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人気のある三角関数 >

証明する (tan(x))/(1+tan^2(x))=cos(x)sin(x)

解

証明する

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

解

真

解答ステップ

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

左側を操作する

\frac{tan ( x )}{1 + tan ^{2} ( x )}

サイン, コサインで表わす

\frac{tan ( x )}{1 + tan ^{2} ( x )}

基本的な三角関数の公式を使用する:

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

= \frac{\frac{s i n ( x )}{c o s ( x )}}{1 + ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}

簡素化

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

\frac{\frac{s i n ( x )}{c o s ( x )}}{1 + ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}

分数の規則を適用する:

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

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

指数の規則を適用する:

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

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

結合

1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} : \frac{cos ^{2} ( x ) + sin ^{2} ( x )}{cos ^{2} ( x )}

1 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

元を分数に変換する:

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

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

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

乗算:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

乗じる

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

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

分数を乗じる:

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

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

共通因数を約分する:

cos (x)

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

= \frac{sin ( x )}{\frac{c o s ^{2} ( x ) + s i n ^{2} ( x )}{c o s ( x )}}

分数の規則を適用する:

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

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

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

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

三角関数の公式を使用して書き換える

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

= \frac{cos ( x ) sin ( x )}{1}

規則を適用

\frac{a}{1} = a

= cos (x) sin (x)

= cos (x) sin (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 2csc(2x)= 1/(sin(x)cos(x))

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

証明する (1+csc(θ))(1-sin(θ))=csc(θ)-sin(θ)

p ro v e (1 + csc (θ)) (1 - sin (θ)) = csc (θ) - sin (θ)

証明する (1-tan(x))/(1-cot(x))=-tan(x)

p ro v e \frac{1 - tan ( x )}{1 - cot ( x )} = - tan (x)

証明する (cos(b))/(sec(b))+(sin(b))/(csc(b))=csc^2(b)-cot^2(b)

p ro v e \frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )} = csc^{2} (b) - cot^{2} (b)

証明する 1+cot^2(-x)=csc^2(x)

p ro v e 1 + cot^{2} (- x) = csc^{2} (x)