ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

cos(2θ)= 1/4

解

cos (2 θ) = \frac{1}{4}

解

θ = \frac{1.31811 \dots}{2} + πn, θ = π - \frac{1.31811 \dots}{2} + πn

+1

度

θ = 37.76124 \dots^{\circ} + 18 0^{\circ} n, θ = 142.23875 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

cos (2 θ) = \frac{1}{4}

三角関数の逆数プロパティを適用する

cos (2 θ) = \frac{1}{4}

以下の一般解

cos (2 θ) = \frac{1}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

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

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

解く

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

2 θ = arccos (\frac{1}{4}) + 2 πn

以下で両辺を割る

2

2 θ = arccos (\frac{1}{4}) + 2 πn

以下で両辺を割る

2

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

簡素化

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

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

解く

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

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

以下で両辺を割る

2

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

以下で両辺を割る

2

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

簡素化

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

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

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

10進法形式で解を証明する

θ = \frac{1.31811 \dots}{2} + πn, θ = π - \frac{1.31811 \dots}{2} + πn

グラフ

人気の例

sec (θ) = \frac{25}{7}

sec(x)=3,(3pi)/2 <= x<= 2pi

sec (x) = 3, \frac{3 π}{2} \leq x \leq 2 π

tan(2x)=cot(3x)

tan (2 x) = cot (3 x)

cos^2(β)-cot^2(β)=cos^2(β)cot^2(β)

cos^{2} (β) - cot^{2} (β) = cos^{2} (β) cot^{2} (β)

cos^3(x)=cos(x),0<= x<2pi

cos^{3} (x) = cos (x), 0 \leq x < 2 π