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受欢迎的三角函数 >

证明 (2+csc(x))/(sec(x))-2cos(x)=cot(x)

解答

证明

\frac{2 + csc ( x )}{sec ( x )} - 2 cos (x) = cot (x)

解答

真

求解步骤

\frac{2 + csc ( x )}{sec ( x )} - 2 cos (x) = cot (x)

调整左侧

\frac{2 + csc ( x )}{sec ( x )} - 2 cos (x)

用 sin, cos 表示

\frac{2 + csc ( x )}{sec ( x )} - 2 cos (x)

使用基本三角恒等式:

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

= \frac{2 + \frac{1}{s i n ( x )}}{sec ( x )} - 2 cos (x)

使用基本三角恒等式:

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

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

化简

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

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

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

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

使用分式法则:

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

= \frac{( 2 + \frac{1}{s i n ( x )} ) cos ( x )}{1}

化简

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

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

将项转换为分式:

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

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

= \frac{\frac{2 s i n ( x ) + 1}{s i n ( x )} cos ( x )}{1}

使用分式法则:

\frac{a}{1} = a

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

分式相乘:

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

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

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

将项转换为分式:

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

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

乘开

(2 sin (x) + 1) cos (x) - 2 cos (x) sin (x) : cos (x)

(2 sin (x) + 1) cos (x) - 2 cos (x) sin (x)

= cos (x) (2 sin (x) + 1) - 2 cos (x) sin (x)

乘开

cos (x) (2 sin (x) + 1) : 2 cos (x) sin (x) + cos (x)

cos (x) (2 sin (x) + 1)

使用分配律:

a (b + c) = ab + a c

a = cos (x), b = 2 sin (x), c = 1

= cos (x) \cdot 2 sin (x) + cos (x) \cdot 1

= 2 cos (x) sin (x) + 1 \cdot cos (x)

乘以:

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

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

= 2 cos (x) sin (x) + cos (x) - 2 cos (x) sin (x)

同类项相加:

2 cos (x) sin (x) - 2 cos (x) sin (x) = 0

= cos (x)

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

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

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

使用基本三角恒等式:

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

= cot (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 (cos(60))/(sin(60))=cot(60)

p ro v e \frac{cos ( 6 0 ^{\circ} )}{sin ( 6 0 ^{\circ} )} = cot (6 0^{\circ})

证明 1/(sec(x)(sec(x)-cos(x)))=cot^2(x)

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

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p ro v e sec (3 0^{\circ}) - sin (3 0^{\circ}) tan (3 0^{\circ}) = cos (3 0^{\circ})

证明 tan(x)*csc^2(x)-tan(x)=cot(x)

p ro v e tan (x) \cdot csc^{2} (x) - tan (x) = cot (x)

证明 2sin^2(2x)+cos(4x)=1

p ro v e 2 sin^{2} (2 x) + cos (4 x) = 1