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人気のある三角関数 >

-5cos(pi/4)(1+ln(10sin(pi/4)))

解

- 5 cos (\frac{π}{4}) (1 + ln (10 sin (\frac{π}{4})))

解

- \frac{5 2 ( 1 + ln ( 5 ) + \frac{1}{2} ln ( 2 ) )}{2}

+1

十進法表記

- 10.45107 \dots

解答ステップ

- 5 cos (\frac{π}{4}) (1 + ln (10 sin (\frac{π}{4})))

次の自明恒等式を使用する:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

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}

= \frac{2}{2}

次の自明恒等式を使用する:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

sin (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} 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{2}{2}

= - 5 \cdot \frac{2}{2} (1 + ln (10 \cdot \frac{2}{2}))

簡素化

- 5 \cdot \frac{2}{2} (1 + ln (10 \cdot \frac{2}{2})) : - \frac{5 2 ( 1 + ln ( 5 ) + \frac{1}{2} ln ( 2 ) )}{2}

- 5 \cdot \frac{2}{2} (1 + ln (10 \cdot \frac{2}{2}))

分数を乗じる:

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

= - \frac{2 \cdot 5 ( 1 + ln ( 10 \cdot \frac{2}{2} ) )}{2}

累乗根の規則を適用する:

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

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

= \frac{5 \cdot 2 ^{\frac{1}{2}} ( ln ( 10 \cdot \frac{2}{2} ) + 1 )}{2}

指数の規則を適用する:

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

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

= \frac{5 ( ln ( 10 \cdot \frac{2}{2} ) + 1 )}{2 ^{1 - \frac{1}{2}}}

数を引く:

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

= \frac{5 ( ln ( 10 \cdot \frac{2}{2} ) + 1 )}{2 ^{\frac{1}{2}}}

累乗根の規則を適用する:

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

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

= - \frac{5 ( ln ( 10 \cdot \frac{2}{2} ) + 1 )}{2}

ln (10 \cdot \frac{2}{2}) = ln (5) + \frac{1}{2} ln (2)

ln (10 \cdot \frac{2}{2})

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

\frac{2}{2}

累乗根の規則を適用する:

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

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

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

指数の規則を適用する:

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

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

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

数を引く:

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

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

累乗根の規則を適用する:

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

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

= \frac{1}{2}

= ln (10 \cdot \frac{1}{2})

乗じる

10 \cdot \frac{1}{2} : 52

10 \cdot \frac{1}{2}

分数を乗じる:

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

= \frac{1 \cdot 10}{2}

数を乗じる:

1 \cdot 10 = 10

= \frac{10}{2}

因数

10 : 2 \cdot 5

因数

10 = 2 \cdot 5

= \frac{2 \cdot 5}{2}

キャンセル

\frac{2 \cdot 5}{2} : 2 \cdot 5

\frac{2 \cdot 5}{2}

累乗根の規則を適用する:

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

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

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

指数の規則を適用する:

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

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

= 5 \cdot 2^{- \frac{1}{2} + 1}

数を引く:

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

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

累乗根の規則を適用する:

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

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

= 52

= 2 \cdot 5

= ln (52)

対数の規則を適用する:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (2 \cdot 5) = ln (5) + ln (2)

= ln (5) + ln (2)

簡素化

ln (2) : \frac{1}{2} ln (2)

ln (2)

書き換え

= ln (2^{\frac{1}{2}})

以下のように仮定して対数の規則

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

を適用する:

x \geq 0

= \frac{1}{2} ln (2)

= ln (5) + \frac{1}{2} ln (2)

= - \frac{5 ( 1 + \frac{1}{2} ln ( 2 ) + ln ( 5 ) )}{2}

有理化する

- \frac{5 ( 1 + \frac{1}{2} ln ( 2 ) + ln ( 5 ) )}{2} : - \frac{5 2 ( 1 + \frac{1}{2} ln ( 2 ) + ln ( 5 ) )}{2}

- \frac{5 ( 1 + \frac{1}{2} ln ( 2 ) + ln ( 5 ) )}{2}

共役で乗じる

\frac{2}{2}

= - \frac{5 ( 1 + ln ( 5 ) + \frac{1}{2} ln ( 2 ) ) 2}{2 2}

22 = 2

22

累乗根の規則を適用する:

a a = a

22 = 2

= 2

= - \frac{5 2 ( 1 + ln ( 5 ) + \frac{1}{2} ln ( 2 ) )}{2}

= - \frac{5 2 ( 1 + \frac{1}{2} ln ( 2 ) + ln ( 5 ) )}{2}

= - \frac{5 2 ( 1 + ln ( 5 ) + \frac{1}{2} ln ( 2 ) )}{2}

人気の例

cos((pi^1)/2) pi/2

cos (\frac{π ^{1}}{2}) \frac{π}{2}

tan((11pi)/6-(5pi)/4)

tan (\frac{11 π}{6} - \frac{5 π}{4})

8 \cdot sin (5 0^{\circ})

2 sin (5 0^{\circ})

2sin(25)cos(25)

2 sin (2 5^{\circ}) cos (2 5^{\circ})