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人気のある三角関数 >

証明する (1-3cos(x)-4cos^2(x))/(sin^2(x))=(1-4cos(x))/(1-cos(x))

解

証明する

\frac{1 - 3 cos ( x ) - 4 cos ^{2} ( x )}{sin ^{2} ( x )} = \frac{1 - 4 cos ( x )}{1 - cos ( x )}

解

真

解答ステップ

\frac{1 - 3 cos ( x ) - 4 cos ^{2} ( x )}{sin ^{2} ( x )} = \frac{1 - 4 cos ( x )}{1 - cos ( x )}

左側を操作する

\frac{1 - 3 cos ( x ) - 4 cos ^{2} ( x )}{sin ^{2} ( x )}

三角関数の公式を使用して書き換える

\frac{1 - 3 cos ( x ) - 4 cos ^{2} ( x )}{sin ^{2} ( x )}

ピタゴラスの公式を使用する:

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

cos^{2} (x) = 1 - sin^{2} (x)

= \frac{1 - 3 cos ( x ) - 4 ( 1 - sin ^{2} ( x ) )}{sin ^{2} ( x )}

拡張

1 - 3 cos (x) - 4 (1 - sin^{2} (x)) : 4 sin^{2} (x) - 3 cos (x) - 3

1 - 3 cos (x) - 4 (1 - sin^{2} (x))

拡張

- 4 (1 - sin^{2} (x)) : - 4 + 4 sin^{2} (x)

- 4 (1 - sin^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

= - 4 \cdot 1 - (- 4) sin^{2} (x)

マイナス・プラスの規則を適用する

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (x)

数を乗じる:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (x)

= 1 - 3 cos (x) - 4 + 4 sin^{2} (x)

簡素化

1 - 3 cos (x) - 4 + 4 sin^{2} (x) : 4 sin^{2} (x) - 3 cos (x) - 3

1 - 3 cos (x) - 4 + 4 sin^{2} (x)

条件のようなグループ

= - 3 cos (x) + 4 sin^{2} (x) + 1 - 4

数を足す/引く:

1 - 4 = - 3

= 4 sin^{2} (x) - 3 cos (x) - 3

= 4 sin^{2} (x) - 3 cos (x) - 3

= \frac{4 sin ^{2} ( x ) - 3 cos ( x ) - 3}{sin ^{2} ( x )}

ピタゴラスの公式を使用する:

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

sin^{2} (x) = 1 - cos^{2} (x)

= \frac{4 ( 1 - cos ^{2} ( x ) ) - 3 cos ( x ) - 3}{1 - cos ^{2} ( x )}

因数

1 - cos^{2} (x) : (1 + cos (x)) (1 - cos (x))

1 - cos^{2} (x)

2乗の差の公式を適用する:

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

1 - cos^{2} (x) = (1 + cos (x)) (1 - cos (x))

= (1 + cos (x)) (1 - cos (x))

= \frac{4 ( 1 + cos ( x ) ) ( 1 - cos ( x ) ) - 3 cos ( x ) - 3}{( 1 + cos ( x ) ) ( 1 - cos ( x ) )}

簡素化

\frac{4 ( 1 + cos ( x ) ) ( 1 - cos ( x ) ) - 3 cos ( x ) - 3}{( 1 + cos ( x ) ) ( 1 - cos ( x ) )} : \frac{- 4 cos ( x ) + 1}{1 - cos ( x )}

\frac{4 ( 1 + cos ( x ) ) ( 1 - cos ( x ) ) - 3 cos ( x ) - 3}{( 1 + cos ( x ) ) ( 1 - cos ( x ) )}

因数

4 (1 + cos (x)) (1 - cos (x)) - 3 cos (x) - 3 : (1 + cos (x)) (- 4 cos (x) + 1)

4 (1 + cos (x)) (1 - cos (x)) - 3 cos (x) - 3

書き換え

= 4 (1 + cos (x)) (1 - cos (x)) - 3 cos (x) - 3 \cdot 1

共通項をくくり出す

3

= 4 (1 + cos (x)) (1 - cos (x)) - 3 (cos (x) + 1)

共通項をくくり出す

(1 + cos (x))

= (1 + cos (x)) (4 (1 - cos (x)) - 3)

拡張

4 (- cos (x) + 1) - 3 : - 4 cos (x) + 1

4 (1 - cos (x)) - 3

拡張

4 (1 - cos (x)) : 4 - 4 cos (x)

4 (1 - cos (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 4, b = 1, c = cos (x)

= 4 \cdot 1 - 4 cos (x)

数を乗じる:

4 \cdot 1 = 4

= 4 - 4 cos (x)

= 4 - 4 cos (x) - 3

簡素化

4 - 4 cos (x) - 3 : - 4 cos (x) + 1

4 - 4 cos (x) - 3

条件のようなグループ

= - 4 cos (x) + 4 - 3

数を足す/引く:

4 - 3 = 1

= - 4 cos (x) + 1

= - 4 cos (x) + 1

= (cos (x) + 1) (- 4 cos (x) + 1)

= \frac{( 1 + cos ( x ) ) ( - 4 cos ( x ) + 1 )}{( 1 + cos ( x ) ) ( 1 - cos ( x ) )}

共通因数を約分する:

1 + cos (x)

= \frac{- 4 cos ( x ) + 1}{1 - cos ( x )}

= \frac{- 4 cos ( x ) + 1}{1 - cos ( x )}

= \frac{- 4 cos ( x ) + 1}{1 - cos ( x )}

= \frac{1 - 4 cos ( x )}{1 - cos ( x )}

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

p ro v e (1 + cos (2 x)) (1 - cos (2 x)) = sin^{2} (2 x)

証明する 1/(sin(x)+1)+1/(csc(x)+1)=1

p ro v e \frac{1}{sin ( x ) + 1} + \frac{1}{csc ( x ) + 1} = 1

証明する sin(t)csc(t)=1

p ro v e sin (t) csc (t) = 1

証明する cos(x)=sin(x+pi/2)

p ro v e cos (x) = sin (x + \frac{π}{2})

証明する sec(2θ)=(sec^2(θ))/(2-sec^2(θ))

p ro v e sec (2 θ) = \frac{sec ^{2} ( θ )}{2 - sec ^{2} ( θ )}