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人気のある三角関数 >

cos^2(a)= 1/5

解

cos^{2} (a) = \frac{1}{5}

解

a = 1.10714 \dots + 2 πn, a = 2 π - 1.10714 \dots + 2 πn, a = 2.03444 \dots + 2 πn, a = - 2.03444 \dots + 2 πn

+1

度

a = 63.43494 \dots^{\circ} + 36 0^{\circ} n, a = 296.56505 \dots^{\circ} + 36 0^{\circ} n, a = 116.56505 \dots^{\circ} + 36 0^{\circ} n, a = - 116.56505 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

cos^{2} (a) = \frac{1}{5}

置換で解く

cos^{2} (a) = \frac{1}{5}

仮定:

cos (a) = u

u^{2} = \frac{1}{5}

u^{2} = \frac{1}{5} : u = \frac{1}{5}, u = - \frac{1}{5}

u^{2} = \frac{1}{5}

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

u = \frac{1}{5}, u = - \frac{1}{5}

代用を戻す

u = cos (a)

cos (a) = \frac{1}{5}, cos (a) = - \frac{1}{5}

cos (a) = \frac{1}{5}, cos (a) = - \frac{1}{5}

cos (a) = \frac{1}{5} : a = arccos (\frac{1}{5}) + 2 πn, a = 2 π - arccos (\frac{1}{5}) + 2 πn

cos (a) = \frac{1}{5}

三角関数の逆数プロパティを適用する

cos (a) = \frac{1}{5}

以下の一般解

cos (a) = \frac{1}{5}

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

a = arccos (\frac{1}{5}) + 2 πn, a = 2 π - arccos (\frac{1}{5}) + 2 πn

a = arccos (\frac{1}{5}) + 2 πn, a = 2 π - arccos (\frac{1}{5}) + 2 πn

cos (a) = - \frac{1}{5} : a = arccos (- \frac{1}{5}) + 2 πn, a = - arccos (- \frac{1}{5}) + 2 πn

cos (a) = - \frac{1}{5}

三角関数の逆数プロパティを適用する

cos (a) = - \frac{1}{5}

以下の一般解

cos (a) = - \frac{1}{5}

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

a = arccos (- \frac{1}{5}) + 2 πn, a = - arccos (- \frac{1}{5}) + 2 πn

a = arccos (- \frac{1}{5}) + 2 πn, a = - arccos (- \frac{1}{5}) + 2 πn

すべての解を組み合わせる

a = arccos (\frac{1}{5}) + 2 πn, a = 2 π - arccos (\frac{1}{5}) + 2 πn, a = arccos (- \frac{1}{5}) + 2 πn, a = - arccos (- \frac{1}{5}) + 2 πn

10進法形式で解を証明する

a = 1.10714 \dots + 2 πn, a = 2 π - 1.10714 \dots + 2 πn, a = 2.03444 \dots + 2 πn, a = - 2.03444 \dots + 2 πn

グラフ

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