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受欢迎的三角函数 >

tan(2t)=1

解答

tan (2 t) = 1

解答

t = \frac{π}{8} + \frac{πn}{2}

+1

度数

t = 22. 5^{\circ} + 9 0^{\circ} n

求解步骤

tan (2 t) = 1

tan (2 t) = 1

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}

2 t = \frac{π}{4} + πn

2 t = \frac{π}{4} + πn

解

2 t = \frac{π}{4} + πn : t = \frac{π}{8} + \frac{πn}{2}

2 t = \frac{π}{4} + πn

两边除以

2

2 t = \frac{π}{4} + πn

两边除以

2

\frac{2 t}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

化简

\frac{2 t}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

化简

\frac{2 t}{2} : t

\frac{2 t}{2}

数字相除:

\frac{2}{2} = 1

= t

化简

\frac{\frac{π}{4}}{2} + \frac{πn}{2} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

使用分式法则:

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

= \frac{π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

t = \frac{π}{8} + \frac{πn}{2}

t = \frac{π}{8} + \frac{πn}{2}

t = \frac{π}{8} + \frac{πn}{2}

t = \frac{π}{8} + \frac{πn}{2}

作图

流行的例子

21=3*2*25sin(x)+9.8*3*2*3*2*0.5

21 = 3 \cdot 2 \cdot 25 sin (x) + 9.8 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 0.5

sin(2x)=((3m-8))/6

sin (2 x) = \frac{( 3 m - 8 )}{6}

cos(x)(18-12sqrt(2))=12+6sqrt(3)

cos (x) (18 - 122) = 12 + 63

cos(2x+pi/3)=(sqrt(3))/2 ,0<= x<= 2pi

cos (2 x + \frac{π}{3}) = \frac{3}{2}, 0 \leq x \leq 2 π

cos(3x+20)=-sin(x-60)

cos (3 x + 20) = - sin (x - 6 0^{\circ})