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人気のある三角関数 >

1+sin(2a)=sin^2(a)

解

1 + sin (2 a) = sin^{2} (a)

解

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn, a = - 0.46364 \dots + πn

+1

度

a = 9 0^{\circ} + 36 0^{\circ} n, a = 27 0^{\circ} + 36 0^{\circ} n, a = - 26.56505 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

1 + sin (2 a) = sin^{2} (a)

両辺から

sin^{2} (a)

を引く

1 + sin (2 a) - sin^{2} (a) = 0

三角関数の公式を使用して書き換える

1 + sin (2 a) - sin^{2} (a)

1 + sin (2 a) = (sin (a) + cos (a))^{2}

1 + sin (2 a)

ピタゴラスの公式を使用する:

1 = cos^{2} (a) + sin^{2} (a)

= (cos^{2} (a) + sin^{2} (a)) + sin (2 a)

2倍角の公式を使用:

sin (2 a) = 2 sin (a) cos (a)

= cos^{2} (a) + sin^{2} (a) + 2 sin (a) cos (a)

完全平方式を適用する:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

sin^{2} (a) + 2 sin (a) cos (a) + cos^{2} (a) = (sin (a) + cos (a))^{2}

= (sin (a) + cos (a))^{2}

= (sin (a) + cos (a))^{2} - sin^{2} (a)

簡素化

(sin (a) + cos (a))^{2} - sin^{2} (a) : 2 sin (a) cos (a) + cos^{2} (a)

(sin (a) + cos (a))^{2} - sin^{2} (a)

(sin (a) + cos (a))^{2} : sin^{2} (a) + 2 sin (a) cos (a) + cos^{2} (a)

完全平方式を適用する:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = sin (a), b = cos (a)

= sin^{2} (a) + 2 sin (a) cos (a) + cos^{2} (a)

= sin^{2} (a) + 2 sin (a) cos (a) + cos^{2} (a) - sin^{2} (a)

類似した元を足す:

sin^{2} (a) - sin^{2} (a) = 0

= 2 sin (a) cos (a) + cos^{2} (a)

= 2 sin (a) cos (a) + cos^{2} (a)

cos^{2} (a) + 2 cos (a) sin (a) = 0

因数

cos^{2} (a) + 2 cos (a) sin (a) : cos (a) (cos (a) + 2 sin (a))

cos^{2} (a) + 2 cos (a) sin (a)

指数の規則を適用する:

a^{b + c} = a^{b} a^{c}

cos^{2} (a) = cos (a) cos (a)

= cos (a) cos (a) + 2 sin (a) cos (a)

共通項をくくり出す

cos (a)

= cos (a) (cos (a) + 2 sin (a))

cos (a) (cos (a) + 2 sin (a)) = 0

各部分を別個に解く

cos (a) = 0 or cos (a) + 2 sin (a) = 0

cos (a) = 0 : a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

cos (a) = 0

以下の一般解

cos (a) = 0

cos (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

cos (a) + 2 sin (a) = 0 : a = arctan (- \frac{1}{2}) + πn

cos (a) + 2 sin (a) = 0

三角関数の公式を使用して書き換える

cos (a) + 2 sin (a) = 0

cos (a), cos (a) \neq = 0

で両辺を割る

\frac{cos ( a ) + 2 sin ( a )}{cos ( a )} = \frac{0}{cos ( a )}

簡素化

1 + \frac{2 sin ( a )}{cos ( a )} = 0

基本的な三角関数の公式を使用する:

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

1 + 2 tan (a) = 0

1 + 2 tan (a) = 0

1

を右側に移動します

1 + 2 tan (a) = 0

両辺から

1

を引く

1 + 2 tan (a) - 1 = 0 - 1

簡素化

2 tan (a) = - 1

2 tan (a) = - 1

以下で両辺を割る

2

2 tan (a) = - 1

以下で両辺を割る

2

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

簡素化

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

tan (x) = - a \Rightarrow x = arctan (- a) + πn

a = arctan (- \frac{1}{2}) + πn

a = arctan (- \frac{1}{2}) + πn

すべての解を組み合わせる

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn, a = arctan (- \frac{1}{2}) + πn

10進法形式で解を証明する

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn, a = - 0.46364 \dots + πn

グラフ

人気の例

((cos^3(a)))/((2cos^2(a)-1))=cos(a)

\frac{( cos ^{3} ( a ) )}{( 2 cos ^{2} ( a ) - 1 )} = cos (a)

cos (x - 4 5^{\circ}) = 0

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2 cos^{2} (x) + 5 sin (x) = 5