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受欢迎的三角函数 >

tan(θ/2)=csc^2(θ)-cot^2(θ)

解答

tan (\frac{θ}{2}) = csc^{2} (θ) - cot^{2} (θ)

解答

θ = \frac{π}{2} + 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

tan (\frac{θ}{2}) = csc^{2} (θ) - cot^{2} (θ)

两边减去

csc^{2} (θ) - cot^{2} (θ)

tan (\frac{θ}{2}) - csc^{2} (θ) + cot^{2} (θ) = 0

使用三角恒等式改写

tan (\frac{θ}{2}) - csc^{2} (θ) + cot^{2} (θ)

使用毕达哥拉斯恒等式:

csc^{2} (x) = 1 + cot^{2} (x)

csc^{2} (x) - cot^{2} (x) = 1

= tan (\frac{θ}{2}) - 1

tan (\frac{θ}{2}) - 1 = 0

将

1

到右边

tan (\frac{θ}{2}) - 1 = 0

两边加上

1

tan (\frac{θ}{2}) - 1 + 1 = 0 + 1

化简

tan (\frac{θ}{2}) = 1

tan (\frac{θ}{2}) = 1

tan (\frac{θ}{2}) = 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}

\frac{θ}{2} = \frac{π}{4} + πn

\frac{θ}{2} = \frac{π}{4} + πn

解

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

\frac{θ}{2} = \frac{π}{4} + πn

在两边乘以

2

\frac{θ}{2} = \frac{π}{4} + πn

在两边乘以

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{4} + 2 πn

化简

\frac{2 θ}{2} = 2 \cdot \frac{π}{4} + 2 πn

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

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

2 \cdot \frac{π}{4} + 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分式相乘:

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

= \frac{π 2}{4}

约分:

2

= \frac{π}{2}

= \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

作图

流行的例子

6cos^2(x)-sin(x)-4=0

6 cos^{2} (x) - sin (x) - 4 = 0

cot^2(x)= 3/2 csc(x)

cot^{2} (x) = \frac{3}{2} csc (x)

2cos(θ)-1=0,0<= θ<= 2pi

2 cos (θ) - 1 = 0, 0 \leq θ \leq 2 π

3-3cos(x)=2sin^2(x)

3 - 3 cos (x) = 2 sin^{2} (x)

2+3sin(x)+sin^2(x)-cos^2(x)=0

2 + 3 sin (x) + sin^{2} (x) - cos^{2} (x) = 0