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受欢迎的三角函数 >

证明 (csc(θ)-cot(θ))(1+sec(θ))=tan(θ)

解答

证明

(csc (θ) - cot (θ)) (1 + sec (θ)) = tan (θ)

解答

真

求解步骤

(csc (θ) - cot (θ)) (1 + sec (θ)) = tan (θ)

调整左侧

(csc (θ) - cot (θ)) (1 + sec (θ))

用 sin, cos 表示

(- cot (θ) + csc (θ)) (1 + sec (θ))

使用基本三角恒等式:

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

= (- \frac{cos ( θ )}{sin ( θ )} + csc (θ)) (1 + sec (θ))

使用基本三角恒等式:

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

= (- \frac{cos ( θ )}{sin ( θ )} + \frac{1}{sin ( θ )}) (1 + sec (θ))

使用基本三角恒等式:

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

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

化简

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

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

化简

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

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

使用法则

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

= \frac{- cos ( θ ) + 1}{sin ( θ )}

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

化简

1 + \frac{1}{cos ( θ )} : \frac{cos ( θ ) + 1}{cos ( θ )}

1 + \frac{1}{cos ( θ )}

将项转换为分式:

1 = \frac{1 cos ( θ )}{cos ( θ )}

= \frac{1 \cdot cos ( θ )}{cos ( θ )} + \frac{1}{cos ( θ )}

因为分母相等，所以合并分式:

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

= \frac{1 \cdot cos ( θ ) + 1}{cos ( θ )}

乘以:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) + 1}{cos ( θ )}

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

分式相乘:

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

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

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

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

使用三角恒等式改写

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

乘开

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

(1 + cos (θ)) (1 - cos (θ))

使用平方差公式:

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

a = 1, b = cos (θ)

= 1^{2} - cos^{2} (θ)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (θ)

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

使用毕达哥拉斯恒等式:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - cos^{2} (θ) = sin^{2} (θ)

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

约分:

sin (θ)

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

使用基本三角恒等式:

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

= tan (θ)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 1/(sin(x)cos(x))=sec(x)csc(x)

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

证明 cot^2(x)+csc^2(x)=1+2cot^2(x)

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

证明 (tan(y)+cot(y))/(csc(y))=sec(y)

p ro v e \frac{tan ( y ) + cot ( y )}{csc ( y )} = sec (y)

证明 cos(3B)=cos(B)(cos^2(B)-3sin^2(B))

p ro v e cos (3 B) = cos (B) (cos^{2} (B) - 3 sin^{2} (B))

证明 tan(2x)= 1/(1-tan(x))-1/(1+tan(x))

p ro v e tan (2 x) = \frac{1}{1 - tan ( x )} - \frac{1}{1 + tan ( x )}