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人気のある三角関数 >

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

解

証明する

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

解

真

解答ステップ

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

左側を操作する

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

三角関数の公式を使用して書き換える

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

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

簡素化

\frac{1 - sin ^{4} ( a )}{1 - sin ^{2} ( a )} : sin^{2} (a) + 1

\frac{1 - sin ^{4} ( a )}{1 - sin ^{2} ( a )}

因数

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

1 - sin^{4} (a)

共通項をくくり出す

- 1

= - (sin^{4} (a) - 1)

因数

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

sin^{4} (a) - 1

sin^{4} (a) - 1

を書き換え

(sin^{2} (a))^{2} - 1^{2}

sin^{4} (a) - 1

1

を書き換え

1^{2}

= sin^{4} (a) - 1^{2}

指数の規則を適用する:

a^{b c} = (a^{b})^{c}

sin^{4} (a) = (sin^{2} (a))^{2}

= (sin^{2} (a))^{2} - 1^{2}

= (sin^{2} (a))^{2} - 1^{2}

2乗の差の公式を適用する:

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

(sin^{2} (a))^{2} - 1^{2} = (sin^{2} (a) + 1) (sin^{2} (a) - 1)

= (sin^{2} (a) + 1) (sin^{2} (a) - 1)

因数

sin^{2} (a) - 1 : (sin (a) + 1) (sin (a) - 1)

sin^{2} (a) - 1

1

を書き換え

1^{2}

= sin^{2} (a) - 1^{2}

2乗の差の公式を適用する:

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

sin^{2} (a) - 1^{2} = (sin (a) + 1) (sin (a) - 1)

= (sin (a) + 1) (sin (a) - 1)

= (sin^{2} (a) + 1) (sin (a) + 1) (sin (a) - 1)

= - (sin^{2} (a) + 1) (sin (a) + 1) (sin (a) - 1)

= - \frac{( sin ^{2} ( a ) + 1 ) ( sin ( a ) + 1 ) ( sin ( a ) - 1 )}{1 - sin ^{2} ( a )}

因数

1 - sin^{2} (a) : - (sin (a) + 1) (sin (a) - 1)

1 - sin^{2} (a)

共通項をくくり出す

- 1

= - (sin^{2} (a) - 1)

因数

sin^{2} (a) - 1 : (sin (a) + 1) (sin (a) - 1)

sin^{2} (a) - 1

1

を書き換え

1^{2}

= sin^{2} (a) - 1^{2}

2乗の差の公式を適用する:

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

sin^{2} (a) - 1^{2} = (sin (a) + 1) (sin (a) - 1)

= (sin (a) + 1) (sin (a) - 1)

= - (sin (a) + 1) (sin (a) - 1)

= \frac{( sin ^{2} ( a ) + 1 ) ( sin ( a ) + 1 ) ( sin ( a ) - 1 )}{( sin ( a ) + 1 ) ( sin ( a ) - 1 )}

キャンセル

\frac{( sin ^{2} ( a ) + 1 ) ( sin ( a ) + 1 ) ( sin ( a ) - 1 )}{( sin ( a ) + 1 ) ( sin ( a ) - 1 )} : sin^{2} (a) + 1

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

共通因数を約分する:

sin (a) + 1

= \frac{( sin ^{2} ( a ) + 1 ) ( sin ( a ) - 1 )}{sin ( a ) - 1}

共通因数を約分する:

sin (a) - 1

= sin^{2} (a) + 1

= sin^{2} (a) + 1

= sin^{2} (a) + 1

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)

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

簡素化

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

1 - cos^{2} (a) + 1

条件のようなグループ

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

数を足す:

1 + 1 = 2

= - cos^{2} (a) + 2

= - cos^{2} (a) + 2

= - cos^{2} (a) + 2

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

p ro v e \frac{sin ( 2 x ) + sin ( 4 x )}{cos ( 2 x ) + cos ( 4 x )} = tan (3 x)

証明する 1/(tan(b))+tan(b)=(sec^2(b))/(tan(b))

p ro v e \frac{1}{tan ( b )} + tan (b) = \frac{sec ^{2} ( b )}{tan ( b )}

証明する cos^3(x)=(cos(x))^3

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

証明する cos(b)(1+tan(b))^2=sec(b)+2sin(b)

p ro v e cos (b) (1 + tan (b))^{2} = sec (b) + 2 sin (b)

証明する csc^2(x)-csc(x)+9=11

p ro v e csc^{2} (x) - csc (x) + 9 = 11