zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

10tan(θ)sec(θ)=10cot(θ)csc(θ)

解答

10 tan (θ) sec (θ) = 10 cot (θ) csc (θ)

解答

θ = \frac{π}{4} + πn

+1

度数

θ = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

10 tan (θ) sec (θ) = 10 cot (θ) csc (θ)

两边减去

10 cot (θ) csc (θ)

10 tan (θ) sec (θ) - 10 cot (θ) csc (θ) = 0

用 sin, cos 表示

- 10 cot (θ) csc (θ) + 10 sec (θ) tan (θ)

使用基本三角恒等式:

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

= - 10 \cdot \frac{cos ( θ )}{sin ( θ )} csc (θ) + 10 sec (θ) tan (θ)

使用基本三角恒等式:

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

= - 10 \cdot \frac{cos ( θ )}{sin ( θ )} \cdot \frac{1}{sin ( θ )} + 10 sec (θ) tan (θ)

使用基本三角恒等式:

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

= - 10 \cdot \frac{cos ( θ )}{sin ( θ )} \cdot \frac{1}{sin ( θ )} + 10 \cdot \frac{1}{cos ( θ )} tan (θ)

使用基本三角恒等式:

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

= - 10 \cdot \frac{cos ( θ )}{sin ( θ )} \cdot \frac{1}{sin ( θ )} + 10 \cdot \frac{1}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )}

化简

- 10 \cdot \frac{cos ( θ )}{sin ( θ )} \cdot \frac{1}{sin ( θ )} + 10 \cdot \frac{1}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )} : \frac{- 10 cos ^{3} ( θ ) + 10 sin ^{3} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

- 10 \cdot \frac{cos ( θ )}{sin ( θ )} \cdot \frac{1}{sin ( θ )} + 10 \cdot \frac{1}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )}

10 \cdot \frac{cos ( θ )}{sin ( θ )} \cdot \frac{1}{sin ( θ )} = \frac{10 cos ( θ )}{sin ^{2} ( θ )}

10 \cdot \frac{cos ( θ )}{sin ( θ )} \cdot \frac{1}{sin ( θ )}

分式相乘:

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

= \frac{cos ( θ ) \cdot 1 \cdot 10}{sin ( θ ) sin ( θ )}

数字相乘:

1 \cdot 10 = 10

= \frac{10 cos ( θ )}{sin ( θ ) sin ( θ )}

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

使用指数法则:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= sin^{2} (θ)

= \frac{10 cos ( θ )}{sin ^{2} ( θ )}

10 \cdot \frac{1}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )} = \frac{10 sin ( θ )}{cos ^{2} ( θ )}

10 \cdot \frac{1}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )}

分式相乘:

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

= \frac{1 \cdot sin ( θ ) \cdot 10}{cos ( θ ) cos ( θ )}

数字相乘:

1 \cdot 10 = 10

= \frac{10 sin ( θ )}{cos ( θ ) cos ( θ )}

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

使用指数法则:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= cos^{2} (θ)

= \frac{10 sin ( θ )}{cos ^{2} ( θ )}

= - \frac{10 cos ( θ )}{sin ^{2} ( θ )} + \frac{10 sin ( θ )}{cos ^{2} ( θ )}

sin^{2} (θ), cos^{2} (θ)

的最小公倍数:

sin^{2} (θ) cos^{2} (θ)

sin^{2} (θ), cos^{2} (θ)

最小公倍数 (LCM)

计算出由出现在

sin^{2} (θ)

或

cos^{2} (θ)

中的因子组成的表达式

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

对于

\frac{10 cos ( θ )}{sin ^{2} ( θ )} :

将分母和分子乘以

cos^{2} (θ)

\frac{10 cos ( θ )}{sin ^{2} ( θ )} = \frac{10 cos ( θ ) cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )} = \frac{10 cos ^{3} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

对于

\frac{10 sin ( θ )}{cos ^{2} ( θ )} :

将分母和分子乘以

sin^{2} (θ)

\frac{10 sin ( θ )}{cos ^{2} ( θ )} = \frac{10 sin ( θ ) sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )} = \frac{10 sin ^{3} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

= - \frac{10 cos ^{3} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )} + \frac{10 sin ^{3} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( 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{- 10 cos ^{3} ( θ ) + 10 sin ^{3} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

= \frac{- 10 cos ^{3} ( θ ) + 10 sin ^{3} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

\frac{- 10 cos ^{3} ( θ ) + 10 sin ^{3} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )} = 0

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

- 10 cos^{3} (θ) + 10 sin^{3} (θ) = 0

分解

- 10 cos^{3} (θ) + 10 sin^{3} (θ) : 10 (sin (θ) - cos (θ)) (sin^{2} (θ) + sin (θ) cos (θ) + cos^{2} (θ))

- 10 cos^{3} (θ) + 10 sin^{3} (θ)

因式分解出通项

10

= 10 (- cos^{3} (θ) + sin^{3} (θ))

分解

sin^{3} (θ) - cos^{3} (θ) : (sin (θ) - cos (θ)) (sin^{2} (θ) + sin (θ) cos (θ) + cos^{2} (θ))

sin^{3} (θ) - cos^{3} (θ)

使用立方差公式:

x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

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

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

= 10 (sin (θ) - cos (θ)) (sin^{2} (θ) + sin (θ) cos (θ) + cos^{2} (θ))

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= 10 (- cos (θ) + sin (θ)) (cos (θ) sin (θ) + 1)

10 (- cos (θ) + sin (θ)) (cos (θ) sin (θ) + 1) = 0

分别求解每个部分

- cos (θ) + sin (θ) = 0 or cos (θ) sin (θ) + 1 = 0

- cos (θ) + sin (θ) = 0 : θ = \frac{π}{4} + πn

- cos (θ) + sin (θ) = 0

使用三角恒等式改写

- cos (θ) + sin (θ) = 0

在两边除以

cos (θ), cos (θ) \neq = 0

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

化简

- 1 + \frac{sin ( θ )}{cos ( θ )} = 0

使用基本三角恒等式:

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

- 1 + tan (θ) = 0

- 1 + tan (θ) = 0

将

1

到右边

- 1 + tan (θ) = 0

两边加上

1

- 1 + tan (θ) + 1 = 0 + 1

化简

tan (θ) = 1

tan (θ) = 1

tan (θ) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

cos (θ) sin (θ) + 1 = 0 :

无解

cos (θ) sin (θ) + 1 = 0

使用三角恒等式改写

cos (θ) sin (θ) + 1

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= 1 + \frac{sin ( 2 θ )}{2}

1 + \frac{sin ( 2 θ )}{2} = 0

将

1

到右边

1 + \frac{sin ( 2 θ )}{2} = 0

两边减去

1

1 + \frac{sin ( 2 θ )}{2} - 1 = 0 - 1

化简

\frac{sin ( 2 θ )}{2} = - 1

\frac{sin ( 2 θ )}{2} = - 1

在两边乘以

2

\frac{sin ( 2 θ )}{2} = - 1

在两边乘以

2

\frac{2 sin ( 2 θ )}{2} = 2 (- 1)

化简

sin (2 θ) = - 2

sin (2 θ) = - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

θ = \frac{π}{4} + πn

作图

流行的例子

arctan(x)=((2*1.8))/([(3170+1.7)-1])

arctan (x) = \frac{( 2 \cdot 1.8 )}{[ ( 3170 + 1.7 ) - 1 ]}

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

cos (4 x) + sin (\frac{x}{2}) = 0

2sin^2(θ)-5sin(θ)=3

2 sin^{2} (θ) - 5 sin (θ) = 3

cos(3x-pi/4)= 1/(sqrt(2)),0<= x<= pi

cos (3 x - \frac{π}{4}) = \frac{1}{2}, 0 \leq x \leq π

4sin^2(x)=3,0<= x<= 108

4 sin^{2} (x) = 3, 0 \leq x \leq 10 8^{\circ}