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受欢迎的三角函数 >

sin(x+pi/6)+cos(x+pi/3)=cos(2x)

解答

sin (x + \frac{π}{6}) + cos (x + \frac{π}{3}) = cos (2 x)

解答

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

+1

度数

x = 0^{\circ} + 24 0^{\circ} n, x = 12 0^{\circ} + 24 0^{\circ} n

求解步骤

sin (x + \frac{π}{6}) + cos (x + \frac{π}{3}) = cos (2 x)

使用三角恒等式改写

sin (x + \frac{π}{6}) + cos (x + \frac{π}{3}) = cos (2 x)

使用三角恒等式改写

sin (x + \frac{π}{6})

使用角和恒等式:

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

= sin (x) cos (\frac{π}{6}) + cos (x) sin (\frac{π}{6})

化简

sin (x) cos (\frac{π}{6}) + cos (x) sin (\frac{π}{6}) : \frac{3}{2} sin (x) + \frac{1}{2} cos (x)

sin (x) cos (\frac{π}{6}) + cos (x) sin (\frac{π}{6})

化简

cos (\frac{π}{6}) : \frac{3}{2}

cos (\frac{π}{6})

使用以下普通恒等式:

cos (\frac{π}{6}) = \frac{3}{2}

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{3}{2}

= \frac{3}{2} sin (x) + sin (\frac{π}{6}) cos (x)

化简

sin (\frac{π}{6}) : \frac{1}{2}

sin (\frac{π}{6})

使用以下普通恒等式:

sin (\frac{π}{6}) = \frac{1}{2}

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}

= \frac{1}{2}

= \frac{3}{2} sin (x) + \frac{1}{2} cos (x)

= \frac{3}{2} sin (x) + \frac{1}{2} cos (x)

使用角和恒等式:

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

= cos (x) cos (\frac{π}{3}) - sin (x) sin (\frac{π}{3})

化简

cos (x) cos (\frac{π}{3}) - sin (x) sin (\frac{π}{3}) : \frac{1}{2} cos (x) - \frac{3}{2} sin (x)

cos (x) cos (\frac{π}{3}) - sin (x) sin (\frac{π}{3})

化简

cos (\frac{π}{3}) : \frac{1}{2}

cos (\frac{π}{3})

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{2}

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{1}{2}

= \frac{1}{2} cos (x) - sin (\frac{π}{3}) sin (x)

化简

sin (\frac{π}{3}) : \frac{3}{2}

sin (\frac{π}{3})

使用以下普通恒等式:

sin (\frac{π}{3}) = \frac{3}{2}

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}

= \frac{3}{2}

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

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

\frac{3}{2} sin (x) + \frac{1}{2} cos (x) + \frac{1}{2} cos (x) - \frac{3}{2} sin (x) = cos (2 x)

化简

\frac{3}{2} sin (x) + \frac{1}{2} cos (x) + \frac{1}{2} cos (x) - \frac{3}{2} sin (x) : cos (x)

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

对同类项分组

= \frac{1}{2} cos (x) + \frac{1}{2} cos (x) + \frac{3}{2} sin (x) - \frac{3}{2} sin (x)

同类项相加:

\frac{1}{2} cos (x) + \frac{1}{2} cos (x) = cos (x)

\frac{1}{2} cos (x) + \frac{1}{2} cos (x)

因式分解出通项

cos (x)

= cos (x) (\frac{1}{2} + \frac{1}{2})

\frac{1}{2} + \frac{1}{2} = 1

\frac{1}{2} + \frac{1}{2}

使用法则

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

= \frac{1 + 1}{2}

整理后得

= 1

= cos (x)

= cos (x) + \frac{3}{2} sin (x) - \frac{3}{2} sin (x)

同类项相加:

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

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

因式分解出通项

sin (x)

= sin (x) (\frac{3}{2} - \frac{3}{2})

\frac{3}{2} - \frac{3}{2} = 0

\frac{3}{2} - \frac{3}{2}

因为分母相等，所以合并分式:

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

= \frac{3 - 3}{2}

分解

3 - 3 : 0

3 - 3

因式分解出通项

3

= 3 (1 - 1)

整理后得

= 0

= \frac{0}{2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

= 0

= cos (x)

cos (x) = cos (2 x)

cos (x) = cos (2 x)

两边减去

cos (2 x)

cos (x) - cos (2 x) = 0

使用三角恒等式改写

- cos (2 x) + cos (x)

使用和差化积恒等式:

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

= - 2 sin (\frac{x + 2 x}{2}) sin (\frac{x - 2 x}{2})

化简

- 2 sin (\frac{x + 2 x}{2}) sin (\frac{x - 2 x}{2}) : 2 sin (\frac{x}{2}) sin (\frac{3 x}{2})

- 2 sin (\frac{x + 2 x}{2}) sin (\frac{x - 2 x}{2})

同类项相加:

x + 2 x = 3 x

= - 2 sin (\frac{3 x}{2}) sin (\frac{x - 2 x}{2})

\frac{x - 2 x}{2} = - \frac{x}{2}

\frac{x - 2 x}{2}

同类项相加:

x - 2 x = - x

= \frac{- x}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{x}{2}

= - 2 sin (\frac{3 x}{2}) sin (- \frac{x}{2})

使用负角恒等式:

sin (- x) = - sin (x)

= - 2 (- sin (\frac{x}{2})) sin (\frac{3 x}{2})

使用法则

- (- a) = a

= 2 sin (\frac{x}{2}) sin (\frac{3 x}{2})

= 2 sin (\frac{x}{2}) sin (\frac{3 x}{2})

2 sin (\frac{3 x}{2}) sin (\frac{x}{2}) = 0

分别求解每个部分

sin (\frac{3 x}{2}) = 0 or sin (\frac{x}{2}) = 0

sin (\frac{3 x}{2}) = 0 : x = \frac{4 πn}{3}, x = \frac{2 π}{3} + \frac{4 πn}{3}

sin (\frac{3 x}{2}) = 0

sin (\frac{3 x}{2}) = 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}

\frac{3 x}{2} = 0 + 2 πn, \frac{3 x}{2} = π + 2 πn

\frac{3 x}{2} = 0 + 2 πn, \frac{3 x}{2} = π + 2 πn

解

\frac{3 x}{2} = 0 + 2 πn : x = \frac{4 πn}{3}

\frac{3 x}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{3 x}{2} = 2 πn

在两边乘以

2

\frac{3 x}{2} = 2 πn

在两边乘以

2

\frac{2 \cdot 3 x}{2} = 2 \cdot 2 πn

化简

3 x = 4 πn

3 x = 4 πn

两边除以

3

3 x = 4 πn

两边除以

3

\frac{3 x}{3} = \frac{4 πn}{3}

化简

x = \frac{4 πn}{3}

x = \frac{4 πn}{3}

解

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

\frac{3 x}{2} = π + 2 πn

在两边乘以

2

\frac{3 x}{2} = π + 2 πn

在两边乘以

2

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

化简

3 x = 2 π + 4 πn

3 x = 2 π + 4 πn

两边除以

3

3 x = 2 π + 4 πn

两边除以

3

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

化简

x = \frac{2 π}{3} + \frac{4 πn}{3}

x = \frac{2 π}{3} + \frac{4 πn}{3}

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

sin (\frac{x}{2}) = 0 : x = 4 πn, x = 2 π + 4 πn

sin (\frac{x}{2}) = 0

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

\frac{x}{2} = 0 + 2 πn, \frac{x}{2} = π + 2 πn

\frac{x}{2} = 0 + 2 πn, \frac{x}{2} = π + 2 πn

解

\frac{x}{2} = 0 + 2 πn : x = 4 πn

\frac{x}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{x}{2} = 2 πn

在两边乘以

2

\frac{x}{2} = 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot 2 πn

化简

x = 4 πn

x = 4 πn

解

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

\frac{x}{2} = π + 2 πn

在两边乘以

2

\frac{x}{2} = π + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 π + 2 \cdot 2 πn

化简

x = 2 π + 4 πn

x = 2 π + 4 πn

x = 4 πn, x = 2 π + 4 πn

合并所有解

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

合并重叠的区间

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

作图

流行的例子

24sin(2t)-24cos(t)=0

24 sin (2 t) - 24 cos (t) = 0

8^{sin^2(x)}=4^{sin(x)-1/8}

8^{s i n^{2} (x)} = 4^{s i n (x) - \frac{1}{8}}

2 cos (2 x) = \frac{1}{2}

(sin(73))/(34)=(sin(d))/(29)

\frac{sin ( 7 3 ^{\circ} )}{34} = \frac{sin ( d )}{29}

solvefor θ,(90^2)/(250(32.2))=tan(θ)

so l v e f or θ, \frac{9 0 ^{2}}{250 ( 32.2 )} = tan (θ)