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受欢迎的三角函数 >

solvefor A,cos(pi/3-A)=2sin(A)

解答

求解

A, cos (\frac{π}{3} - A) = 2 sin (A)

解答

A = 0.41528 \dots + πn

+1

度数

A = 23.79397 \dots^{\circ} + 18 0^{\circ} n

求解步骤

cos (\frac{π}{3} - A) = 2 sin (A)

使用三角恒等式改写

cos (\frac{π}{3} - A) = 2 sin (A)

使用三角恒等式改写

cos (\frac{π}{3} - A)

使用角差恒等式:

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

= cos (\frac{π}{3}) cos (A) + sin (\frac{π}{3}) sin (A)

化简

cos (\frac{π}{3}) cos (A) + sin (\frac{π}{3}) sin (A) : \frac{1}{2} cos (A) + \frac{3}{2} sin (A)

cos (\frac{π}{3}) cos (A) + sin (\frac{π}{3}) sin (A)

化简

cos (\frac{π}{3}) : \frac{1}{2}

cos (\frac{π}{3})

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{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{1}{2}

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

化简

sin (\frac{π}{3}) : \frac{3}{2}

sin (\frac{π}{3})

使用以下普通恒等式:

sin (\frac{π}{3}) = \frac{3}{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{3}{2}

= \frac{1}{2} cos (A) + \frac{3}{2} sin (A)

= \frac{1}{2} cos (A) + \frac{3}{2} sin (A)

\frac{1}{2} cos (A) + \frac{3}{2} sin (A) = 2 sin (A)

\frac{1}{2} cos (A) + \frac{3}{2} sin (A) = 2 sin (A)

两边减去

2 sin (A)

\frac{1}{2} cos (A) + \frac{3 - 4}{2} sin (A) = 0

化简

\frac{1}{2} cos (A) + \frac{3 - 4}{2} sin (A) : \frac{cos ( A ) + ( 3 - 4 ) sin ( A )}{2}

\frac{1}{2} cos (A) + \frac{3 - 4}{2} sin (A)

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

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

分式相乘:

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

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

乘以:

1 \cdot cos (A) = cos (A)

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

\frac{3 - 4}{2} sin (A) = \frac{( 3 - 4 ) sin ( A )}{2}

\frac{3 - 4}{2} sin (A)

分式相乘:

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

= \frac{( 3 - 4 ) sin ( A )}{2}

= \frac{cos ( A )}{2} + \frac{( 3 - 4 ) sin ( A )}{2}

使用法则

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

= \frac{cos ( A ) + ( 3 - 4 ) sin ( A )}{2}

\frac{cos ( A ) + ( 3 - 4 ) sin ( A )}{2} = 0

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

cos (A) + (3 - 4) sin (A) = 0

使用三角恒等式改写

cos (A) + (3 - 4) sin (A) = 0

在两边除以

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

\frac{cos ( A ) + ( 3 - 4 ) sin ( A )}{cos ( A )} = \frac{0}{cos ( A )}

化简

1 + \frac{3 sin ( A )}{cos ( A )} - \frac{4 sin ( A )}{cos ( A )} = 0

使用基本三角恒等式:

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

1 + (- 4 + 3) tan (A) = 0

1 + (- 4 + 3) tan (A) = 0

将

1

到右边

1 + (- 4 + 3) tan (A) = 0

两边减去

1

1 + (- 4 + 3) tan (A) - 1 = 0 - 1

化简

(- 4 + 3) tan (A) = - 1

(- 4 + 3) tan (A) = - 1

两边除以

- 4 + 3

(- 4 + 3) tan (A) = - 1

两边除以

- 4 + 3

\frac{( - 4 + 3 ) tan ( A )}{- 4 + 3} = \frac{- 1}{- 4 + 3}

化简

\frac{( - 4 + 3 ) tan ( A )}{- 4 + 3} = \frac{- 1}{- 4 + 3}

化简

\frac{( - 4 + 3 ) tan ( A )}{- 4 + 3} : tan (A)

\frac{( - 4 + 3 ) tan ( A )}{- 4 + 3}

约分:

- 4 + 3

= tan (A)

化简

\frac{- 1}{- 4 + 3} : \frac{4 + 3}{13}

\frac{- 1}{- 4 + 3}

使用分式法则:

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

- 4 + 3 = - (4 - 3)

= \frac{1}{4 - 3}

\frac{1}{4 - 3}

有理化:

\frac{4 + 3}{13}

\frac{1}{4 - 3}

乘以共轭根式

\frac{4 + 3}{4 + 3}

= \frac{1 \cdot ( 4 + 3 )}{( 4 - 3 ) ( 4 + 3 )}

1 \cdot (4 + 3) = 4 + 3

(4 - 3) (4 + 3) = 13

(4 - 3) (4 + 3)

使用平方差公式:

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

a = 4, b = 3

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

化简

4^{2} - (3)^{2} : 13

4^{2} - (3)^{2}

4^{2} = 16

4^{2}

4^{2} = 16

= 16

(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

= 16 - 3

数字相减:

16 - 3 = 13

= 13

= 13

= \frac{4 + 3}{13}

= \frac{4 + 3}{13}

tan (A) = \frac{4 + 3}{13}

tan (A) = \frac{4 + 3}{13}

tan (A) = \frac{4 + 3}{13}

使用反三角函数性质

tan (A) = \frac{4 + 3}{13}

tan (A) = \frac{4 + 3}{13}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

A = arctan (\frac{4 + 3}{13}) + πn

A = arctan (\frac{4 + 3}{13}) + πn

以小数形式表示解

A = 0.41528 \dots + πn

作图

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