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人気のある三角関数 >

証明する cot(2θ)=(cot^2(θ)-1)/(2cot(θ))

解

証明する

cot (2 θ) = \frac{cot ^{2} ( θ ) - 1}{2 cot ( θ )}

解

真

解答ステップ

cot (2 θ) = \frac{cot ^{2} ( θ ) - 1}{2 cot ( θ )}

右側を操作する

\frac{cot ^{2} ( θ ) - 1}{2 cot ( θ )}

サイン, コサインで表わす

\frac{- 1 + cot ^{2} ( θ )}{2 cot ( θ )}

基本的な三角関数の公式を使用する:

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

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

簡素化

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

\frac{- 1 + ( \frac{c o s ( θ )}{s i n ( θ )} ) ^{2}}{2 \cdot \frac{c o s ( θ )}{s i n ( θ )}}

乗じる

2 \cdot \frac{cos ( θ )}{sin ( θ )} : \frac{2 cos ( θ )}{sin ( θ )}

2 \cdot \frac{cos ( θ )}{sin ( θ )}

分数を乗じる:

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

= \frac{cos ( θ ) \cdot 2}{sin ( θ )}

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

指数の規則を適用する:

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

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

分数の規則を適用する:

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

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

結合

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

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

元を分数に変換する:

1 = \frac{1 sin ^{2} ( θ )}{sin ^{2} ( θ )}

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

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

乗算:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

乗じる

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

\frac{- sin ^{2} ( θ ) + cos ^{2} ( θ )}{sin ^{2} ( θ )} sin (θ)

分数を乗じる:

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

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

共通因数を約分する:

sin (θ)

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

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

分数の規則を適用する:

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

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

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

= \frac{cos ^{2} ( θ ) - sin ^{2} ( θ )}{2 cos ( θ ) sin ( θ )}

三角関数の公式を使用して書き換える

\frac{cos ^{2} ( θ ) - sin ^{2} ( θ )}{2 cos ( θ ) sin ( θ )}

2倍角の公式を使用:

2 sin (x) cos (x) = sin (2 x)

= \frac{cos ^{2} ( θ ) - sin ^{2} ( θ )}{sin ( 2 θ )}

2倍角の公式を使用:

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

= \frac{cos ( 2 θ )}{sin ( 2 θ )}

= \frac{cos ( 2 θ )}{sin ( 2 θ )}

基本的な三角関数の公式を使用する:

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

= cot (2 θ)

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

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

証明する 2/(cot(x)+tan(x))=sin(2x)

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

証明する (1-cos(θ))(1+cos(θ))=sin^2(θ)

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

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

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

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)