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受欢迎的三角函数 >

sin^2(2θ)=1,-180<= θ<= 180

解答

sin^{2} (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

解答

θ = 4 5^{\circ}, θ = - 13 5^{\circ}, θ = 13 5^{\circ}, θ = - 4 5^{\circ}

+1

弧度

θ = \frac{π}{4}, θ = - \frac{3 π}{4}, θ = \frac{3 π}{4}, θ = - \frac{π}{4}

求解步骤

sin^{2} (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

用替代法求解

sin^{2} (2 θ) = 1

令

: sin (2 θ) = u

u^{2} = 1

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = sin (2 θ)

代回

sin (2 θ) = 1, sin (2 θ) = - 1

sin (2 θ) = 1, sin (2 θ) = - 1

sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 4 5^{\circ}, θ = - 13 5^{\circ}

sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

sin (2 θ) = 1

的通解

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ = 9 0^{\circ} + 36 0^{\circ} n

2 θ = 9 0^{\circ} + 36 0^{\circ} n

解

2 θ = 9 0^{\circ} + 36 0^{\circ} n : θ = 4 5^{\circ} + 18 0^{\circ} n

2 θ = 9 0^{\circ} + 36 0^{\circ} n

两边除以

2

2 θ = 9 0^{\circ} + 36 0^{\circ} n

两边除以

2

\frac{2 θ}{2} = \frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

化简

\frac{2 θ}{2} = \frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 4 5^{\circ} + 18 0^{\circ} n

\frac{9 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{9 0 ^{\circ}}{2} = 4 5^{\circ}

\frac{9 0 ^{\circ}}{2}

使用分式法则:

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

= \frac{18 0 ^{\circ}}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= 4 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

数字相除:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

θ = 4 5^{\circ} + 18 0^{\circ} n

在

- 18 0^{\circ} \leq θ \leq 18 0^{\circ}

范围内的解

θ = 4 5^{\circ}, θ = - 13 5^{\circ}

sin (2 θ) = - 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 13 5^{\circ}, θ = - 4 5^{\circ}

sin (2 θ) = - 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

sin (2 θ) = - 1

的通解

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ = 27 0^{\circ} + 36 0^{\circ} n

2 θ = 27 0^{\circ} + 36 0^{\circ} n

解

2 θ = 27 0^{\circ} + 36 0^{\circ} n : θ = 13 5^{\circ} + 18 0^{\circ} n

2 θ = 27 0^{\circ} + 36 0^{\circ} n

两边除以

2

2 θ = 27 0^{\circ} + 36 0^{\circ} n

两边除以

2

\frac{2 θ}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

化简

\frac{2 θ}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 13 5^{\circ} + 18 0^{\circ} n

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{27 0 ^{\circ}}{2} = 13 5^{\circ}

\frac{27 0 ^{\circ}}{2}

使用分式法则:

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

= \frac{54 0 ^{\circ}}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= 13 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

数字相除:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

θ = 13 5^{\circ} + 18 0^{\circ} n

在

- 18 0^{\circ} \leq θ \leq 18 0^{\circ}

范围内的解

θ = 13 5^{\circ}, θ = - 4 5^{\circ}

合并所有解

θ = 4 5^{\circ}, θ = - 13 5^{\circ}, θ = 13 5^{\circ}, θ = - 4 5^{\circ}

作图

流行的例子

cos (A) = \frac{1}{3}

100=(50)/(sin(x/2))

100 = \frac{50}{sin ( \frac{x}{2} )}

tan(2θ)=(-2)/(1-1)

tan (2 θ) = \frac{- 2}{1 - 1}

3 - 3 cos (x) = 3

cos(x)= 5/6 ,(3pi)/2 <= x<= 2pi

cos (x) = \frac{5}{6}, \frac{3 π}{2} \leq x \leq 2 π