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受欢迎的三角函数 >

sin(4y)+sin(6y)+cos(y)=0,0<= y<= pi/2

解答

sin (4 y) + sin (6 y) + cos (y) = 0, 0 \leq y \leq \frac{π}{2}

解答

y = \frac{π}{2}, y = \frac{7 π}{30}, y = \frac{11 π}{30}

+1

度数

y = 9 0^{\circ}, y = 4 2^{\circ}, y = 6 6^{\circ}

求解步骤

sin (4 y) + sin (6 y) + cos (y) = 0, 0 \leq y \leq \frac{π}{2}

使用三角恒等式改写

cos (y) + sin (4 y) + sin (6 y)

使用和差化积恒等式:

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

= cos (y) + 2 sin (\frac{4 y + 6 y}{2}) cos (\frac{4 y - 6 y}{2})

2 sin (\frac{4 y + 6 y}{2}) cos (\frac{4 y - 6 y}{2}) = 2 cos (y) sin (5 y)

2 sin (\frac{4 y + 6 y}{2}) cos (\frac{4 y - 6 y}{2})

\frac{4 y + 6 y}{2} = 5 y

\frac{4 y + 6 y}{2}

同类项相加:

4 y + 6 y = 10 y

= \frac{10 y}{2}

数字相除:

\frac{10}{2} = 5

= 5 y

= 2 sin (5 y) cos (\frac{4 y - 6 y}{2})

\frac{4 y - 6 y}{2} = - y

\frac{4 y - 6 y}{2}

同类项相加:

4 y - 6 y = - 2 y

= \frac{- 2 y}{2}

使用分式法则:

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

= - \frac{2 y}{2}

数字相除:

\frac{2}{2} = 1

= - y

= 2 sin (5 y) cos (- y)

使用负角恒等式:

cos (- x) = cos (x)

= 2 cos (y) sin (5 y)

= cos (y) + 2 cos (y) sin (5 y)

cos (y) + 2 cos (y) sin (5 y) = 0

分解

cos (y) + 2 cos (y) sin (5 y) : cos (y) (2 sin (5 y) + 1)

cos (y) + 2 cos (y) sin (5 y)

因式分解出通项

cos (y)

= cos (y) (1 + 2 sin (5 y))

cos (y) (2 sin (5 y) + 1) = 0

分别求解每个部分

cos (y) = 0 or 2 sin (5 y) + 1 = 0

cos (y) = 0, 0 \leq y \leq \frac{π}{2} : y = \frac{π}{2}

cos (y) = 0, 0 \leq y \leq \frac{π}{2}

cos (y) = 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}

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

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

在

0 \leq y \leq \frac{π}{2}

范围内的解

y = \frac{π}{2}

2 sin (5 y) + 1 = 0, 0 \leq y \leq \frac{π}{2} : y = \frac{7 π}{30}, y = \frac{11 π}{30}

2 sin (5 y) + 1 = 0, 0 \leq y \leq \frac{π}{2}

将

1

到右边

2 sin (5 y) + 1 = 0

两边减去

1

2 sin (5 y) + 1 - 1 = 0 - 1

化简

2 sin (5 y) = - 1

2 sin (5 y) = - 1

两边除以

2

2 sin (5 y) = - 1

两边除以

2

\frac{2 sin ( 5 y )}{2} = \frac{- 1}{2}

化简

sin (5 y) = - \frac{1}{2}

sin (5 y) = - \frac{1}{2}

sin (5 y) = - \frac{1}{2}

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}

5 y = \frac{7 π}{6} + 2 πn, 5 y = \frac{11 π}{6} + 2 πn

5 y = \frac{7 π}{6} + 2 πn, 5 y = \frac{11 π}{6} + 2 πn

解

5 y = \frac{7 π}{6} + 2 πn : y = \frac{7 π}{30} + \frac{2 πn}{5}

5 y = \frac{7 π}{6} + 2 πn

两边除以

5

5 y = \frac{7 π}{6} + 2 πn

两边除以

5

\frac{5 y}{5} = \frac{\frac{7 π}{6}}{5} + \frac{2 πn}{5}

化简

\frac{5 y}{5} = \frac{\frac{7 π}{6}}{5} + \frac{2 πn}{5}

化简

\frac{5 y}{5} : y

\frac{5 y}{5}

数字相除:

\frac{5}{5} = 1

= y

化简

\frac{\frac{7 π}{6}}{5} + \frac{2 πn}{5} : \frac{7 π}{30} + \frac{2 πn}{5}

\frac{\frac{7 π}{6}}{5} + \frac{2 πn}{5}

\frac{\frac{7 π}{6}}{5} = \frac{7 π}{30}

\frac{\frac{7 π}{6}}{5}

使用分式法则:

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

= \frac{7 π}{6 \cdot 5}

数字相乘:

6 \cdot 5 = 30

= \frac{7 π}{30}

= \frac{7 π}{30} + \frac{2 πn}{5}

y = \frac{7 π}{30} + \frac{2 πn}{5}

y = \frac{7 π}{30} + \frac{2 πn}{5}

y = \frac{7 π}{30} + \frac{2 πn}{5}

解

5 y = \frac{11 π}{6} + 2 πn : y = \frac{11 π}{30} + \frac{2 πn}{5}

5 y = \frac{11 π}{6} + 2 πn

两边除以

5

5 y = \frac{11 π}{6} + 2 πn

两边除以

5

\frac{5 y}{5} = \frac{\frac{11 π}{6}}{5} + \frac{2 πn}{5}

化简

\frac{5 y}{5} = \frac{\frac{11 π}{6}}{5} + \frac{2 πn}{5}

化简

\frac{5 y}{5} : y

\frac{5 y}{5}

数字相除:

\frac{5}{5} = 1

= y

化简

\frac{\frac{11 π}{6}}{5} + \frac{2 πn}{5} : \frac{11 π}{30} + \frac{2 πn}{5}

\frac{\frac{11 π}{6}}{5} + \frac{2 πn}{5}

\frac{\frac{11 π}{6}}{5} = \frac{11 π}{30}

\frac{\frac{11 π}{6}}{5}

使用分式法则:

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

= \frac{11 π}{6 \cdot 5}

数字相乘:

6 \cdot 5 = 30

= \frac{11 π}{30}

= \frac{11 π}{30} + \frac{2 πn}{5}

y = \frac{11 π}{30} + \frac{2 πn}{5}

y = \frac{11 π}{30} + \frac{2 πn}{5}

y = \frac{11 π}{30} + \frac{2 πn}{5}

y = \frac{7 π}{30} + \frac{2 πn}{5}, y = \frac{11 π}{30} + \frac{2 πn}{5}

在

0 \leq y \leq \frac{π}{2}

范围内的解

y = \frac{7 π}{30}, y = \frac{11 π}{30}

合并所有解

y = \frac{π}{2}, y = \frac{7 π}{30}, y = \frac{11 π}{30}

作图

流行的例子

cos(θ)=-(sqrt(35))/6

cos (θ) = - \frac{35}{6}

5cos(x)+2sin(x)=0

5 cos (x) + 2 sin (x) = 0

5cos(x)+2sin(x)=2

5 cos (x) + 2 sin (x) = 2

sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

4sin(x)+3cos(x)=4

4 sin (x) + 3 cos (x) = 4