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 >

sin(piy)=cos(1)sin(pi/6)-1/2

解答

sin (π y) = cos (1) sin (\frac{π}{6}) - \frac{1}{2}

解答

y = \frac{- 0.23192 \dots}{π} + 2 n, y = 1 + \frac{0.23192 \dots}{π} + 2 n

+1

度数

y = - 4.22975 \dots^{\circ} + 114.59155 \dots^{\circ} n, y = 61.52553 \dots^{\circ} + 114.59155 \dots^{\circ} n

求解步骤

sin (π y) = cos (1) sin (\frac{π}{6}) - \frac{1}{2}

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

使用以下普通恒等式:

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

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{1}{2}

= \frac{1}{2}

sin (π y) = cos (1) \frac{1}{2} - \frac{1}{2}

使用反三角函数性质

sin (π y) = cos (1) \frac{1}{2} - \frac{1}{2}

sin (π y) = cos (1) \frac{1}{2} - \frac{1}{2}

的通解

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

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn, π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn, π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

解

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn : y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn

两边除以

π

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn

两边除以

π

\frac{π y}{π} = \frac{arcsin ( cos ( 1 ) \frac{1}{2} - \frac{1}{2} )}{π} + \frac{2 πn}{π}

化简

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n

解

π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn : y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

两边除以

π

π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

两边除以

π

\frac{π y}{π} = \frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π}

化简

\frac{π y}{π} = \frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π}

化简

\frac{π y}{π} : y

\frac{π y}{π}

约分:

π

= y

化简

\frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π} : 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

\frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π}

使用法则

\frac{a}{a} = 1

\frac{π}{π} = 1

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

消掉

\frac{2 πn}{π} : 2 n

\frac{2 πn}{π}

约分:

π

= 2 n

= 1 + \frac{arcsin ( \frac{1}{2} - \frac{1}{2} cos ( 1 ) )}{π} + 2 n

y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n, y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

以小数形式表示解

y = \frac{- 0.23192 \dots}{π} + 2 n, y = 1 + \frac{0.23192 \dots}{π} + 2 n

作图

流行的例子

(2sin(x)-1)(cos(x)-sqrt(2))=0

(2 sin (x) - 1) (cos (x) - 2) = 0

sin (a) = \frac{9}{41}

sin (x) = \frac{10}{300}

cos(216)=cos^2(a)-sin^2(a)

cos (21 6^{\circ}) = cos^{2} (a) - sin^{2} (a)

csc(-x)-1=cot^2(x)

csc (- x) - 1 = cot^{2} (x)