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受欢迎的三角函数 >

-sqrt(3)+3=2sqrt(6)cos(θ)

解答

- 3 + 3 = 26 cos (θ)

解答

θ = 1.30899 \dots + 2 πn, θ = 2 π - 1.30899 \dots + 2 πn

+1

度数

θ = 7 5^{\circ} + 36 0^{\circ} n, θ = 28 5^{\circ} + 36 0^{\circ} n

求解步骤

- 3 + 3 = 26 cos (θ)

交换两边

26 cos (θ) = - 3 + 3

两边除以

26

26 cos (θ) = - 3 + 3

两边除以

26

\frac{2 6 cos ( θ )}{2 6} = - \frac{3}{2 6} + \frac{3}{2 6}

化简

\frac{2 6 cos ( θ )}{2 6} = - \frac{3}{2 6} + \frac{3}{2 6}

化简

\frac{2 6 cos ( θ )}{2 6} : cos (θ)

\frac{2 6 cos ( θ )}{2 6}

数字相除:

\frac{2}{2} = 1

= \frac{6 cos ( θ )}{6}

约分:

6

= cos (θ)

化简

- \frac{3}{2 6} + \frac{3}{2 6} : \frac{2 ( - 1 + 3 )}{4}

- \frac{3}{2 6} + \frac{3}{2 6}

使用法则

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

= \frac{- 3 + 3}{2 6}

分解

- 3 + 3 : 3 (- 1 + 3)

- 3 + 3

3 = 33

= - 3 + 33

因式分解出通项

3

= 3 (- 1 + 3)

= \frac{3 ( - 1 + 3 )}{2 6}

分解

6 : 23

因式分解

6 = 2 \cdot 3

= 2 \cdot 3

使用根式运算法则:

n ab = n a n b

= 23

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

约分:

3

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

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

有理化:

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

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

乘以共轭根式

\frac{2}{2}

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

222 = 2^{2}

222

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

同类项相加:

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

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

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

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

数字相加:

1 + 1 = 2

= 2^{2}

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

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

2^{2} = 4

= \frac{2 ( 3 - 1 )}{4}

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

使用反三角函数性质

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn, θ = 2 π - arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn

θ = arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn, θ = 2 π - arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn

以小数形式表示解

θ = 1.30899 \dots + 2 πn, θ = 2 π - 1.30899 \dots + 2 πn

作图

流行的例子

sin(pi/2-x)=-1,0<= x<= 360

sin (\frac{π}{2} - x) = - 1, 0^{\circ} \leq x \leq 36 0^{\circ}

2 sin (5 x) = - 1

4sin^2(x)+8cos(x)-7=0

4 sin^{2} (x) + 8 cos (x) - 7 = 0

6(cos(b))^2+cos(b)-1=0

6 (cos (b))^{2} + cos (b) - 1 = 0

sin(θ)=(1.33)/(2.42)

sin (θ) = \frac{1.33}{2.42}