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受欢迎的三角函数 >

sin(x+pi/4)=sqrt(2)cos(x+pi/4)

解答

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

解答

x = \frac{0.33983 \dots}{2} + πn

+1

度数

x = 9.73561 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

两边进行平方

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

使用三角恒等式改写

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

使用三角恒等式改写

sin (x + \frac{π}{4})

使用角和恒等式:

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

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

化简

sin (x) cos (\frac{π}{4}) + cos (x) sin (\frac{π}{4}) : \frac{2 sin ( x ) + 2 cos ( x )}{2}

sin (x) cos (\frac{π}{4}) + cos (x) sin (\frac{π}{4})

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

sin (x) cos (\frac{π}{4})

化简

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{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{2}{2}

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

分式相乘:

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

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

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

cos (x) sin (\frac{π}{4})

化简

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{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{2}{2}

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

分式相乘:

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

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

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

使用法则

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

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

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

使用角和恒等式:

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

= cos (x) cos (\frac{π}{4}) - sin (x) sin (\frac{π}{4})

化简

cos (x) cos (\frac{π}{4}) - sin (x) sin (\frac{π}{4}) : \frac{2 cos ( x ) - 2 sin ( x )}{2}

cos (x) cos (\frac{π}{4}) - sin (x) sin (\frac{π}{4})

cos (x) cos (\frac{π}{4}) = \frac{2 cos ( x )}{2}

cos (x) cos (\frac{π}{4})

化简

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{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{2}{2}

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

分式相乘:

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

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

sin (x) sin (\frac{π}{4}) = \frac{2 sin ( x )}{2}

sin (x) sin (\frac{π}{4})

化简

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{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{2}{2}

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

分式相乘:

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

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

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

使用法则

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

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

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

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

化简

(\frac{2 sin ( x ) + 2 cos ( x )}{2})^{2} : \frac{( sin ( x ) + cos ( x ) ) ^{2}}{2}

(\frac{2 sin ( x ) + 2 cos ( x )}{2})^{2}

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

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

因式分解出通项

2

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

消掉

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

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

使用根式运算法则:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

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

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{1}} = \frac{1}{2 ^{1 - \frac{1}{2}}}

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

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

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

使用根式运算法则:

a^{\frac{1}{n}} = n a

2^{\frac{1}{2}} = 2

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

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

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

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

化简

(2 \frac{2 cos ( x ) - 2 sin ( x )}{2})^{2} : (cos (x) - sin (x))^{2}

(2 \frac{2 cos ( x ) - 2 sin ( x )}{2})^{2}

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

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

因式分解出通项

2

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

消掉

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

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

使用根式运算法则:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

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

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{1}} = \frac{1}{2 ^{1 - \frac{1}{2}}}

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

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

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

使用根式运算法则:

a^{\frac{1}{n}} = n a

2^{\frac{1}{2}} = 2

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

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

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

乘

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

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

分式相乘:

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

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

约分:

2

= cos (x) - sin (x)

= (cos (x) - sin (x))^{2}

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

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

两边减去

(cos (x) - sin (x))^{2}

\frac{- sin ^{2} ( x ) + 6 cos ( x ) sin ( x ) - cos ^{2} ( x )}{2} = 0

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

- sin^{2} (x) + 6 cos (x) sin (x) - cos^{2} (x) = 0

使用三角恒等式改写

6 cos (x) sin (x) - 1

使用倍角公式:

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

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

= - 1 + 6 \cdot \frac{sin ( 2 x )}{2}

- 1 + 6 \cdot \frac{sin ( 2 x )}{2} = 0

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

6 \cdot \frac{sin ( 2 x )}{2}

分式相乘:

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

= \frac{sin ( 2 x ) \cdot 6}{2}

数字相除:

\frac{6}{2} = 3

= 3 sin (2 x)

- 1 + 3 sin (2 x) = 0

将

1

到右边

- 1 + 3 sin (2 x) = 0

两边加上

1

- 1 + 3 sin (2 x) + 1 = 0 + 1

化简

3 sin (2 x) = 1

3 sin (2 x) = 1

两边除以

3

3 sin (2 x) = 1

两边除以

3

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

化简

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

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

使用反三角函数性质

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

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

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 x = arcsin (\frac{1}{3}) + 2 πn, 2 x = π - arcsin (\frac{1}{3}) + 2 πn

2 x = arcsin (\frac{1}{3}) + 2 πn, 2 x = π - arcsin (\frac{1}{3}) + 2 πn

解

2 x = arcsin (\frac{1}{3}) + 2 πn : x = \frac{arcsin ( \frac{1}{3} )}{2} + πn

2 x = arcsin (\frac{1}{3}) + 2 πn

两边除以

2

2 x = arcsin (\frac{1}{3}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arcsin ( \frac{1}{3} )}{2} + πn

x = \frac{arcsin ( \frac{1}{3} )}{2} + πn

解

2 x = π - arcsin (\frac{1}{3}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

2 x = π - arcsin (\frac{1}{3}) + 2 πn

两边除以

2

2 x = π - arcsin (\frac{1}{3}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

x = \frac{arcsin ( \frac{1}{3} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

\frac{arcsin ( \frac{1}{3} )}{2} + πn

的解:

真

\frac{arcsin ( \frac{1}{3} )}{2} + πn

代入

n = 1

\frac{arcsin ( \frac{1}{3} )}{2} + π 1

对于

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

代入

x = \frac{arcsin ( \frac{1}{3} )}{2} + π 1

sin (\frac{arcsin ( \frac{1}{3} )}{2} + π 1 + \frac{π}{4}) = 2 cos (\frac{arcsin ( \frac{1}{3} )}{2} + π 1 + \frac{π}{4})

整理后得

- 0.81649 \dots = - 0.81649 \dots

\Rightarrow 真

检验

\frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

的解:

假

\frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

代入

n = 1

\frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + π 1

对于

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

代入

x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + π 1

sin (\frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + π 1 + \frac{π}{4}) = 2 cos (\frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + π 1 + \frac{π}{4})

整理后得

- 0.81649 \dots = 0.81649 \dots

\Rightarrow 假

x = \frac{arcsin ( \frac{1}{3} )}{2} + πn

以小数形式表示解

x = \frac{0.33983 \dots}{2} + πn

作图

流行的例子

5 = 7 cos (\frac{π}{3} t)

6 cos^{2} (x) - 4 = 0

csc (x) = 3.5

sin(x)=0,sin(x)=0

sin (x) = 0, sin (x) = 0

cos(x)=-7/9 , pi/2 <x<pi

cos (x) = - \frac{7}{9}, \frac{π}{2} < x < π