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受欢迎的三角函数 >

solvefor x,sin(x)cos(x)=sin(x+p)cos(x+p)

解答

求解

x, sin (x) cos (x) = sin (x + p) cos (x + p)

解答

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

求解步骤

sin (x) cos (x) = sin (x + p) cos (x + p)

两边减去

sin (x + p) cos (x + p)

sin (x) cos (x) - sin (x + p) cos (x + p) = 0

使用三角恒等式改写

- cos (p + x) sin (p + x) + cos (x) sin (x)

使用倍角公式:

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

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

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

使用法则

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

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

使用和差化积恒等式:

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

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

化简

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

\frac{2 sin ( \frac{2 x - ( p + x ) \cdot 2}{2} ) cos ( \frac{2 x + ( p + x ) \cdot 2}{2} )}{2}

数字相除:

\frac{2}{2} = 1

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

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

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

分解

2 x - 2 (x + p) : - 2 p

2 x - 2 (x + p)

因式分解出通项

2

= 2 (x - (p + x))

乘开

x - (x + p) : - p

x - (p + x)

- (p + x) : - p - x

- (p + x)

打开括号

= - p - x

使用加减运算法则

+ (- a) = - a

= - p - x

= x - p - x

化简

x - p - x : - p

x - p - x

对同类项分组

= x - x - p

同类项相加:

x - x = 0

= - p

= - p

= 2 (- p)

整理后得

= - 2 p

= - \frac{2 p}{2}

数字相除:

\frac{2}{2} = 1

= - p

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

\frac{2 x + 2 ( x + p )}{2} = 2 x + p

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

分解

2 x + 2 (x + p) : 2 (2 x + p)

2 x + 2 (x + p)

因式分解出通项

2

= 2 (x + p + x)

整理后得

= 2 (2 x + p)

= \frac{2 ( 2 x + p )}{2}

数字相除:

\frac{2}{2} = 1

= 2 x + p

= sin (- p) cos (2 x + p)

= sin (- p) cos (2 x + p)

sin (- p) cos (2 x + p) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

cos (2 x + p) = 0

cos (2 x + p) = 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}

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

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

解

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

2 x + p = \frac{π}{2} + 2 πn

将

p

到右边

2 x + p = \frac{π}{2} + 2 πn

两边减去

p

2 x + p - p = \frac{π}{2} + 2 πn - p

化简

2 x = \frac{π}{2} + 2 πn - p

2 x = \frac{π}{2} + 2 πn - p

两边除以

2

2 x = \frac{π}{2} + 2 πn - p

两边除以

2

\frac{2 x}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} - \frac{p}{2}

化简

\frac{2 x}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} - \frac{p}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} - \frac{p}{2} : \frac{π}{4} + πn - \frac{p}{2}

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} - \frac{p}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn - \frac{p}{2}

x = \frac{π}{4} + πn - \frac{p}{2}

x = \frac{π}{4} + πn - \frac{p}{2}

x = \frac{π}{4} + πn - \frac{p}{2}

解

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

2 x + p = \frac{3 π}{2} + 2 πn

将

p

到右边

2 x + p = \frac{3 π}{2} + 2 πn

两边减去

p

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

化简

2 x = \frac{3 π}{2} + 2 πn - p

2 x = \frac{3 π}{2} + 2 πn - p

两边除以

2

2 x = \frac{3 π}{2} + 2 πn - p

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} - \frac{p}{2} : \frac{3 π}{4} + πn - \frac{p}{2}

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

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

使用分式法则:

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

= \frac{3 π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn - \frac{p}{2}

x = \frac{3 π}{4} + πn - \frac{p}{2}

x = \frac{3 π}{4} + πn - \frac{p}{2}

x = \frac{3 π}{4} + πn - \frac{p}{2}

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

作图

流行的例子

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sin (θ) = \frac{3}{5}, cos (2 θ)

11^2=6^2+7^2-2*6*7*cos(x)

1 1^{2} = 6^{2} + 7^{2} - 2 \cdot 6 \cdot 7 \cdot cos (x)

2cos^2(x)+cos(x)=1,0<= x<= 2pi

2 cos^{2} (x) + cos (x) = 1, 0 \leq x \leq 2 π

sin(θ/2)=(7.83)/(2(4.35))

sin (\frac{θ}{2}) = \frac{7.83}{2 ( 4.35 )}

-4sin(c)-2=sin(c)+1

- 4 sin (c) - 2 = sin (c) + 1