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人気のある三角関数 >

solvefor α,sin(α)-sin(β)=cos(β)-cos(α)

解

解く

α, sin (α) - sin (β) = cos (β) - cos (α)

解

α = arcsin (\frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}, α = π + arcsin (- \frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}

解答ステップ

sin (α) - sin (β) = cos (β) - cos (α)

両辺から

cos (β) - cos (α)

を引く

sin (α) - sin (β) - cos (β) + cos (α) = 0

三角関数の公式を使用して書き換える

sin (α) - sin (β) - cos (β) + cos (α)

恒等式を証明する:

sin (x) + cos (x) = 2 sin (x + \frac{π}{4})

sin (α) + cos (α)

書き換え

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

次の自明恒等式を使用する:

cos (\frac{π}{4}) = \frac{1}{2}

次の自明恒等式を使用する:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (α) + sin (\frac{π}{4}) cos (α))

角の和の公式を使用する:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (α + \frac{π}{4})

= - cos (β) - sin (β) + 2 sin (α + \frac{π}{4})

- cos (β) - sin (β) + 2 sin (α + \frac{π}{4}) = 0

cos (β)

を右側に移動します

- cos (β) - sin (β) + 2 sin (α + \frac{π}{4}) = 0

両辺に

cos (β)

を足す

- cos (β) - sin (β) + 2 sin (α + \frac{π}{4}) + cos (β) = 0 + cos (β)

簡素化

- sin (β) + 2 sin (α + \frac{π}{4}) = cos (β)

- sin (β) + 2 sin (α + \frac{π}{4}) = cos (β)

sin (β)

を右側に移動します

- sin (β) + 2 sin (α + \frac{π}{4}) = cos (β)

両辺に

sin (β)

を足す

- sin (β) + 2 sin (α + \frac{π}{4}) + sin (β) = cos (β) + sin (β)

簡素化

2 sin (α + \frac{π}{4}) = cos (β) + sin (β)

2 sin (α + \frac{π}{4}) = cos (β) + sin (β)

以下で両辺を割る

2

2 sin (α + \frac{π}{4}) = cos (β) + sin (β)

以下で両辺を割る

2

\frac{2 sin ( α + \frac{π}{4} )}{2} = \frac{cos ( β )}{2} + \frac{sin ( β )}{2}

簡素化

\frac{2 sin ( α + \frac{π}{4} )}{2} = \frac{cos ( β )}{2} + \frac{sin ( β )}{2}

簡素化

\frac{2 sin ( α + \frac{π}{4} )}{2} : sin (α + \frac{π}{4})

\frac{2 sin ( α + \frac{π}{4} )}{2}

共通因数を約分する:

2

= sin (α + \frac{π}{4})

簡素化

\frac{cos ( β )}{2} + \frac{sin ( β )}{2} : \frac{2 ( cos ( β ) + sin ( β ) )}{2}

\frac{cos ( β )}{2} + \frac{sin ( β )}{2}

規則を適用

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

= \frac{cos ( β ) + sin ( β )}{2}

共役で乗じる

\frac{2}{2}

= \frac{( cos ( β ) + sin ( β ) ) 2}{2 2}

22 = 2

22

累乗根の規則を適用する:

a a = a

22 = 2

= 2

= \frac{2 ( cos ( β ) + sin ( β ) )}{2}

sin (α + \frac{π}{4}) = \frac{2 ( cos ( β ) + sin ( β ) )}{2}

sin (α + \frac{π}{4}) = \frac{2 ( cos ( β ) + sin ( β ) )}{2}

sin (α + \frac{π}{4}) = \frac{2 ( cos ( β ) + sin ( β ) )}{2}

三角関数の逆数プロパティを適用する

sin (α + \frac{π}{4}) = \frac{2 ( cos ( β ) + sin ( β ) )}{2}

以下の一般解

sin (α + \frac{π}{4}) = \frac{2 ( cos ( β ) + sin ( β ) )}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

α + \frac{π}{4} = arcsin (\frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn, α + \frac{π}{4} = π + arcsin (- \frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn

α + \frac{π}{4} = arcsin (\frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn, α + \frac{π}{4} = π + arcsin (- \frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn

解く

α + \frac{π}{4} = arcsin (\frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn : α = arcsin (\frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}

α + \frac{π}{4} = arcsin (\frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn

簡素化

arcsin (\frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn : arcsin (\frac{cos ( β ) + sin ( β )}{2}) + 2 πn

arcsin (\frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn

\frac{2 ( cos ( β ) + sin ( β ) )}{2} = \frac{cos ( β ) + sin ( β )}{2}

\frac{2 ( cos ( β ) + sin ( β ) )}{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}} ( cos ( β ) + sin ( β ) )}{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{cos ( β ) + sin ( β )}{2 ^{1 - \frac{1}{2}}}

数を引く:

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

= \frac{cos ( β ) + sin ( β )}{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{cos ( β ) + sin ( β )}{2}

= arcsin (\frac{cos ( β ) + sin ( β )}{2}) + 2 πn

α + \frac{π}{4} = arcsin (\frac{cos ( β ) + sin ( β )}{2}) + 2 πn

\frac{π}{4}

を右側に移動します

α + \frac{π}{4} = arcsin (\frac{cos ( β ) + sin ( β )}{2}) + 2 πn

両辺から

\frac{π}{4}

を引く

α + \frac{π}{4} - \frac{π}{4} = arcsin (\frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}

簡素化

α = arcsin (\frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}

α = arcsin (\frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}

解く

α + \frac{π}{4} = π + arcsin (- \frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn : α = π + arcsin (- \frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}

α + \frac{π}{4} = π + arcsin (- \frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn

簡素化

π + arcsin (- \frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn : π + arcsin (- \frac{cos ( β ) + sin ( β )}{2}) + 2 πn

π + arcsin (- \frac{2 ( cos ( β ) + sin ( β ) )}{2}) + 2 πn

\frac{2 ( cos ( β ) + sin ( β ) )}{2} = \frac{cos ( β ) + sin ( β )}{2}

\frac{2 ( cos ( β ) + sin ( β ) )}{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}} ( cos ( β ) + sin ( β ) )}{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{cos ( β ) + sin ( β )}{2 ^{1 - \frac{1}{2}}}

数を引く:

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

= \frac{cos ( β ) + sin ( β )}{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{cos ( β ) + sin ( β )}{2}

= π + arcsin (- \frac{cos ( β ) + sin ( β )}{2}) + 2 πn

α + \frac{π}{4} = π + arcsin (- \frac{cos ( β ) + sin ( β )}{2}) + 2 πn

\frac{π}{4}

を右側に移動します

α + \frac{π}{4} = π + arcsin (- \frac{cos ( β ) + sin ( β )}{2}) + 2 πn

両辺から

\frac{π}{4}

を引く

α + \frac{π}{4} - \frac{π}{4} = π + arcsin (- \frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}

簡素化

α = π + arcsin (- \frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}

α = π + arcsin (- \frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}

α = arcsin (\frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}, α = π + arcsin (- \frac{cos ( β ) + sin ( β )}{2}) + 2 πn - \frac{π}{4}

グラフ

人気の例

sin(2x)=sqrt(3)

sin (2 x) = 3

cos (2 x + 15) = 0.3

(sin^2(x))/(cos(x))=16.33

\frac{sin ^{2} ( x )}{cos ( x )} = 16.33

sin(2pix)cos(pix)+cos(2pix)sin(pix)=-1

sin (2 π x) cos (π x) + cos (2 π x) sin (π x) = - 1

-4pi^2sin(2pix)=0

- 4 π^{2} sin (2 π x) = 0