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受欢迎的三角函数 >

sin(a)tan(a)>0

解答

sin (a) tan (a) > 0

解答

2 πn < a < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < a < 2 π + 2 πn

+2

间隔符号

(2 πn, \frac{π}{2} + 2 πn) \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn)

十进制

2 πn < a < 1.57079 \dots + 2 πn or 4.71238 \dots + 2 πn < a < 6.28318 \dots + 2 πn

求解步骤

sin (a) tan (a) > 0

sin (a) tan (a)

的周期:

2 π

sin (a) tan (a)

包含以下函数及对应周期:

sin (a)

的周期为

2 π

复合周期为:

= 2 π

用 sin, cos 表示

sin (a) tan (a) > 0

使用基本三角恒等式:

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

sin (a) \frac{sin ( a )}{cos ( a )} > 0

sin (a) \frac{sin ( a )}{cos ( a )} > 0

化简

sin (a) \frac{sin ( a )}{cos ( a )} : \frac{sin ^{2} ( a )}{cos ( a )}

sin (a) \frac{sin ( a )}{cos ( a )}

分式相乘:

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

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

sin (a) sin (a) = sin^{2} (a)

sin (a) sin (a)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

sin (a) sin (a) = sin^{1 + 1} (a)

= sin^{1 + 1} (a)

数字相加:

1 + 1 = 2

= sin^{2} (a)

= \frac{sin ^{2} ( a )}{cos ( a )}

\frac{sin ^{2} ( a )}{cos ( a )} > 0

确定

0 \leq a < 2 π

时

\frac{sin ^{2} ( a )}{cos ( a )}

的零点和无定义点

要找到零点，将不等式设置为零

\frac{sin ^{2} ( a )}{cos ( a )} = 0

\frac{sin ^{2} ( a )}{cos ( a )} = 0, 0 \leq a < 2 π : a = 0, a = π

\frac{sin ^{2} ( a )}{cos ( a )} = 0, 0 \leq a < 2 π

使用三角恒等式改写

\frac{sin ^{2} ( a )}{cos ( a )} = 0

在两边除以

cos (a), cos (a) \neq = 0

\frac{\frac{s i n ^{2} ( a )}{c o s ( a )}}{cos ( a )} = \frac{0}{cos ( a )}

化简

\frac{sin ^{2} ( a )}{cos ^{2} ( a )} = 0

使用基本三角恒等式:

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

tan^{2} (a) = 0

tan^{2} (a) = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

tan (a) = 0

tan (a) = 0

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

a = 0 + πn

a = 0 + πn

解

a = 0 + πn : a = πn

a = 0 + πn

0 + πn = πn

a = πn

a = πn

在

0 \leq a < 2 π

范围内的解

a = 0, a = π

确定无定义点:

a = \frac{π}{2}, a = \frac{3 π}{2}

找到分母的零解

cos (a) = 0

cos (a) = 0

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

在

0 \leq a < 2 π

范围内的解

a = \frac{π}{2}, a = \frac{3 π}{2}

0, \frac{π}{2}, π, \frac{3 π}{2}

确定区间

0 < a < \frac{π}{2}, \frac{π}{2} < a < π, π < a < \frac{3 π}{2}, \frac{3 π}{2} < a < 2 π

总结如下表：

sin^{2} (a) cos (a) \frac{s i n ^{2} ( a )}{c o s ( a )} a = 0 0 + 0 0 < a < \frac{π}{2} + + + a = \frac{π}{2} + 0 未定义 \frac{π}{2} < a < π + - - a = π 0 - 0 π < a < \frac{3 π}{2} + - - a = \frac{3 π}{2} + 0 未定义 \frac{3 π}{2} < a < 2 π + + + a = 2 π 0 + 0

确定满足所需条件的区间：

> 0

0 < a < \frac{π}{2} or \frac{3 π}{2} < a < 2 π

使用周期

sin (a) tan (a)

2 πn < a < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < a < 2 π + 2 πn

流行的例子

tan (x) > cot (x)

- 6 sin (2 x - 30) > 0

cos(θ)>0,tan(θ)<0

cos (θ) > 0, tan (θ) < 0

2sin^2(x)+3sin(x)>= 2,0<= x<= pi/2

2 sin^{2} (x) + 3 sin (x) \geq 2, 0 \leq x \leq \frac{π}{2}

cos (\frac{π}{4} x) > 0