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人気のある三角関数 >

証明する (2sin(x)+cos(x))^2+(2cos(x)-sin(x))^2=5

解

証明する

(2 sin (x) + cos (x))^{2} + (2 cos (x) - sin (x))^{2} = 5

解

真

解答ステップ

(2 sin (x) + cos (x))^{2} + (2 cos (x) - sin (x))^{2} = 5

左側を操作する

(2 sin (x) + cos (x))^{2} + (2 cos (x) - sin (x))^{2}

因数

(- sin (x) + 2 cos (x))^{2} + (cos (x) + 2 sin (x))^{2} : 5 (sin^{2} (x) + cos^{2} (x))

(- sin (x) + 2 cos (x))^{2} + (cos (x) + 2 sin (x))^{2}

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

完全平方式を適用する:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = - sin (x), b = 2 cos (x)

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

簡素化

(- sin (x))^{2} + 2 (- sin (x)) \cdot 2 cos (x) + (2 cos (x))^{2} : sin^{2} (x) - 4 sin (x) cos (x) + 4 cos^{2} (x)

(- sin (x))^{2} + 2 (- sin (x)) \cdot 2 cos (x) + (2 cos (x))^{2}

括弧を削除する:

(- a) = - a

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

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

(- sin (x))^{2}

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

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

= sin^{2} (x)

2 sin (x) \cdot 2 cos (x) = 4 sin (x) cos (x)

2 sin (x) \cdot 2 cos (x)

数を乗じる:

2 \cdot 2 = 4

= 4 sin (x) cos (x)

(2 cos (x))^{2} = 4 cos^{2} (x)

(2 cos (x))^{2}

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} cos^{2} (x)

2^{2} = 4

= 4 cos^{2} (x)

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

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

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

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

完全平方式を適用する:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = cos (x), b = 2 sin (x)

= cos^{2} (x) + 2 cos (x) \cdot 2 sin (x) + (2 sin (x))^{2}

簡素化

cos^{2} (x) + 2 cos (x) \cdot 2 sin (x) + (2 sin (x))^{2} : cos^{2} (x) + 4 cos (x) sin (x) + 4 sin^{2} (x)

cos^{2} (x) + 2 cos (x) \cdot 2 sin (x) + (2 sin (x))^{2}

2 cos (x) \cdot 2 sin (x) = 4 cos (x) sin (x)

2 cos (x) \cdot 2 sin (x)

数を乗じる:

2 \cdot 2 = 4

= 4 cos (x) sin (x)

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

(2 sin (x))^{2}

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

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

2^{2} = 4

= 4 sin^{2} (x)

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

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

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

簡素化

sin^{2} (x) - 4 sin (x) cos (x) + 4 cos^{2} (x) + cos^{2} (x) + 4 cos (x) sin (x) + 4 sin^{2} (x) : 5 sin^{2} (x) + 5 cos^{2} (x)

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

類似した元を足す:

- 4 sin (x) cos (x) + 4 cos (x) sin (x) = 0

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

類似した元を足す:

4 cos^{2} (x) + cos^{2} (x) = 5 cos^{2} (x)

= sin^{2} (x) + 5 cos^{2} (x) + 4 sin^{2} (x)

類似した元を足す:

sin^{2} (x) + 4 sin^{2} (x) = 5 sin^{2} (x)

= 5 sin^{2} (x) + 5 cos^{2} (x)

= 5 sin^{2} (x) + 5 cos^{2} (x)

共通項をくくり出す

5

= 5 (sin^{2} (x) + cos^{2} (x))

= (cos^{2} (x) + sin^{2} (x)) \cdot 5

三角関数の公式を使用して書き換える

(cos^{2} (x) + sin^{2} (x)) \cdot 5

ピタゴラスの公式を使用する:

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

= 1 \cdot 5

簡素化

= 5

= 5

両辺を同じ形式にできることを証明した

\Rightarrow 真

人気の例

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