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受欢迎的三角函数 >

2tan(θ)cos^2(θ)=1

解答

2 tan (θ) cos^{2} (θ) = 1

解答

θ = \frac{π}{4} + πn

+1

度数

θ = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

2 tan (θ) cos^{2} (θ) = 1

两边减去

1

2 cos^{2} (θ) tan (θ) - 1 = 0

使用三角恒等式改写

- 1 + 2 cos^{2} (θ) tan (θ)

使用基本三角恒等式:

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

= - 1 + 2 cos^{2} (θ) \frac{sin ( θ )}{cos ( θ )}

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

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

分式相乘:

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

= \frac{sin ( θ ) \cdot 2 cos ^{2} ( θ )}{cos ( θ )}

约分:

cos (θ)

= 2 sin (θ) cos (θ)

= - 1 + 2 sin (θ) cos (θ)

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

= - 1 + sin (2 θ)

- 1 + sin (2 θ) = 0

将

1

到右边

- 1 + sin (2 θ) = 0

两边加上

1

- 1 + sin (2 θ) + 1 = 0 + 1

化简

sin (2 θ) = 1

sin (2 θ) = 1

sin (2 θ) = 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ = \frac{π}{2} + 2 πn

2 θ = \frac{π}{2} + 2 πn

解

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

2 θ = \frac{π}{2} + 2 πn

两边除以

2

2 θ = \frac{π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

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

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

作图

流行的例子

sin (θ) = - \frac{3}{4}

4cos^2(θ)-4cos(θ)+1=0

4 cos^{2} (θ) - 4 cos (θ) + 1 = 0

3sin(x)=3cos(x)

3 sin (x) = 3 cos (x)

solvefor y,x=-3|sin(y)|

so l v e f or y, x = - 3 ∣ sin (y) ∣

tan(x)-cot(x)=0

tan (x) - cot (x) = 0