ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

証明する tan^8(x)=tan^6(x)sec^2(x)-tan^6(x)

解

証明する

tan^{8} (x) = tan^{6} (x) sec^{2} (x) - tan^{6} (x)

解

真

解答ステップ

tan^{8} (x) = tan^{6} (x) sec^{2} (x) - tan^{6} (x)

右側を操作する

tan^{6} (x) sec^{2} (x) - tan^{6} (x)

三角関数の公式を使用して書き換える

tan^{6} (x) sec^{2} (x) - tan^{6} (x)

ピタゴラスの公式を使用する:

sec^{2} (x) = tan^{2} (x) + 1

= tan^{6} (x) (tan^{2} (x) + 1) - tan^{6} (x)

簡素化

tan^{6} (x) (tan^{2} (x) + 1) - tan^{6} (x) : tan^{8} (x)

tan^{6} (x) (tan^{2} (x) + 1) - tan^{6} (x)

拡張

tan^{6} (x) (tan^{2} (x) + 1) : tan^{8} (x) + tan^{6} (x)

tan^{6} (x) (tan^{2} (x) + 1)

分配法則を適用する:

a (b + c) = ab + a c

a = tan^{6} (x), b = tan^{2} (x), c = 1

= tan^{6} (x) tan^{2} (x) + tan^{6} (x) \cdot 1

= tan^{6} (x) tan^{2} (x) + 1 \cdot tan^{6} (x)

簡素化

tan^{6} (x) tan^{2} (x) + 1 \cdot tan^{6} (x) : tan^{8} (x) + tan^{6} (x)

tan^{6} (x) tan^{2} (x) + 1 \cdot tan^{6} (x)

tan^{6} (x) tan^{2} (x) = tan^{8} (x)

tan^{6} (x) tan^{2} (x)

指数の規則を適用する:

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

tan^{6} (x) tan^{2} (x) = tan^{6 + 2} (x)

= tan^{6 + 2} (x)

数を足す:

6 + 2 = 8

= tan^{8} (x)

1 \cdot tan^{6} (x) = tan^{6} (x)

1 \cdot tan^{6} (x)

乗算:

1 \cdot tan^{6} (x) = tan^{6} (x)

= tan^{6} (x)

= tan^{8} (x) + tan^{6} (x)

= tan^{8} (x) + tan^{6} (x)

= tan^{8} (x) + tan^{6} (x) - tan^{6} (x)

類似した元を足す:

tan^{6} (x) - tan^{6} (x) = 0

= tan^{8} (x)

= tan^{8} (x)

= tan^{8} (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 2(sin(2y)cos(2y))=sin(4y)

p ro v e 2 (sin (2 y) cos (2 y)) = sin (4 y)

証明する ((tan^2(t)))/(sec(t))=sec(t)-cos(t)

p ro v e \frac{( tan ^{2} ( t ) )}{sec ( t )} = sec (t) - cos (t)

証明する-sin(-1)=sin(1)

p ro v e - sin (- 1) = sin (1)

証明する sin^2(y)cos^2(y)=(1-cos(4x))/8

p ro v e sin^{2} (y) cos^{2} (y) = \frac{1 - cos ( 4 x )}{8}

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

p ro v e \frac{sin ( 2 x )}{1 - cos ( 2 x )} = \frac{cos ( x )}{sin ( x )}