)

使用指数法则:

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

sin (t) sin^{3} (t) = sin^{1 + 3} (t)

= sin^{1 + 3} (t)

数字相加:

1 + 3 = 4

= sin^{4} (t)

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

= \frac{- sin ^{4} ( t ) + cos ^{2} ( t )}{sin ^{3} ( t )}

= \frac{- sin ^{4} ( t ) + cos ^{2} ( t )}{sin ^{3} ( t )}

\frac{cos ^{2} ( t ) - sin ^{4} ( t )}{sin ^{3} ( t )} = 0

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

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

分解

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

cos^{2} (t) - sin^{4} (t)

使用指数法则:

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

sin^{4} (t) = (sin^{2} (t))^{2}

= cos^{2} (t) - (sin^{2} (t))^{2}

使用平方差公式:

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

cos^{2} (t) - (sin^{2} (t))^{2} = (cos (t) + sin^{2} (t)) (cos (t) - sin^{2} (t))

= (cos (t) + sin^{2} (t)) (cos (t) - sin^{2} (t))

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

分别求解每个部分

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

cos (t) + sin^{2} (t) = 0 : t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (t) + sin^{2} (t) = 0

使用三角恒等式改写

cos (t) + sin^{2} (t)

使用毕达哥拉斯恒等式:

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

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

= cos (t) + 1 - cos^{2} (t)

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

用替代法求解

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

令

: cos (t) = u

1 + u - u^{2} = 0

1 + u - u^{2} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

1 + u - u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 4 (- 1) \cdot 1 = 5

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 ( - 1 )}, u_{2} = \frac{- 1 - 5}{2 ( - 1 )}

u = \frac{- 1 + 5}{2 ( - 1 )} : - \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

使用分式法则:

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

= - \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 ( - 1 )} : \frac{1 + 5}{2}

\frac{- 1 - 5}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - 5}{- 2}

使用分式法则:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

二次方程组的解是:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

u = cos (t)

代回

cos (t) = - \frac{- 1 + 5}{2}, cos (t) = \frac{1 + 5}{2}

cos (t) = - \frac{- 1 + 5}{2}, cos (t) = \frac{1 + 5}{2}

cos (t) = - \frac{- 1 + 5}{2} : t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (t) = - \frac{- 1 + 5}{2}

使用反三角函数性质

cos (t) = - \frac{- 1 + 5}{2}

cos (t) = - \frac{- 1 + 5}{2}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (t) = \frac{1 + 5}{2} :

无解

cos (t) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (t) - sin^{2} (t) = 0 : t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (t) - sin^{2} (t) = 0

使用三角恒等式改写

cos (t) - sin^{2} (t)

使用毕达哥拉斯恒等式:

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

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

= cos (t) - (1 - cos^{2} (t))

- (1 - cos^{2} (t)) : - 1 + cos^{2} (t)

- (1 - cos^{2} (t))

打开括号

= - (1) - (- cos^{2} (t))

使用加减运算法则

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

= - 1 + cos^{2} (t)

= cos (t) - 1 + cos^{2} (t)

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

用替代法求解

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

令

: cos (t) = u

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

改写成标准形式

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

u^{2} + u - 1 = 0

使用求根公式求解

u^{2} + u - 1 = 0

二次方程求根公式:

若

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

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

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

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

二次方程组的解是:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

u = cos (t)

代回

cos (t) = \frac{- 1 + 5}{2}, cos (t) = \frac{- 1 - 5}{2}

cos (t) = \frac{- 1 + 5}{2}, cos (t) = \frac{- 1 - 5}{2}

cos (t) = \frac{- 1 + 5}{2} : t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (t) = \frac{- 1 + 5}{2}

使用反三角函数性质

cos (t) = \frac{- 1 + 5}{2}

cos (t) = \frac{- 1 + 5}{2}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (t) = \frac{- 1 - 5}{2} :

无解

cos (t) = \frac{- 1 - 5}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

合并所有解

t = arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = - arccos (- \frac{- 1 + 5}{2}) + 2 πn, t = arccos (\frac{- 1 + 5}{2}) + 2 πn, t = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

以小数形式表示解

t = 2.23703 \dots + 2 πn, t = - 2.23703 \dots + 2 πn, t = 0.90455 \dots + 2 πn, t = 2 π - 0.90455 \dots + 2 πn

作图

流行的例子

\frac{6}{12} = cos (θ)

sin(a)=(6.88)/(6.9)

sin (a) = \frac{6.88}{6.9}

sin (x) = cos (3 2^{\circ})

solvefor x,y=cos(4x)

so l v e f or x, y = cos (4 x)

-0.002=sin(2pi*(3)-500pi(0)+x)

- 0.002 = sin (2 π \cdot (3) - 500 π (0) + x)