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人気のある三角関数 >

sec^2(θ)cot(θ)=8

解

sec^{2} (θ) cot (θ) = 8

解

θ = \frac{0.25268 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.25268 \dots}{2} + πn

+1

度

θ = 7.23875 \dots^{\circ} + 18 0^{\circ} n, θ = 82.76124 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

sec^{2} (θ) cot (θ) = 8

両辺から

8

を引く

sec^{2} (θ) cot (θ) - 8 = 0

サイン, コサインで表わす

- 8 + cot (θ) sec^{2} (θ)

基本的な三角関数の公式を使用する:

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

= - 8 + \frac{cos ( θ )}{sin ( θ )} sec^{2} (θ)

基本的な三角関数の公式を使用する:

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

= - 8 + \frac{cos ( θ )}{sin ( θ )} (\frac{1}{cos ( θ )})^{2}

簡素化

- 8 + \frac{cos ( θ )}{sin ( θ )} (\frac{1}{cos ( θ )})^{2} : \frac{- 8 cos ( θ ) sin ( θ ) + 1}{cos ( θ ) sin ( θ )}

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

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

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

分数を乗じる:

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

= \frac{cos ( θ ) ( \frac{1}{c o s ( θ )} ) ^{2}}{sin ( θ )}

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

(\frac{1}{cos ( θ )})^{2}

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( θ )}

規則を適用

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= \frac{\frac{1}{c o s ^{2} ( θ )} cos ( θ )}{sin ( θ )}

乗じる

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

cos (θ) \frac{1}{cos ^{2} ( θ )}

分数を乗じる:

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

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

乗算:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ )}{cos ^{2} ( θ )}

共通因数を約分する:

cos (θ)

= \frac{1}{cos ( θ )}

= \frac{\frac{1}{c o s ( θ )}}{sin ( θ )}

分数の規則を適用する:

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

= \frac{1}{cos ( θ ) sin ( θ )}

= - 8 + \frac{1}{cos ( θ ) sin ( θ )}

元を分数に変換する:

8 = \frac{8 cos ( θ ) sin ( θ )}{cos ( θ ) sin ( θ )}

= - \frac{8 cos ( θ ) sin ( θ )}{cos ( θ ) sin ( θ )} + \frac{1}{cos ( θ ) sin ( θ )}

分母が等しいので, 分数を組み合わせる:

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

= \frac{- 8 cos ( θ ) sin ( θ ) + 1}{cos ( θ ) sin ( θ )}

= \frac{- 8 cos ( θ ) sin ( θ ) + 1}{cos ( θ ) sin ( θ )}

\frac{1 - 8 cos ( θ ) sin ( θ )}{cos ( θ ) sin ( θ )} = 0

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

1 - 8 cos (θ) sin (θ) = 0

三角関数の公式を使用して書き換える

1 - 8 cos (θ) sin (θ)

2倍角の公式を使用:

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

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

= 1 - 8 \cdot \frac{sin ( 2 θ )}{2}

1 - 8 \cdot \frac{sin ( 2 θ )}{2} = 0

8 \cdot \frac{sin ( 2 θ )}{2} = 4 sin (2 θ)

8 \cdot \frac{sin ( 2 θ )}{2}

分数を乗じる:

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

= \frac{sin ( 2 θ ) \cdot 8}{2}

数を割る:

\frac{8}{2} = 4

= 4 sin (2 θ)

1 - 4 sin (2 θ) = 0

1

を右側に移動します

1 - 4 sin (2 θ) = 0

両辺から

1

を引く

1 - 4 sin (2 θ) - 1 = 0 - 1

簡素化

- 4 sin (2 θ) = - 1

- 4 sin (2 θ) = - 1

以下で両辺を割る

- 4

- 4 sin (2 θ) = - 1

以下で両辺を割る

- 4

\frac{- 4 sin ( 2 θ )}{- 4} = \frac{- 1}{- 4}

簡素化

sin (2 θ) = \frac{1}{4}

sin (2 θ) = \frac{1}{4}

三角関数の逆数プロパティを適用する

sin (2 θ) = \frac{1}{4}

以下の一般解

sin (2 θ) = \frac{1}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 θ = arcsin (\frac{1}{4}) + 2 πn, 2 θ = π - arcsin (\frac{1}{4}) + 2 πn

2 θ = arcsin (\frac{1}{4}) + 2 πn, 2 θ = π - arcsin (\frac{1}{4}) + 2 πn

解く

2 θ = arcsin (\frac{1}{4}) + 2 πn : θ = \frac{arcsin ( \frac{1}{4} )}{2} + πn

2 θ = arcsin (\frac{1}{4}) + 2 πn

以下で両辺を割る

2

2 θ = arcsin (\frac{1}{4}) + 2 πn

以下で両辺を割る

2

\frac{2 θ}{2} = \frac{arcsin ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

簡素化

θ = \frac{arcsin ( \frac{1}{4} )}{2} + πn

θ = \frac{arcsin ( \frac{1}{4} )}{2} + πn

解く

2 θ = π - arcsin (\frac{1}{4}) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

2 θ = π - arcsin (\frac{1}{4}) + 2 πn

以下で両辺を割る

2

2 θ = π - arcsin (\frac{1}{4}) + 2 πn

以下で両辺を割る

2

\frac{2 θ}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

簡素化

θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

θ = \frac{arcsin ( \frac{1}{4} )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

10進法形式で解を証明する

θ = \frac{0.25268 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.25268 \dots}{2} + πn

グラフ

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