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受欢迎的三角函数 >

证明 (1+sec(x))(csc(x)-cot(x))=tan(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

(1 + sec (x)) (csc (x) - cot (x))

用 sin, cos 表示

(- cot (x) + csc (x)) (1 + sec (x))

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

化简

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

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

使用法则

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

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

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

化简

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

1 + \frac{1}{cos ( x )}

将项转换为分式:

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

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

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 ( x ) + 1}{cos ( x )}

乘以:

1 \cdot cos (x) = cos (x)

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

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

分式相乘:

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

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

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

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

使用三角恒等式改写

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

乘开

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

(1 + cos (x)) (1 - cos (x))

使用平方差公式:

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

a = 1, b = cos (x)

= 1^{2} - cos^{2} (x)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (x)

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

使用毕达哥拉斯恒等式:

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

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

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

约分:

sin (x)

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

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

使用基本三角恒等式:

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

= tan (x)

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 2-csc^2(z)=1-cot^2(z)

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

证明 csc(pi/2-u)=sec(u)

p ro v e csc (\frac{π}{2} - u) = sec (u)

证明-2sin(k-l)sin(k+l)=cos(2k)-cos(2l)

p ro v e - 2 sin (k - l) sin (k + l) = cos (2 k) - cos (2 l)

证明 cos(θ-60)+cos(θ+60)=cos(θ)

p ro v e cos (θ - 6 0^{\circ}) + cos (θ + 6 0^{\circ}) = cos (θ)

证明 tan(2A)=(2tan(A))/(1-tan^2(A))

p ro v e tan (2 A) = \frac{2 tan ( A )}{1 - tan ^{2} ( A )}