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受欢迎的三角函数 >

cos^2(β)-cot^2(β)=cos^2(β)cot^2(β)

解答

cos^{2} (β) - cot^{2} (β) = cos^{2} (β) cot^{2} (β)

解答

β = \frac{π}{2} + πn

+1

度数

β = 9 0^{\circ} + 18 0^{\circ} n

求解步骤

cos^{2} (β) - cot^{2} (β) = cos^{2} (β) cot^{2} (β)

两边减去

cos^{2} (β) cot^{2} (β)

cos^{2} (β) - cot^{2} (β) - cos^{2} (β) cot^{2} (β) = 0

用 sin, cos 表示

cos^{2} (β) - cot^{2} (β) - cos^{2} (β) cot^{2} (β)

使用基本三角恒等式:

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

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

化简

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

cos^{2} (β) - (\frac{cos ( β )}{sin ( β )})^{2} - cos^{2} (β) (\frac{cos ( β )}{sin ( β )})^{2}

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

(\frac{cos ( β )}{sin ( β )})^{2}

使用指数法则:

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

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

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

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

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

(\frac{cos ( β )}{sin ( β )})^{2}

使用指数法则:

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

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

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

分式相乘:

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

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

cos^{2} (β) cos^{2} (β) = cos^{4} (β)

cos^{2} (β) cos^{2} (β)

使用指数法则:

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

cos^{2} (β) cos^{2} (β) = cos^{2 + 2} (β)

= cos^{2 + 2} (β)

数字相加:

2 + 2 = 4

= cos^{4} (β)

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

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

合并分式

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

使用法则

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

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

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

将项转换为分式:

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

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

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

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

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

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

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

分解

- cos^{2} (β) - cos^{4} (β) + cos^{2} (β) sin^{2} (β) : - cos^{2} (β) (1 + cos^{2} (β) - sin^{2} (β))

- cos^{2} (β) - cos^{4} (β) + cos^{2} (β) sin^{2} (β)

使用指数法则:

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

cos^{4} (β) = cos^{2} (β) cos^{2} (β)

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

因式分解出通项

- cos^{2} (β)

= - cos^{2} (β) (1 + cos^{2} (β) - sin^{2} (β))

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

分别求解每个部分

cos^{2} (β) = 0 or 1 + cos^{2} (β) - sin^{2} (β) = 0

cos^{2} (β) = 0 : β = \frac{π}{2} + 2 πn, β = \frac{3 π}{2} + 2 πn

cos^{2} (β) = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

cos (β) = 0

cos (β) = 0

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}

β = \frac{π}{2} + 2 πn, β = \frac{3 π}{2} + 2 πn

β = \frac{π}{2} + 2 πn, β = \frac{3 π}{2} + 2 πn

1 + cos^{2} (β) - sin^{2} (β) = 0 : β = \frac{π}{2} + πn

1 + cos^{2} (β) - sin^{2} (β) = 0

使用三角恒等式改写

1 + cos^{2} (β) - sin^{2} (β)

使用倍角公式:

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

= 1 + cos (2 β)

1 + cos (2 β) = 0

将

1

到右边

1 + cos (2 β) = 0

两边减去

1

1 + cos (2 β) - 1 = 0 - 1

化简

cos (2 β) = - 1

cos (2 β) = - 1

cos (2 β) = - 1

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}

2 β = π + 2 πn

2 β = π + 2 πn

解

2 β = π + 2 πn : β = \frac{π}{2} + πn

2 β = π + 2 πn

两边除以

2

2 β = π + 2 πn

两边除以

2

\frac{2 β}{2} = \frac{π}{2} + \frac{2 πn}{2}

化简

β = \frac{π}{2} + πn

β = \frac{π}{2} + πn

β = \frac{π}{2} + πn

合并所有解

β = \frac{π}{2} + 2 πn, β = \frac{3 π}{2} + 2 πn, β = \frac{π}{2} + πn

合并重叠的区间

β = \frac{π}{2} + πn

作图

流行的例子

cos^3(x)=cos(x),0<= x<2pi

cos^{3} (x) = cos (x), 0 \leq x < 2 π

sin (r) = 0.58

sec (α) = \frac{9}{5}

cos^2(x)+sin(x)= 5/4

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

(8.87)/(9.8)=sin(α)-0.1sqrt(1-sin^2(α))

\frac{8.87}{9.8} = sin (α) - 0.1 1 - sin^{2} (α)