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 (1+csc(b))/(cot(b)+cos(b))=sec(b)

解答

证明

\frac{1 + csc ( b )}{cot ( b ) + cos ( b )} = sec (b)

解答

真

求解步骤

\frac{1 + csc ( b )}{cot ( b ) + cos ( b )} = sec (b)

调整左侧

\frac{1 + csc ( b )}{cot ( b ) + cos ( b )}

用 sin, cos 表示

\frac{1 + csc ( b )}{cos ( b ) + cot ( b )}

使用基本三角恒等式:

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

= \frac{1 + \frac{1}{s i n ( b )}}{cos ( b ) + cot ( b )}

使用基本三角恒等式:

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

= \frac{1 + \frac{1}{s i n ( b )}}{cos ( b ) + \frac{c o s ( b )}{s i n ( b )}}

化简

\frac{1 + \frac{1}{s i n ( b )}}{cos ( b ) + \frac{c o s ( b )}{s i n ( b )}} : \frac{1}{cos ( b )}

\frac{1 + \frac{1}{s i n ( b )}}{cos ( b ) + \frac{c o s ( b )}{s i n ( b )}}

化简

cos (b) + \frac{cos ( b )}{sin ( b )} : \frac{cos ( b ) sin ( b ) + cos ( b )}{sin ( b )}

cos (b) + \frac{cos ( b )}{sin ( b )}

将项转换为分式:

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

= \frac{cos ( b ) sin ( b )}{sin ( b )} + \frac{cos ( b )}{sin ( b )}

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{cos ( b ) sin ( b ) + cos ( b )}{sin ( b )}

= \frac{1 + \frac{1}{s i n ( b )}}{\frac{c o s ( b ) s i n ( b ) + c o s ( b )}{s i n ( b )}}

化简

1 + \frac{1}{sin ( b )} : \frac{sin ( b ) + 1}{sin ( b )}

1 + \frac{1}{sin ( b )}

将项转换为分式:

1 = \frac{1 sin ( b )}{sin ( b )}

= \frac{1 \cdot sin ( b )}{sin ( b )} + \frac{1}{sin ( b )}

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 sin ( b ) + 1}{sin ( b )}

乘以:

1 \cdot sin (b) = sin (b)

= \frac{sin ( b ) + 1}{sin ( b )}

= \frac{\frac{s i n ( b ) + 1}{s i n ( b )}}{\frac{c o s ( b ) s i n ( b ) + c o s ( b )}{s i n ( b )}}

分式相除:

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

= \frac{( sin ( b ) + 1 ) sin ( b )}{sin ( b ) ( cos ( b ) sin ( b ) + cos ( b ) )}

约分:

sin (b)

= \frac{sin ( b ) + 1}{cos ( b ) sin ( b ) + cos ( b )}

因式分解出通项

cos (b)

= \frac{sin ( b ) + 1}{cos ( b ) ( sin ( b ) + 1 )}

约分:

sin (b) + 1

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

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

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

使用三角恒等式改写

使用基本三角恒等式:

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

\frac{1}{\frac{1}{s e c ( b )}}

化简

\frac{1}{\frac{1}{s e c ( b )}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sec ( b )}{1}

使用法则

\frac{a}{1} = a

= sec (b)

sec (b)

sec (b)

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+tan^2(θ))/(tan(θ))=sec(θ)csc(θ)

p ro v e \frac{1 + tan ^{2} ( θ )}{tan ( θ )} = sec (θ) csc (θ)

证明 sin(4a)=2sin(2a)cos(2a)

p ro v e sin (4 a) = 2 sin (2 a) cos (2 a)

证明 sin(2D)=2cot(D)sin^2(D)

p ro v e sin (2 D) = 2 cot (D) sin^{2} (D)

证明 csc(x)*tan(x)+sec(x)=2sec(x)

p ro v e csc (x) \cdot tan (x) + sec (x) = 2 sec (x)

证明 (cos(x)+sin(x))^2=1+2cos(x)sin(x)

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