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 >

cos(θ)= 2/5 \land tan(θ)<0

解答

cos (θ) = \frac{2}{5} and tan (θ) < 0

解答

对所有 θ \in R 为假

求解步骤

cos (θ) = \frac{2}{5} and tan (θ) < 0

cos (θ) = \frac{2}{5} : θ = 1.15927 \dots + 2 πn, θ = 2 π - 1.15927 \dots + 2 πn

cos (θ) = \frac{2}{5}

使用反三角函数性质

cos (θ) = \frac{2}{5}

cos (θ) = \frac{2}{5}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{2}{5}) + 2 πn, θ = 2 π - arccos (\frac{2}{5}) + 2 πn

θ = arccos (\frac{2}{5}) + 2 πn, θ = 2 π - arccos (\frac{2}{5}) + 2 πn

以小数形式表示解

θ = 1.15927 \dots + 2 πn, θ = 2 π - 1.15927 \dots + 2 πn

tan (θ) < 0 : - \frac{π}{2} + πn < θ < πn

tan (θ) < 0

若

tan (x) < a

，则

- \frac{π}{2} + πn < x < arctan (a) + πn

- \frac{π}{2} + πn < θ < arctan (0) + πn

化简

arctan (0) : 0

arctan (0)

使用以下普通恒等式:

arctan (0) = 0

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= 0

- \frac{π}{2} + πn < θ < 0 + πn

化简

- \frac{π}{2} + πn < θ < πn

合并区间

(θ = 1.15927 \dots + 2 πn or θ = 2 π - 1.15927 \dots + 2 πn) and - \frac{π}{2} + πn < θ < πn

合并重叠的区间

对所有 θ \in R 为假

流行的例子

sin(θ)=-1/8 \land sec(θ)<0

sin (θ) = - \frac{1}{8} and sec (θ) < 0

csc(θ)=-5/4 \land cos(θ)>0

csc (θ) = - \frac{5}{4} and cos (θ) > 0

sin(θ)<0\land tan(θ)<0

sin (θ) < 0 and tan (θ) < 0

csc(θ)=4\land cot(θ)<0

csc (θ) = 4 and cot (θ) < 0

-(sqrt(2))/2 <sin(x/2)<(sqrt(2))/2

- \frac{2}{2} < sin (\frac{x}{2}) < \frac{2}{2}