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(α-30))/1 =(sin(60))/3

解

\frac{cos ( α - 3 0 ^{\circ} )}{1} = \frac{sin ( 6 0 ^{\circ} )}{3}

解

α = 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}, α = 36 0^{\circ} - 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}

+1

ラジアン

α = 1.27795 \dots + \frac{π}{6} + 2 πn, α = 2 π - 1.27795 \dots + \frac{π}{6} + 2 πn

解答ステップ

\frac{cos ( α - 3 0 ^{\circ} )}{1} = \frac{sin ( 6 0 ^{\circ} )}{3}

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

次の自明恒等式を使用する:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

sin (x)

36 0^{\circ} n

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

= \frac{3}{2}

= \frac{3}{2}

\frac{cos ( α - 3 0 ^{\circ} )}{1} = \frac{\frac{3}{2}}{3}

規則を適用

\frac{a}{1} = a

cos (α - 3 0^{\circ}) = \frac{\frac{3}{2}}{3}

分数の規則を適用する:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

cos (α - 3 0^{\circ}) = \frac{3}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

cos (α - 3 0^{\circ}) = \frac{3}{6}

三角関数の逆数プロパティを適用する

cos (α - 3 0^{\circ}) = \frac{3}{6}

以下の一般解

cos (α - 3 0^{\circ}) = \frac{3}{6}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n, α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n, α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

解く

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n : α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n

簡素化

arccos (\frac{3}{6}) + 36 0^{\circ} n : arccos (\frac{1}{2 3}) + 36 0^{\circ} n

arccos (\frac{3}{6}) + 36 0^{\circ} n

\frac{3}{6} = \frac{1}{2 3}

\frac{3}{6}

因数

6 : 2 \cdot 3

因数

6 = 2 \cdot 3

= \frac{3}{2 \cdot 3}

キャンセル

\frac{3}{2 \cdot 3} : \frac{1}{2 3}

\frac{3}{2 \cdot 3}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}}}{2 \cdot 3}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{2 \cdot 3 ^{- \frac{1}{2} + 1}}

数を引く:

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

= \frac{1}{2 \cdot 3 ^{\frac{1}{2}}}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{1}{2 3}

= \frac{1}{2 3}

= arccos (\frac{1}{2 3}) + 36 0^{\circ} n

α - 3 0^{\circ} = arccos (\frac{1}{2 3}) + 36 0^{\circ} n

3 0^{\circ}

を右側に移動します

α - 3 0^{\circ} = arccos (\frac{1}{2 3}) + 36 0^{\circ} n

両辺に

3 0^{\circ}

を足す

α - 3 0^{\circ} + 3 0^{\circ} = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

簡素化

α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

解く

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n : α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

簡素化

36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n : 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

\frac{3}{6} = \frac{1}{2 3}

\frac{3}{6}

因数

6 : 2 \cdot 3

因数

6 = 2 \cdot 3

= \frac{3}{2 \cdot 3}

キャンセル

\frac{3}{2 \cdot 3} : \frac{1}{2 3}

\frac{3}{2 \cdot 3}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}}}{2 \cdot 3}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{2 \cdot 3 ^{- \frac{1}{2} + 1}}

数を引く:

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

= \frac{1}{2 \cdot 3 ^{\frac{1}{2}}}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{1}{2 3}

= \frac{1}{2 3}

= 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

3 0^{\circ}

を右側に移動します

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

両辺に

3 0^{\circ}

を足す

α - 3 0^{\circ} + 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

簡素化

α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}, α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

10進法形式で解を証明する

α = 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}, α = 36 0^{\circ} - 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}

グラフ

人気の例

tan(x)tan(7x)=1,x

tan (x) tan (7 x) = 1, x

3sec^2(x)+tan(x)-6=0

3 sec^{2} (x) + tan (x) - 6 = 0

2sin(x)+cot(x)=csc(x)

2 sin (x) + cot (x) = csc (x)

8cos^2(x)+2cos(x)-3=0

8 cos^{2} (x) + 2 cos (x) - 3 = 0

cos(A)= 1/(sqrt(65)),cos(B)= 1/(sqrt(5))

cos (A) = \frac{1}{65}, cos (B) = \frac{1}{5}