zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

证明 (cos(3x)-cos(5x))/(sin(3x)+sin(5x))=tan(x)

解答

证明

\frac{cos ( 3 x ) - cos ( 5 x )}{sin ( 3 x ) + sin ( 5 x )} = tan (x)

解答

真

求解步骤

\frac{cos ( 3 x ) - cos ( 5 x )}{sin ( 3 x ) + sin ( 5 x )} = tan (x)

调整左侧

\frac{cos ( 3 x ) - cos ( 5 x )}{sin ( 3 x ) + sin ( 5 x )}

使用三角恒等式改写

\frac{cos ( 3 x ) - cos ( 5 x )}{sin ( 3 x ) + sin ( 5 x )}

使用和差化积恒等式:

sin (s) + sin (t) = 2 sin (\frac{s + t}{2}) cos (\frac{s - t}{2})

= \frac{cos ( 3 x ) - cos ( 5 x )}{2 sin ( \frac{3 x + 5 x}{2} ) cos ( \frac{3 x - 5 x}{2} )}

化简

\frac{cos ( 3 x ) - cos ( 5 x )}{2 sin ( \frac{3 x + 5 x}{2} ) cos ( \frac{3 x - 5 x}{2} )} : \frac{cos ( 3 x ) - cos ( 5 x )}{2 cos ( x ) sin ( 4 x )}

\frac{cos ( 3 x ) - cos ( 5 x )}{2 sin ( \frac{3 x + 5 x}{2} ) cos ( \frac{3 x - 5 x}{2} )}

2 sin (\frac{3 x + 5 x}{2}) cos (\frac{3 x - 5 x}{2}) = 2 sin (\frac{8 x}{2}) cos (- \frac{2 x}{2})

2 sin (\frac{3 x + 5 x}{2}) cos (\frac{3 x - 5 x}{2})

同类项相加:

3 x + 5 x = 8 x

= 2 sin (\frac{8 x}{2}) cos (\frac{3 x - 5 x}{2})

\frac{3 x - 5 x}{2} = - \frac{2 x}{2}

\frac{3 x - 5 x}{2}

同类项相加:

3 x - 5 x = - 2 x

= \frac{- 2 x}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 x}{2}

= 2 sin (\frac{8 x}{2}) cos (- \frac{2 x}{2})

= \frac{cos ( 3 x ) - cos ( 5 x )}{2 sin ( \frac{8 x}{2} ) cos ( - \frac{2 x}{2} )}

使用负角恒等式:

cos (- x) = cos (x)

= \frac{cos ( 3 x ) - cos ( 5 x )}{2 cos ( \frac{2 x}{2} ) sin ( \frac{8 x}{2} )}

数字相除:

\frac{2}{2} = 1

= \frac{cos ( 3 x ) - cos ( 5 x )}{2 cos ( x ) sin ( \frac{8 x}{2} )}

数字相除:

\frac{8}{2} = 4

= \frac{cos ( 3 x ) - cos ( 5 x )}{2 cos ( x ) sin ( 4 x )}

= \frac{cos ( 3 x ) - cos ( 5 x )}{2 cos ( x ) sin ( 4 x )}

使用和差化积恒等式:

cos (s) - cos (t) = - 2 sin (\frac{s + t}{2}) sin (\frac{s - t}{2})

= \frac{- 2 sin ( \frac{3 x + 5 x}{2} ) sin ( \frac{3 x - 5 x}{2} )}{2 cos ( x ) sin ( 4 x )}

化简

\frac{- 2 sin ( \frac{3 x + 5 x}{2} ) sin ( \frac{3 x - 5 x}{2} )}{2 cos ( x ) sin ( 4 x )} : \frac{sin ( x )}{cos ( x )}

\frac{- 2 sin ( \frac{3 x + 5 x}{2} ) sin ( \frac{3 x - 5 x}{2} )}{2 cos ( x ) sin ( 4 x )}

- 2 sin (\frac{3 x + 5 x}{2}) sin (\frac{3 x - 5 x}{2}) = - 2 sin (\frac{8 x}{2}) sin (- \frac{2 x}{2})

- 2 sin (\frac{3 x + 5 x}{2}) sin (\frac{3 x - 5 x}{2})

同类项相加:

3 x + 5 x = 8 x

= - 2 sin (\frac{8 x}{2}) sin (\frac{3 x - 5 x}{2})

\frac{3 x - 5 x}{2} = - \frac{2 x}{2}

\frac{3 x - 5 x}{2}

同类项相加:

3 x - 5 x = - 2 x

= \frac{- 2 x}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 x}{2}

= - 2 sin (\frac{8 x}{2}) sin (- \frac{2 x}{2})

= \frac{- 2 sin ( \frac{8 x}{2} ) sin ( - \frac{2 x}{2} )}{2 cos ( x ) sin ( 4 x )}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 sin ( \frac{8 x}{2} ) sin ( - \frac{2 x}{2} )}{2 cos ( x ) sin ( 4 x )}

数字相除:

\frac{2}{2} = 1

= - \frac{sin ( \frac{8 x}{2} ) sin ( - \frac{2 x}{2} )}{cos ( x ) sin ( 4 x )}

使用负角恒等式:

sin (- x) = - sin (x)

= - \frac{( - sin ( \frac{2 x}{2} ) ) sin ( \frac{8 x}{2} )}{cos ( x ) sin ( 4 x )}

去除括号:

(- a) = - a

= - \frac{- sin ( \frac{2 x}{2} ) sin ( \frac{8 x}{2} )}{cos ( x ) sin ( 4 x )}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - (- \frac{sin ( \frac{2 x}{2} ) sin ( \frac{8 x}{2} )}{cos ( x ) sin ( 4 x )})

使用法则

- (- a) = a

= \frac{sin ( \frac{2 x}{2} ) sin ( \frac{8 x}{2} )}{cos ( x ) sin ( 4 x )}

数字相除:

\frac{2}{2} = 1

= \frac{sin ( x ) sin ( \frac{8 x}{2} )}{cos ( x ) sin ( 4 x )}

数字相除:

\frac{8}{2} = 4

= \frac{sin ( x ) sin ( 4 x )}{cos ( x ) sin ( 4 x )}

约分:

sin (4 x)

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

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

证明 tan(θ/2)=(1-cos(θ))/(sin(θ))

p ro v e tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{sin ( θ )}

证明 1/(2cos(x)sin(x))=csc(2x)

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

证明 cos(x+pi/2)=sin(x)

p ro v e cos (x + \frac{π}{2}) = sin (x)

证明 tan(pi/2-x)sin(x)=cos(x)

p ro v e tan (\frac{π}{2} - x) sin (x) = cos (x)