zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

1/321200^2sin(2θ)=3000

解答

\frac{1}{32} \cdot 120 0^{2} \cdot sin (2 θ) = 3000

解答

θ = \frac{0.06671 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.06671 \dots}{2} + πn

+1

弧度

θ = 0.03335 \dots + πn, θ = \frac{π}{2} - \frac{0.06671 \dots}{2} + πn

求解步骤

\frac{1}{32} \cdot 120 0^{2} sin (2 θ) = 3000

化简

\frac{1}{32} \cdot 120 0^{2} : 45000

\frac{1}{32} \cdot 120 0^{2}

分式相乘:

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

= \frac{1 \cdot 120 0 ^{2}}{32}

乘以:

1 \cdot 120 0^{2} = 120 0^{2}

= \frac{120 0 ^{2}}{32}

分解

120 0^{2} : 2^{8} \cdot 5^{4} \cdot 3^{2}

因式分解

1200 = 2^{4} \cdot 5^{2} \cdot 3

= (2^{4} \cdot 5^{2} \cdot 3)^{2}

使用指数法则:

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

= 3^{2} (2^{4})^{2} (5^{2})^{2}

化简

(2^{4})^{2} : 2^{8}

(2^{4})^{2}

使用指数法则:

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

= 2^{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= 2^{8}

化简

(5^{2})^{2} : 5^{4}

(5^{2})^{2}

使用指数法则:

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

= 5^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= 5^{4}

= 2^{8} \cdot 5^{4} \cdot 3^{2}

分解

32 : 2^{5}

因式分解

32 = 2^{5}

= \frac{2 ^{8} \cdot 5 ^{4} \cdot 3 ^{2}}{2 ^{5}}

消掉

\frac{2 ^{8} \cdot 3 ^{2} \cdot 5 ^{4}}{2 ^{5}} : 5^{4} \cdot 2^{3} \cdot 3^{2}

\frac{2 ^{8} \cdot 3 ^{2} \cdot 5 ^{4}}{2 ^{5}}

使用指数法则:

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

\frac{2 ^{8}}{2 ^{5}} = 2^{8 - 5}

= 5^{4} \cdot 3^{2} \cdot 2^{8 - 5}

数字相减:

8 - 5 = 3

= 5^{4} \cdot 2^{3} \cdot 3^{2}

= 5^{4} \cdot 2^{3} \cdot 3^{2}

5^{4} = 625

= 2^{3} \cdot 3^{2} \cdot 625

2^{3} = 8

= 3^{2} \cdot 625 \cdot 8

3^{2} = 9

= 625 \cdot 8 \cdot 9

数字相乘:

625 \cdot 8 \cdot 9 = 45000

= 45000

45000 sin (2 θ) = 3000

两边除以

45000

45000 sin (2 θ) = 3000

两边除以

45000

\frac{45000 sin ( 2 θ )}{45000} = \frac{3000}{45000}

化简

sin (2 θ) = \frac{1}{15}

sin (2 θ) = \frac{1}{15}

使用反三角函数性质

sin (2 θ) = \frac{1}{15}

sin (2 θ) = \frac{1}{15}

的通解

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

2 θ = arcsin (\frac{1}{15}) + 2 πn, 2 θ = π - arcsin (\frac{1}{15}) + 2 πn

2 θ = arcsin (\frac{1}{15}) + 2 πn, 2 θ = π - arcsin (\frac{1}{15}) + 2 πn

解

2 θ = arcsin (\frac{1}{15}) + 2 πn : θ = \frac{arcsin ( \frac{1}{15} )}{2} + πn

2 θ = arcsin (\frac{1}{15}) + 2 πn

两边除以

2

2 θ = arcsin (\frac{1}{15}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arcsin ( \frac{1}{15} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arcsin ( \frac{1}{15} )}{2} + πn

θ = \frac{arcsin ( \frac{1}{15} )}{2} + πn

解

2 θ = π - arcsin (\frac{1}{15}) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{15} )}{2} + πn

2 θ = π - arcsin (\frac{1}{15}) + 2 πn

两边除以

2

2 θ = π - arcsin (\frac{1}{15}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{15} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{15} )}{2} + πn

θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{15} )}{2} + πn

θ = \frac{arcsin ( \frac{1}{15} )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{15} )}{2} + πn

以小数形式表示解

θ = \frac{0.06671 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.06671 \dots}{2} + πn

作图

流行的例子

1-cos^2(a)=sin^2(2a)

1 - cos^{2} (a) = sin^{2} (2 a)

4cos(2x)=-2sin(x)

4 cos (2 x) = - 2 sin (x)

tan^2(θ)+2tan(θ)=0,tan(2θ)+2tan(θ)=0

tan^{2} (θ) + 2 tan (θ) = 0, tan (2 θ) + 2 tan (θ) = 0

2cot(x)csc(x)-csc^2(x)=0

2 cot (x) csc (x) - csc^{2} (x) = 0

8tan(a)+3=2tan(a)+3

8 tan (a) + 3 = 2 tan (a) + 3