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人気のある三角関数 >

tan(35)tan(56-x)=1

解

tan (3 5^{\circ}) tan (5 6^{\circ} - x) = 1

解

x = - 18 0^{\circ} n + 5 6^{\circ} - 0.95993 \dots

+1

ラジアン

x = \frac{14 π}{45} - 0.95993 \dots - πn

解答ステップ

tan (3 5^{\circ}) tan (5 6^{\circ} - x) = 1

以下で両辺を割る

tan (3 5^{\circ})

tan (3 5^{\circ}) tan (5 6^{\circ} - x) = 1

以下で両辺を割る

tan (3 5^{\circ})

\frac{tan ( 3 5 ^{\circ} ) tan ( 5 6 ^{\circ} - x )}{tan ( 3 5 ^{\circ} )} = \frac{1}{tan ( 3 5 ^{\circ} )}

簡素化

tan (5 6^{\circ} - x) = \frac{1}{tan ( 3 5 ^{\circ} )}

tan (5 6^{\circ} - x) = \frac{1}{tan ( 3 5 ^{\circ} )}

三角関数の逆数プロパティを適用する

tan (5 6^{\circ} - x) = \frac{1}{tan ( 3 5 ^{\circ} )}

以下の一般解

tan (5 6^{\circ} - x) = \frac{1}{tan ( 3 5 ^{\circ} )}

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

5 6^{\circ} - x = arctan (\frac{1}{tan ( 3 5 ^{\circ} )}) + 18 0^{\circ} n

5 6^{\circ} - x = arctan (\frac{1}{tan ( 3 5 ^{\circ} )}) + 18 0^{\circ} n

解く

5 6^{\circ} - x = arctan (\frac{1}{tan ( 3 5 ^{\circ} )}) + 18 0^{\circ} n : x = - 18 0^{\circ} n + 5 6^{\circ} - arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

5 6^{\circ} - x = arctan (\frac{1}{tan ( 3 5 ^{\circ} )}) + 18 0^{\circ} n

5 6^{\circ}

を右側に移動します

5 6^{\circ} - x = arctan (\frac{1}{tan ( 3 5 ^{\circ} )}) + 18 0^{\circ} n

両辺から

5 6^{\circ}

を引く

5 6^{\circ} - x - 5 6^{\circ} = arctan (\frac{1}{tan ( 3 5 ^{\circ} )}) + 18 0^{\circ} n - 5 6^{\circ}

簡素化

- x = arctan (\frac{1}{tan ( 3 5 ^{\circ} )}) + 18 0^{\circ} n - 5 6^{\circ}

- x = arctan (\frac{1}{tan ( 3 5 ^{\circ} )}) + 18 0^{\circ} n - 5 6^{\circ}

以下で両辺を割る

- 1

- x = arctan (\frac{1}{tan ( 3 5 ^{\circ} )}) + 18 0^{\circ} n - 5 6^{\circ}

以下で両辺を割る

- 1

\frac{- x}{- 1} = \frac{arctan ( \frac{1}{t a n ( 3 5 ^{\circ} )} )}{- 1} + \frac{18 0 ^{\circ} n}{- 1} - \frac{5 6 ^{\circ}}{- 1}

簡素化

\frac{- x}{- 1} = \frac{arctan ( \frac{1}{t a n ( 3 5 ^{\circ} )} )}{- 1} + \frac{18 0 ^{\circ} n}{- 1} - \frac{5 6 ^{\circ}}{- 1}

簡素化

\frac{- x}{- 1} : x

\frac{- x}{- 1}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

= \frac{x}{1}

規則を適用

\frac{a}{1} = a

= x

簡素化

\frac{arctan ( \frac{1}{t a n ( 3 5 ^{\circ} )} )}{- 1} + \frac{18 0 ^{\circ} n}{- 1} - \frac{5 6 ^{\circ}}{- 1} : - 18 0^{\circ} n + 5 6^{\circ} - arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

\frac{arctan ( \frac{1}{t a n ( 3 5 ^{\circ} )} )}{- 1} + \frac{18 0 ^{\circ} n}{- 1} - \frac{5 6 ^{\circ}}{- 1}

条件のようなグループ

= \frac{18 0 ^{\circ} n}{- 1} - \frac{5 6 ^{\circ}}{- 1} + \frac{arctan ( \frac{1}{t a n ( 3 5 ^{\circ} )} )}{- 1}

\frac{18 0 ^{\circ} n}{- 1} = - 18 0^{\circ} n

\frac{18 0 ^{\circ} n}{- 1}

分数の規則を適用する:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{18 0 ^{\circ} n}{1}

規則を適用

\frac{a}{1} = a

= - 18 0^{\circ} n

= - 18 0^{\circ} n - \frac{5 6 ^{\circ}}{- 1} + \frac{arctan ( \frac{1}{t a n ( 3 5 ^{\circ} )} )}{- 1}

\frac{5 6 ^{\circ}}{- 1} = - 5 6^{\circ}

\frac{5 6 ^{\circ}}{- 1}

分数の規則を適用する:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{5 6 ^{\circ}}{1}

分数の規則を適用する:

\frac{a}{1} = a

\frac{5 6 ^{\circ}}{1} = 5 6^{\circ}

= - 5 6^{\circ}

\frac{arctan ( \frac{1}{t a n ( 3 5 ^{\circ} )} )}{- 1} = - arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

\frac{arctan ( \frac{1}{t a n ( 3 5 ^{\circ} )} )}{- 1}

分数の規則を適用する:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{arctan ( \frac{1}{t a n ( 3 5 ^{\circ} )} )}{1}

分数の規則を適用する:

\frac{a}{1} = a

\frac{arctan ( \frac{1}{t a n ( 3 5 ^{\circ} )} )}{1} = arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

= - arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

= - 18 0^{\circ} n - (- 5 6^{\circ}) - arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

規則を適用

- (- a) = a

= - 18 0^{\circ} n + 5 6^{\circ} - arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

x = - 18 0^{\circ} n + 5 6^{\circ} - arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

x = - 18 0^{\circ} n + 5 6^{\circ} - arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

x = - 18 0^{\circ} n + 5 6^{\circ} - arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

x = - 18 0^{\circ} n + 5 6^{\circ} - arctan (\frac{1}{tan ( 3 5 ^{\circ} )})

10進法形式で解を証明する

x = - 18 0^{\circ} n + 5 6^{\circ} - 0.95993 \dots

グラフ

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