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人気のある三角関数 >

cos^2(a)= 1/2

解

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

解

a = 0.78539 \dots + 2 πn, a = 2 π - 0.78539 \dots + 2 πn, a = 2.35619 \dots + 2 πn, a = - 2.35619 \dots + 2 πn

+1

度

a = 4 5^{\circ} + 36 0^{\circ} n, a = 31 5^{\circ} + 36 0^{\circ} n, a = 13 5^{\circ} + 36 0^{\circ} n, a = - 13 5^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (a) = u

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

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

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

x^{2} = f (a)

の場合, 解は

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

u = \frac{1}{2}, u = - \frac{1}{2}

代用を戻す

u = cos (a)

cos (a) = \frac{1}{2}, cos (a) = - \frac{1}{2}

cos (a) = \frac{1}{2}, cos (a) = - \frac{1}{2}

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

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

cos (a) = - \frac{1}{2}

三角関数の逆数プロパティを適用する

cos (a) = - \frac{1}{2}

以下の一般解

cos (a) = - \frac{1}{2}

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

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

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

すべての解を組み合わせる

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

10進法形式で解を証明する

a = 0.78539 \dots + 2 πn, a = 2 π - 0.78539 \dots + 2 πn, a = 2.35619 \dots + 2 πn, a = - 2.35619 \dots + 2 πn

グラフ

人気の例

cos (15 θ) = 0

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sin(3x)=(sqrt(2))/2 ,0<= x<2pi

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sin^2((2x)/3)-1=0

sin^{2} (\frac{2 x}{3}) - 1 = 0