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受欢迎的三角函数 >

sin(A)=(-4)/5 \land cos(A)>0,cos(A)

解答

sin (A) = \frac{- 4}{5} and cos (A) > 0, cos (A)

解答

A = - 0.92729 \dots + 2 πn

+1

间隔符号

A = - 0.92729 \dots + 2 πn

求解步骤

sin (A) = \frac{- 4}{5} and cos (A) > 0

sin (A) = \frac{- 4}{5} : A = - 0.92729 \dots + 2 πn, A = π + 0.92729 \dots + 2 πn

sin (A) = \frac{- 4}{5}

使用分式法则:

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

sin (A) = - \frac{4}{5}

使用反三角函数性质

sin (A) = - \frac{4}{5}

sin (A) = - \frac{4}{5}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

A = arcsin (- \frac{4}{5}) + 2 πn, A = π + arcsin (\frac{4}{5}) + 2 πn

A = arcsin (- \frac{4}{5}) + 2 πn, A = π + arcsin (\frac{4}{5}) + 2 πn

以小数形式表示解

A = - 0.92729 \dots + 2 πn, A = π + 0.92729 \dots + 2 πn

cos (A) > 0 : - \frac{π}{2} + 2 πn < A < \frac{π}{2} + 2 πn

cos (A) > 0

对于

cos (x) > a

，若

- 1 \leq a < 1

，则

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (0) + 2 πn < A < arccos (0) + 2 πn

化简

- arccos (0) : - \frac{π}{2}

- arccos (0)

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2}

化简

arccos (0) : \frac{π}{2}

arccos (0)

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

- \frac{π}{2} + 2 πn < A < \frac{π}{2} + 2 πn

合并区间

(A = π + 0.92729 \dots + 2 πn or A = - 0.92729 \dots + 2 πn) and - \frac{π}{2} + 2 πn < A < \frac{π}{2} + 2 πn

合并重叠的区间

A = - 0.92729 \dots + 2 πn

流行的例子

sin(θ)= 2/5 \land sec(θ)>0

sin (θ) = \frac{2}{5} and sec (θ) > 0

csc(θ)<0\land cos(θ)>0

csc (θ) < 0 and cos (θ) > 0

sin(2x)0<= pi/2 \land 0-pi/2 <0

sin (2 x) 0 \leq \frac{π}{2} and 0 - \frac{π}{2} < 0

3-3(0)^2<= g(0)<= 3cos(0)

3 - 3 (0)^{2} \leq g (0) \leq 3 cos (0)

0<sin(x)cos(x)<sqrt(2)

0 < sin (x) cos (x) < 2