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受欢迎的三角函数 >

solvefor x,cos(x+y)=-cos(x)

解答

求解

x, cos (x + y) = - cos (x)

解答

x = \frac{π}{2} + 2 πn - \frac{y}{2}, x = \frac{3 π}{2} + 2 πn - \frac{y}{2}

求解步骤

cos (x + y) = - cos (x)

两边减去

- cos (x)

cos (x + y) + cos (x) = 0

使用三角恒等式改写

cos (x + y) + cos (x)

使用和差化积恒等式:

cos (s) + cos (t) = 2 cos (\frac{s + t}{2}) cos (\frac{s - t}{2})

= 2 cos (\frac{x + y + x}{2}) cos (\frac{x + y - x}{2})

化简

2 cos (\frac{x + y + x}{2}) cos (\frac{x + y - x}{2}) : 2 cos (\frac{y}{2}) cos (\frac{2 x + y}{2})

2 cos (\frac{x + y + x}{2}) cos (\frac{x + y - x}{2})

x + y + x = 2 x + y

x + y + x

对同类项分组

= x + x + y

同类项相加:

x + x = 2 x

= 2 x + y

= 2 cos (\frac{2 x + y}{2}) cos (\frac{x - x + y}{2})

x + y - x = y

x + y - x

对同类项分组

= x - x + y

同类项相加:

x - x = 0

= y

= 2 cos (\frac{y}{2}) cos (\frac{2 x + y}{2})

= 2 cos (\frac{y}{2}) cos (\frac{2 x + y}{2})

2 cos (\frac{y}{2}) cos (\frac{2 x + y}{2}) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

cos (\frac{2 x + y}{2}) = 0

cos (\frac{2 x + y}{2}) = 0

的通解

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

\frac{2 x + y}{2} = \frac{π}{2} + 2 πn, \frac{2 x + y}{2} = \frac{3 π}{2} + 2 πn

\frac{2 x + y}{2} = \frac{π}{2} + 2 πn, \frac{2 x + y}{2} = \frac{3 π}{2} + 2 πn

解

\frac{2 x + y}{2} = \frac{π}{2} + 2 πn : x = \frac{π}{2} + 2 πn - \frac{y}{2}

\frac{2 x + y}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{2 x + y}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{2 ( 2 x + y )}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

化简

\frac{2 ( 2 x + y )}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

化简

\frac{2 ( 2 x + y )}{2} : 2 x + y

\frac{2 ( 2 x + y )}{2}

数字相除:

\frac{2}{2} = 1

= 2 x + y

化简

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

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

分式相乘:

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

= \frac{π 2}{2}

约分:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

2 x + y = π + 4 πn

2 x + y = π + 4 πn

2 x + y = π + 4 πn

将

y

到右边

2 x + y = π + 4 πn

两边减去

y

2 x + y - y = π + 4 πn - y

化简

2 x = π + 4 πn - y

2 x = π + 4 πn - y

两边除以

2

2 x = π + 4 πn - y

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{4 πn}{2} - \frac{y}{2}

化简

x = \frac{π}{2} + 2 πn - \frac{y}{2}

x = \frac{π}{2} + 2 πn - \frac{y}{2}

解

\frac{2 x + y}{2} = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{2} + 2 πn - \frac{y}{2}

\frac{2 x + y}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{2 x + y}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{2 ( 2 x + y )}{2} = 2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

化简

\frac{2 ( 2 x + y )}{2} = 2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

化简

\frac{2 ( 2 x + y )}{2} : 2 x + y

\frac{2 ( 2 x + y )}{2}

数字相除:

\frac{2}{2} = 1

= 2 x + y

化简

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn : 3 π + 4 πn

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

分式相乘:

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

= \frac{3 π 2}{2}

约分:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

2 x + y = 3 π + 4 πn

2 x + y = 3 π + 4 πn

2 x + y = 3 π + 4 πn

将

y

到右边

2 x + y = 3 π + 4 πn

两边减去

y

2 x + y - y = 3 π + 4 πn - y

化简

2 x = 3 π + 4 πn - y

2 x = 3 π + 4 πn - y

两边除以

2

2 x = 3 π + 4 πn - y

两边除以

2

\frac{2 x}{2} = \frac{3 π}{2} + \frac{4 πn}{2} - \frac{y}{2}

化简

x = \frac{3 π}{2} + 2 πn - \frac{y}{2}

x = \frac{3 π}{2} + 2 πn - \frac{y}{2}

x = \frac{π}{2} + 2 πn - \frac{y}{2}, x = \frac{3 π}{2} + 2 πn - \frac{y}{2}

作图

流行的例子

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4 + 2 tan^{2} (x) = 5 tan^{2} (x) + 3

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so l v e f or θ, sec (θ) = 2

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