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受欢迎的三角函数 >

cos(5t)cos(3t)= 1/2+sin(-5t)sin(3t)

解答

cos (5 t) cos (3 t) = \frac{1}{2} + sin (- 5 t) sin (3 t)

解答

t = \frac{π}{6} + πn, t = \frac{5 π}{6} + πn

+1

度数

t = 3 0^{\circ} + 18 0^{\circ} n, t = 15 0^{\circ} + 18 0^{\circ} n

求解步骤

cos (5 t) cos (3 t) = \frac{1}{2} + sin (- 5 t) sin (3 t)

使用三角恒等式改写

cos (5 t) cos (3 t) = \frac{1}{2} + sin (- 5 t) sin (3 t)

cos (5 t) cos (3 t) = \frac{1}{2} + (- sin (5 t)) sin (3 t)

化简

\frac{1}{2} + (- sin (5 t)) sin (3 t) : \frac{1}{2} - sin (5 t) sin (3 t)

\frac{1}{2} + (- sin (5 t)) sin (3 t)

去除括号:

(- a) = - a

= \frac{1}{2} - sin (5 t) sin (3 t)

cos (5 t) cos (3 t) = \frac{1}{2} - sin (5 t) sin (3 t)

cos (5 t) cos (3 t) = \frac{1}{2} - sin (5 t) sin (3 t)

两边减去

\frac{1}{2} - sin (5 t) sin (3 t)

cos (5 t) cos (3 t) - \frac{1}{2} + sin (5 t) sin (3 t) = 0

化简

cos (5 t) cos (3 t) - \frac{1}{2} + sin (5 t) sin (3 t) : \frac{2 cos ( 5 t ) cos ( 3 t ) - 1 + 2 sin ( 5 t ) sin ( 3 t )}{2}

cos (5 t) cos (3 t) - \frac{1}{2} + sin (5 t) sin (3 t)

将项转换为分式:

cos (5 t) cos (3 t) = \frac{cos ( 5 t ) cos ( 3 t ) 2}{2}, sin (5 t) sin (3 t) = \frac{sin ( 5 t ) sin ( 3 t ) 2}{2}

= \frac{cos ( 5 t ) cos ( 3 t ) \cdot 2}{2} - \frac{1}{2} + \frac{sin ( 5 t ) sin ( 3 t ) \cdot 2}{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{cos ( 5 t ) cos ( 3 t ) \cdot 2 - 1 + sin ( 5 t ) sin ( 3 t ) \cdot 2}{2}

\frac{2 cos ( 5 t ) cos ( 3 t ) - 1 + 2 sin ( 5 t ) sin ( 3 t )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 cos (5 t) cos (3 t) - 1 + 2 sin (5 t) sin (3 t) = 0

使用三角恒等式改写

- 1 + 2 cos (3 t) cos (5 t) + 2 sin (3 t) sin (5 t)

使用积化和差恒等式:

cos (s) cos (t) = \frac{1}{2} (cos (s - t) + cos (s + t))

= - 1 + 2 \cdot \frac{1}{2} (cos (3 t - 5 t) + cos (3 t + 5 t)) + 2 sin (3 t) sin (5 t)

2 \cdot \frac{1}{2} (cos (3 t - 5 t) + cos (3 t + 5 t)) = cos (2 t) + cos (8 t)

2 \cdot \frac{1}{2} (cos (3 t - 5 t) + cos (3 t + 5 t))

分式相乘:

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

= \frac{1 \cdot 2 ( cos ( 3 t - 5 t ) + cos ( 3 t + 5 t ) )}{2}

约分:

2

= 1 \cdot (cos (3 t - 5 t) + cos (3 t + 5 t))

同类项相加:

3 t - 5 t = - 2 t

= 1 \cdot (cos (- 2 t) + cos (3 t + 5 t))

同类项相加:

3 t + 5 t = 8 t

= 1 \cdot (cos (- 2 t) + cos (8 t))

整理后得

= cos (- 2 t) + cos (8 t)

使用负角恒等式:

cos (- x) = cos (x)

= cos (2 t) + cos (8 t)

= - 1 + cos (2 t) + cos (8 t) + 2 sin (3 t) sin (5 t)

使用积化和差恒等式:

sin (s) sin (t) = \frac{1}{2} (cos (s - t) - cos (s + t))

= - 1 + cos (2 t) + cos (8 t) + 2 \cdot \frac{1}{2} (cos (3 t - 5 t) - cos (3 t + 5 t))

化简

- 1 + cos (2 t) + cos (8 t) + 2 \cdot \frac{1}{2} (cos (3 t - 5 t) - cos (3 t + 5 t)) : 2 cos (2 t) - 1

- 1 + cos (2 t) + cos (8 t) + 2 \cdot \frac{1}{2} (cos (3 t - 5 t) - cos (3 t + 5 t))

2 \cdot \frac{1}{2} (cos (3 t - 5 t) - cos (3 t + 5 t)) = cos (2 t) - cos (8 t)

2 \cdot \frac{1}{2} (cos (3 t - 5 t) - cos (3 t + 5 t))

分式相乘:

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

= \frac{1 \cdot 2 ( cos ( 3 t - 5 t ) - cos ( 3 t + 5 t ) )}{2}

约分:

2

= 1 \cdot (cos (3 t - 5 t) - cos (3 t + 5 t))

同类项相加:

3 t - 5 t = - 2 t

= 1 \cdot (cos (- 2 t) - cos (3 t + 5 t))

同类项相加:

3 t + 5 t = 8 t

= 1 \cdot (cos (- 2 t) - cos (8 t))

整理后得

= cos (- 2 t) - cos (8 t)

使用负角恒等式:

cos (- x) = cos (x)

= cos (2 t) - cos (8 t)

= - 1 + cos (2 t) + cos (8 t) + cos (2 t) - cos (8 t)

对同类项分组

= cos (2 t) + cos (8 t) + cos (2 t) - cos (8 t) - 1

同类项相加:

cos (2 t) + cos (2 t) = 2 cos (2 t)

= 2 cos (2 t) + cos (8 t) - cos (8 t) - 1

同类项相加:

cos (8 t) - cos (8 t) = 0

= 2 cos (2 t) - 1

= 2 cos (2 t) - 1

- 1 + 2 cos (2 t) = 0

将

1

到右边

- 1 + 2 cos (2 t) = 0

两边加上

1

- 1 + 2 cos (2 t) + 1 = 0 + 1

化简

2 cos (2 t) = 1

2 cos (2 t) = 1

两边除以

2

2 cos (2 t) = 1

两边除以

2

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

化简

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

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

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

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}

2 t = \frac{π}{3} + 2 πn, 2 t = \frac{5 π}{3} + 2 πn

2 t = \frac{π}{3} + 2 πn, 2 t = \frac{5 π}{3} + 2 πn

解

2 t = \frac{π}{3} + 2 πn : t = \frac{π}{6} + πn

2 t = \frac{π}{3} + 2 πn

两边除以

2

2 t = \frac{π}{3} + 2 πn

两边除以

2

\frac{2 t}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 t}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 t}{2} : t

\frac{2 t}{2}

数字相除:

\frac{2}{2} = 1

= t

化简

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

使用分式法则:

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

= \frac{π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

t = \frac{π}{6} + πn

t = \frac{π}{6} + πn

t = \frac{π}{6} + πn

解

2 t = \frac{5 π}{3} + 2 πn : t = \frac{5 π}{6} + πn

2 t = \frac{5 π}{3} + 2 πn

两边除以

2

2 t = \frac{5 π}{3} + 2 πn

两边除以

2

\frac{2 t}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 t}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 t}{2} : t

\frac{2 t}{2}

数字相除:

\frac{2}{2} = 1

= t

化简

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

使用分式法则:

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

= \frac{5 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

t = \frac{5 π}{6} + πn

t = \frac{5 π}{6} + πn

t = \frac{5 π}{6} + πn

t = \frac{π}{6} + πn, t = \frac{5 π}{6} + πn

作图

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