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受欢迎的三角函数 >

cos(x+60)=-sin(x)

解答

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

解答

x = - 1.30899 \dots + 18 0^{\circ} n

+1

弧度

x = - 1.30899 \dots + πn

求解步骤

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

使用三角恒等式改写

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

使用三角恒等式改写

cos (x + 6 0^{\circ})

使用角和恒等式:

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

= cos (x) cos (6 0^{\circ}) - sin (x) sin (6 0^{\circ})

化简

cos (x) cos (6 0^{\circ}) - sin (x) sin (6 0^{\circ}) : \frac{1}{2} cos (x) - \frac{3}{2} sin (x)

cos (x) cos (6 0^{\circ}) - sin (x) sin (6 0^{\circ})

化简

cos (6 0^{\circ}) : \frac{1}{2}

cos (6 0^{\circ})

使用以下普通恒等式:

cos (6 0^{\circ}) = \frac{1}{2}

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 (6 0^{\circ}) sin (x)

化简

sin (6 0^{\circ}) : \frac{3}{2}

sin (6 0^{\circ})

使用以下普通恒等式:

sin (6 0^{\circ}) = \frac{3}{2}

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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{1}{2} cos (x) - \frac{3}{2} sin (x) = - sin (x)

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

两边减去

- sin (x)

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

化简

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

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

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

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

分式相乘:

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

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

乘以:

1 \cdot cos (x) = cos (x)

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

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

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

分式相乘:

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

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

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

使用法则

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

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

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

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

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

使用三角恒等式改写

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

在两边除以

cos (x), cos (x) \neq = 0

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

化简

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

使用基本三角恒等式:

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

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

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

将

1

到右边

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

两边减去

1

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

化简

(2 - 3) tan (x) = - 1

(2 - 3) tan (x) = - 1

两边除以

2 - 3

(2 - 3) tan (x) = - 1

两边除以

2 - 3

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

化简

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

化简

\frac{( 2 - 3 ) tan ( x )}{2 - 3} : tan (x)

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

约分:

2 - 3

= tan (x)

化简

\frac{- 1}{2 - 3} : - 2 - 3

\frac{- 1}{2 - 3}

使用分式法则:

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

= - \frac{1}{2 - 3}

- \frac{1}{2 - 3}

有理化:

- 2 - 3

- \frac{1}{2 - 3}

乘以共轭根式

\frac{2 + 3}{2 + 3}

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

1 \cdot (2 + 3) = 2 + 3

(2 - 3) (2 + 3) = 1

(2 - 3) (2 + 3)

使用平方差公式:

(a - b) (a + b) = a^{2} - b^{2}

a = 2, b = 3

= 2^{2} - (3)^{2}

化简

2^{2} - (3)^{2} : 1

2^{2} - (3)^{2}

2^{2} = 4

2^{2}

2^{2} = 4

= 4

(3)^{2} = 3

(3)^{2}

使用根式运算法则:

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

= (3^{\frac{1}{2}})^{2}

使用指数法则:

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

= 3^{\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

= 3

= 4 - 3

数字相减:

4 - 3 = 1

= 1

= 1

= - \frac{2 + 3}{1}

使用法则

\frac{a}{1} = a

= - (2 + 3)

打开括号

= - (2) - (3)

使用加减运算法则

+ (- a) = - a

= - 2 - 3

= - 2 - 3

tan (x) = - 2 - 3

tan (x) = - 2 - 3

tan (x) = - 2 - 3

使用反三角函数性质

tan (x) = - 2 - 3

tan (x) = - 2 - 3

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + 18 0^{\circ} n

x = arctan (- 2 - 3) + 18 0^{\circ} n

x = arctan (- 2 - 3) + 18 0^{\circ} n

以小数形式表示解

x = - 1.30899 \dots + 18 0^{\circ} n

作图

流行的例子

sin(40+x)=cos(5x+10)

sin (4 0^{\circ} + x) = cos (5 x + 10)

cos (a) = \frac{5}{8}

tan(θ)sin^2(θ)=tan(θ)

tan (θ) sin^{2} (θ) = tan (θ)

cos (θ) = \frac{5}{8}

cot(θ)+sqrt(3)=0,0<= θ<= 2pi

cot (θ) + 3 = 0, 0 \leq θ \leq 2 π