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人気のある三角関数 >

証明する (cos^2(A))/(cos(A)-sin(A))+(sin(A))/(1-cot(A))=sin(A)+cos(A)

解

証明する

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - cot ( A )} = sin (A) + cos (A)

解

真

解答ステップ

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - cot ( A )} = sin (A) + cos (A)

左側を操作する

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

サイン, コサインで表わす

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

基本的な三角関数の公式を使用する:

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

= \frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - \frac{c o s ( A )}{s i n ( A )}}

簡素化

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - \frac{c o s ( A )}{s i n ( A )}} : sin (A) + cos (A)

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - \frac{c o s ( A )}{s i n ( A )}}

\frac{sin ( A )}{1 - \frac{c o s ( A )}{s i n ( A )}} = \frac{sin ^{2} ( A )}{sin ( A ) - cos ( A )}

\frac{sin ( A )}{1 - \frac{c o s ( A )}{s i n ( A )}}

結合

1 - \frac{cos ( A )}{sin ( A )} : \frac{sin ( A ) - cos ( A )}{sin ( A )}

1 - \frac{cos ( A )}{sin ( A )}

元を分数に変換する:

1 = \frac{1 sin ( A )}{sin ( A )}

= \frac{1 \cdot sin ( A )}{sin ( A )} - \frac{cos ( A )}{sin ( A )}

分母が等しいので, 分数を組み合わせる:

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

= \frac{1 \cdot sin ( A ) - cos ( A )}{sin ( A )}

乗算:

1 \cdot sin (A) = sin (A)

= \frac{sin ( A ) - cos ( A )}{sin ( A )}

= \frac{sin ( A )}{\frac{s i n ( A ) - c o s ( A )}{s i n ( A )}}

分数の規則を適用する:

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

= \frac{sin ( A ) sin ( A )}{sin ( A ) - cos ( A )}

sin (A) sin (A) = sin^{2} (A)

sin (A) sin (A)

指数の規則を適用する:

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

sin (A) sin (A) = sin^{1 + 1} (A)

= sin^{1 + 1} (A)

数を足す:

1 + 1 = 2

= sin^{2} (A)

= \frac{sin ^{2} ( A )}{sin ( A ) - cos ( A )}

= \frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ^{2} ( A )}{sin ( A ) - cos ( A )}

以下の最小公倍数:

cos (A) - sin (A), sin (A) - cos (A) : - (cos (A) - sin (A))

cos (A) - sin (A), sin (A) - cos (A)

最小公倍数 (LCM)

sin (A) - cos (A) = - (cos (A) - sin (A))

= cos (A) - sin (A), - (cos (A) - sin (A))

cos (A) - sin (A)

または以下のいずれかに現れる因数で構成された式を計算する:

sin (A) - cos (A)

= - (cos (A) - sin (A))

LCMに基づいて分数を調整する

該当する分母を乗じてLCMに変えるために

必要な量で各分子を乗じる

- (cos (A) - sin (A))

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )}

の場合

:

分母と分子に以下を乗じる:

- 1

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} = \frac{cos ^{2} ( A ) ( - 1 )}{( cos ( A ) - sin ( A ) ) ( - 1 )} = \frac{- cos ^{2} ( A )}{- ( cos ( A ) - sin ( A ) )}

= \frac{- cos ^{2} ( A )}{- ( cos ( A ) - sin ( A ) )} + \frac{sin ^{2} ( A )}{- ( cos ( A ) - sin ( A ) )}

分母が等しいので, 分数を組み合わせる:

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

= \frac{- cos ^{2} ( A ) + sin ^{2} ( A )}{- ( cos ( A ) - sin ( A ) )}

分数の規則を適用する:

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

= - \frac{- cos ^{2} ( A ) + sin ^{2} ( A )}{( cos ( A ) - sin ( A ) )}

キャンセル

\frac{- cos ^{2} ( A ) + sin ^{2} ( A )}{( cos ( A ) - sin ( A ) )} : - sin (A) - cos (A)

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

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

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

= \frac{( sin ( A ) + cos ( A ) ) ( sin ( A ) - cos ( A ) )}{( cos ( A ) - sin ( A ) )}

キャンセル

\frac{( sin ( A ) + cos ( A ) ) ( sin ( A ) - cos ( A ) )}{( cos ( A ) - sin ( A ) )} : - (sin (A) + cos (A))

\frac{( sin ( A ) + cos ( A ) ) ( sin ( A ) - cos ( A ) )}{( cos ( A ) - sin ( A ) )}

cos (A) - sin (A) = - (sin (A) - cos (A))

= \frac{( sin ( A ) + cos ( A ) ) ( sin ( A ) - cos ( A ) )}{- ( sin ( A ) - cos ( A ) )}

改良

= - \frac{( sin ( A ) + cos ( A ) ) ( sin ( A ) - cos ( A ) )}{sin ( A ) - cos ( A )}

共通因数を約分する:

sin (A) - cos (A)

= - (sin (A) + cos (A))

= - (sin (A) + cos (A))

否定

- (sin (A) + cos (A)) = - sin (A) - cos (A)

= - sin (A) - cos (A)

= - (- sin (A) - cos (A))

括弧を分配する

= - (- sin (A)) - (- cos (A))

マイナス・プラスの規則を適用する

- (- a) = a

= sin (A) + cos (A)

= sin (A) + cos (A)

= cos (A) + sin (A)

= sin (A) + cos (A)

両辺を同じ形式にできることを証明した

\Rightarrow 真

人気の例

証明する cos(x-pi/6)-sin(x+pi/3)=0

p ro v e cos (x - \frac{π}{6}) - sin (x + \frac{π}{3}) = 0

証明する cot^2(α)+1= 1/(sin^2(α))

p ro v e cot^{2} (α) + 1 = \frac{1}{sin ^{2} ( α )}

証明する sin(3x)=3sin(x)cos^2(x)-sin^3(x)

p ro v e sin (3 x) = 3 sin (x) cos^{2} (x) - sin^{3} (x)

証明する (cos(3x)-cos(5x))/(sin(3x)+sin(5x))=tan(x)

p ro v e \frac{cos ( 3 x ) - cos ( 5 x )}{sin ( 3 x ) + sin ( 5 x )} = tan (x)

証明する cos^2(x)= 1/2+1/2 cos(2x)

p ro v e cos^{2} (x) = \frac{1}{2} + \frac{1}{2} cos (2 x)