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受欢迎的三角函数 >

cos^2(x)<sin^2(x)

解答

cos^{2} (x) < sin^{2} (x)

解答

\frac{π}{4} + πn < x < \frac{3 π}{4} + πn

+2

间隔符号

(\frac{π}{4} + πn, \frac{3 π}{4} + πn)

十进制

0.78539 \dots + πn < x < 2.35619 \dots + πn

求解步骤

cos^{2} (x) < sin^{2} (x)

将

sin^{2} (x)

para o lado esquerdo

cos^{2} (x) < sin^{2} (x)

两边减去

sin^{2} (x)

cos^{2} (x) - sin^{2} (x) < sin^{2} (x) - sin^{2} (x)

cos^{2} (x) - sin^{2} (x) < 0

cos^{2} (x) - sin^{2} (x) < 0

利用以下特性:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

cos (2 x) < 0

对于

cos (x) < a

，若

- 1 < a \leq 1

，则

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

arccos (0) + 2 πn < 2 x < 2 π - arccos (0) + 2 πn

若

a < u < b

，则

a < u and u < b

arccos (0) + 2 πn < 2 x and 2 x < 2 π - arccos (0) + 2 πn

arccos (0) + 2 πn < 2 x : x > \frac{π}{4} + πn

arccos (0) + 2 πn < 2 x

交换两边

2 x > arccos (0) + 2 πn

化简

arccos (0) + 2 πn : \frac{π}{2} + 2 πn

arccos (0) + 2 πn

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} + 2 πn

2 x > \frac{π}{2} + 2 πn

两边除以

2

2 x > \frac{π}{2} + 2 πn

两边除以

2

\frac{2 x}{2} > \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} > \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x > \frac{π}{4} + πn

x > \frac{π}{4} + πn

x > \frac{π}{4} + πn

2 x < 2 π - arccos (0) + 2 πn : x < \frac{3 π}{4} + πn

2 x < 2 π - arccos (0) + 2 πn

化简

2 π - arccos (0) + 2 πn : 2 π - \frac{π}{2} + 2 πn

2 π - arccos (0) + 2 πn

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}

= 2 π - \frac{π}{2} + 2 πn

2 x < 2 π - \frac{π}{2} + 2 πn

两边除以

2

2 x < 2 π - \frac{π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{2 π}{2} - \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : π - \frac{π}{4} + πn

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

\frac{2 π}{2} = π

\frac{2 π}{2}

数字相除:

\frac{2}{2} = 1

= π

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= π - \frac{π}{4} + πn

x < π - \frac{π}{4} + πn

x < π - \frac{π}{4} + πn

化简

π - \frac{π}{4} : \frac{3 π}{4}

π - \frac{π}{4}

将项转换为分式:

π = \frac{π 4}{4}

= \frac{π 4}{4} - \frac{π}{4}

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{π 4 - π}{4}

同类项相加:

4 π - π = 3 π

= \frac{3 π}{4}

x < \frac{3 π}{4} + πn

x < \frac{3 π}{4} + πn

合并区间

x > \frac{π}{4} + πn and x < \frac{3 π}{4} + πn

合并重叠的区间

\frac{π}{4} + πn < x < \frac{3 π}{4} + πn

流行的例子

sin(x-45)> 1/2 sqrt(3),0<= x<= 360

sin (x - 4 5^{\circ}) > \frac{1}{2} 3, 0^{\circ} \leq x \leq 36 0^{\circ}

2 sin^{2} (x) > - 1

sin(θ)+cos(θ)+1>0

sin (θ) + cos (θ) + 1 > 0

-cos(x)-4sin(2x)<0

- cos (x) - 4 sin (2 x) < 0

50sin(-pi/2 x-pi/2)>=-15

50 sin (- \frac{π}{2} x - \frac{π}{2}) \geq - 15