三角学示例

sec (x) = \frac{13}{12}

$\sec (x) = \frac{13}{12}$ ,

cot (x) = \frac{12}{5}

$\cot (x) = \frac{12}{5}$

解题步骤 1

利用

sec (x) = \frac{13}{12}

$\sec (x) = \frac{13}{12}$ 确定三角形的已知边。

sec (x) = \frac{1}{cos (x)} = \frac{hyp}{adj} = \frac{13}{12}

$\sec (x) = \frac{1}{\cos (x)} = \frac{hyp}{adj} = \frac{13}{12}$

解题步骤 2

利用

cot (x) = \frac{12}{5}

$\cot (x) = \frac{12}{5}$ 确定三角形的已知边。

cot (x) = \frac{1}{tan (x)} = \frac{adj}{opp} = \frac{12}{5}

$\cot (x) = \frac{1}{\tan (x)} = \frac{adj}{opp} = \frac{12}{5}$

解题步骤 3

使用

hyp = 13

$hyp = 13$ 和

opp = 5

$opp = 5$ 求正弦函数的值。

sin (x) = \frac{opp}{hyp} = \frac{5}{12}

$\sin (x) = \frac{opp}{hyp} = \frac{5}{12}$

解题步骤 4

使用

hyp = 13

$hyp = 13$ 和

adj = 12

$adj = 12$ 求余弦函数的值。

cos (x) = \frac{adj}{hyp} = \frac{12}{13}

$\cos (x) = \frac{adj}{hyp} = \frac{12}{13}$

解题步骤 5

使用

opp = 5

$opp = 5$ 和

adj = 12

$adj = 12$ 求正切函数的值。

tan (x) = \frac{opp}{adj} = \frac{5}{12}

$\tan (x) = \frac{opp}{adj} = \frac{5}{12}$

解题步骤 6

使用

opp = 5

$opp = 5$ 和

adj = 12

$adj = 12$ 求余割函数的值。

csc (x) = \frac{1}{sin (x)} = \frac{hyp}{opp} = \frac{13}{5}

$\csc (x) = \frac{1}{\sin (x)} = \frac{hyp}{opp} = \frac{13}{5}$

解题步骤 7

求得的三角函数如下：

sin (x) = \frac{5}{12}

$\sin (x) = \frac{5}{12}$

cos (x) = \frac{12}{13}

$\cos (x) = \frac{12}{13}$

tan (x) = \frac{5}{12}

$\tan (x) = \frac{5}{12}$

cot (x) = \frac{12}{5}

$\cot (x) = \frac{12}{5}$

sec (x) = \frac{13}{12}

$\sec (x) = \frac{13}{12}$

csc (x) = \frac{13}{5}

$\csc (x) = \frac{13}{5}$

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三角学 示例