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受欢迎的三角函数 >

cos(θ)csc(θ)=sin(θ)sec(θ)

解答

cos (θ) csc (θ) = sin (θ) sec (θ)

解答

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

+1

度数

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

cos (θ) csc (θ) = sin (θ) sec (θ)

两边减去

sin (θ) sec (θ)

cos (θ) csc (θ) - sin (θ) sec (θ) = 0

用 sin, cos 表示

cos (θ) csc (θ) - sec (θ) sin (θ)

使用基本三角恒等式:

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

= cos (θ) \frac{1}{sin ( θ )} - sec (θ) sin (θ)

使用基本三角恒等式:

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

= cos (θ) \frac{1}{sin ( θ )} - \frac{1}{cos ( θ )} sin (θ)

化简

cos (θ) \frac{1}{sin ( θ )} - \frac{1}{cos ( θ )} sin (θ) : \frac{cos ^{2} ( θ ) - sin ^{2} ( θ )}{sin ( θ ) cos ( θ )}

cos (θ) \frac{1}{sin ( θ )} - \frac{1}{cos ( θ )} sin (θ)

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

cos (θ) \frac{1}{sin ( θ )}

分式相乘:

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

= \frac{1 \cdot cos ( θ )}{sin ( θ )}

乘以:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

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

\frac{1}{cos ( θ )} sin (θ)

分式相乘:

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

= \frac{1 \cdot sin ( θ )}{cos ( θ )}

乘以:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ )}{cos ( θ )}

= \frac{cos ( θ )}{sin ( θ )} - \frac{sin ( θ )}{cos ( θ )}

sin (θ), cos (θ)

的最小公倍数:

sin (θ) cos (θ)

sin (θ), cos (θ)

最小公倍数 (LCM)

计算出由出现在

sin (θ)

或

cos (θ)

中的因子组成的表达式

= sin (θ) cos (θ)

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

sin (θ) cos (θ)

对于

\frac{cos ( θ )}{sin ( θ )} :

将分母和分子乘以

cos (θ)

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

对于

\frac{sin ( θ )}{cos ( θ )} :

将分母和分子乘以

sin (θ)

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

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

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

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

\frac{cos ^{2} ( θ ) - sin ^{2} ( θ )}{cos ( θ ) sin ( θ )} = 0

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

cos^{2} (θ) - sin^{2} (θ) = 0

使用三角恒等式改写

cos^{2} (θ) - sin^{2} (θ)

使用倍角公式:

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

= cos (2 θ)

cos (2 θ) = 0

cos (2 θ) = 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}

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

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

解

2 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{4} + πn

2 θ = \frac{π}{2} + 2 πn

两边除以

2

2 θ = \frac{π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

解

2 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{4} + πn

2 θ = \frac{3 π}{2} + 2 πn

两边除以

2

2 θ = \frac{3 π}{2} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

使用分式法则:

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

= \frac{3 π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

作图

流行的例子

64cos^2(θ)=49,(0<= ,θ<= 360)

64 cos^{2} (θ) = 49, (0^{\circ} \leq, θ \leq 36 0^{\circ})

solvefor y,x=4sec(y)

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

0 = 4 cos (3 θ)

sin(θ)= 3/5 ,sin(2θ)

sin (θ) = \frac{3}{5}, sin (2 θ)

solvefor x, 1/6 =cos^2(x)

so l v e f or x, \frac{1}{6} = cos^{2} (x)