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人気のある三角関数 >

cos^2(t)= 1/2

解

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

解

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

+1

度

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

解答ステップ

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

置換で解く

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

仮定:

cos (t) = 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 (t)

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

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

すべての解を組み合わせる

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

10進法形式で解を証明する

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

グラフ

人気の例

2cos^2(θ)-1=sin(θ),0<= θ<= 180

2 cos^{2} (θ) - 1 = sin (θ), 0^{\circ} \leq θ \leq 18 0^{\circ}

cos^4(x)-sin^4(x)= 1/2

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50sin(45)=600sin(θ)

50 sin (4 5^{\circ}) = 600 sin (θ)