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受欢迎的三角函数 >

cos^2(θ)+2sin(θ)cos(θ)-sin^2(θ)=0

解答

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

解答

θ = πn - \frac{π}{8}, θ = πn + \frac{π}{2} - \frac{π}{8}

+1

度数

θ = - 22. 5^{\circ} + 18 0^{\circ} n, θ = 67. 5^{\circ} + 18 0^{\circ} n

求解步骤

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

使用三角恒等式改写

cos^{2} (θ) - sin^{2} (θ) + 2 cos (θ) sin (θ)

使用倍角公式:

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

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

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

使用三角恒等式改写

sin (2 θ) + cos (2 θ)

sin (2 θ) + cos (2 θ) = 2 sin (2 θ + \frac{π}{4})

sin (2 θ) + cos (2 θ)

改写为

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

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (2 θ) + sin (\frac{π}{4}) cos (2 θ))

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (2 θ + \frac{π}{4})

= 2 sin (2 θ + \frac{π}{4})

2 sin (2 θ + \frac{π}{4}) = 0

两边除以

2

2 sin (2 θ + \frac{π}{4}) = 0

两边除以

2

\frac{2 sin ( 2 θ + \frac{π}{4} )}{2} = \frac{0}{2}

化简

sin (2 θ + \frac{π}{4}) = 0

sin (2 θ + \frac{π}{4}) = 0

sin (2 θ + \frac{π}{4}) = 0

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}{4} = 0 + 2 πn, 2 θ + \frac{π}{4} = π + 2 πn

2 θ + \frac{π}{4} = 0 + 2 πn, 2 θ + \frac{π}{4} = π + 2 πn

解

2 θ + \frac{π}{4} = 0 + 2 πn : θ = πn - \frac{π}{8}

2 θ + \frac{π}{4} = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ + \frac{π}{4} = 2 πn

将

\frac{π}{4}

到右边

2 θ + \frac{π}{4} = 2 πn

两边减去

\frac{π}{4}

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

化简

2 θ = 2 πn - \frac{π}{4}

2 θ = 2 πn - \frac{π}{4}

两边除以

2

2 θ = 2 πn - \frac{π}{4}

两边除以

2

\frac{2 θ}{2} = \frac{2 πn}{2} - \frac{\frac{π}{4}}{2}

化简

\frac{2 θ}{2} = \frac{2 πn}{2} - \frac{\frac{π}{4}}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

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

\frac{2 πn}{2} - \frac{\frac{π}{4}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\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}

= πn - \frac{π}{8}

θ = πn - \frac{π}{8}

θ = πn - \frac{π}{8}

θ = πn - \frac{π}{8}

解

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

2 θ + \frac{π}{4} = π + 2 πn

将

\frac{π}{4}

到右边

2 θ + \frac{π}{4} = π + 2 πn

两边减去

\frac{π}{4}

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

化简

2 θ = π + 2 πn - \frac{π}{4}

2 θ = π + 2 πn - \frac{π}{4}

两边除以

2

2 θ = π + 2 πn - \frac{π}{4}

两边除以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

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

\frac{π}{2} + \frac{2 πn}{2} - \frac{\frac{π}{4}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\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}{2} + πn - \frac{π}{8}

对同类项分组

= πn + \frac{π}{2} - \frac{π}{8}

θ = πn + \frac{π}{2} - \frac{π}{8}

θ = πn + \frac{π}{2} - \frac{π}{8}

θ = πn + \frac{π}{2} - \frac{π}{8}

θ = πn - \frac{π}{8}, θ = πn + \frac{π}{2} - \frac{π}{8}

作图

流行的例子

tan(x)+tan^2(x)=0

tan (x) + tan^{2} (x) = 0

sin(2θ)+2sin(θ)=0

sin (2 θ) + 2 sin (θ) = 0

sin (x + \frac{π}{2}) = 1

cos (x π) = - 1

tan (x) = \frac{2}{- 2}