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受欢迎的三角函数 >

sec(θ)csc(θ)=1

解答

sec (θ) csc (θ) = 1

解答

θ \in R 无解

求解步骤

sec (θ) csc (θ) = 1

两边减去

1

sec (θ) csc (θ) - 1 = 0

用 sin, cos 表示

- 1 + csc (θ) sec (θ)

使用基本三角恒等式:

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

= - 1 + \frac{1}{sin ( θ )} sec (θ)

使用基本三角恒等式:

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

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

化简

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

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

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

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

分式相乘:

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

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

数字相乘:

1 \cdot 1 = 1

= \frac{1}{sin ( θ ) cos ( θ )}

= - 1 + \frac{1}{sin ( θ ) cos ( θ )}

将项转换为分式:

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

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

乘以:

1 \cdot sin (θ) = sin (θ)

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

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

\frac{1 - cos ( θ ) sin ( θ )}{cos ( θ ) sin ( θ )} = 0

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

1 - cos (θ) sin (θ) = 0

使用三角恒等式改写

1 - cos (θ) sin (θ)

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

2 (- \frac{sin ( 2 θ )}{2}) = 2 (- 1)

化简

- sin (2 θ) = - 2

- sin (2 θ) = - 2

两边除以

- 1

- sin (2 θ) = - 2

两边除以

- 1

\frac{- sin ( 2 θ )}{- 1} = \frac{- 2}{- 1}

化简

sin (2 θ) = 2

sin (2 θ) = 2

- 1 \leq sin (x) \leq 1

θ \in R 无解

作图

流行的例子

证明 cos(a+270)=sin(a)

p ro v e cos (a + 27 0^{\circ}) = sin (a)

1+sin(θ)=2-sin(θ)

1 + sin (θ) = 2 - sin (θ)

-4cos^2(θ)+cos(θ)=9cos(θ)+3

- 4 cos^{2} (θ) + cos (θ) = 9 cos (θ) + 3

3+4cot(x)=5csc(x)

3 + 4 cot (x) = 5 csc (x)

(tan(x)-1)(sec(x)+1)=0

(tan (x) - 1) (sec (x) + 1) = 0