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受欢迎的三角函数 >

(sin^2(a)+1)/((1+tan^2(a)))=1

解答

\frac{sin ^{2} ( a ) + 1}{( 1 + tan ^{2} ( a ) )} = 1

解答

a = 2 πn, a = π + 2 πn

+1

度数

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

\frac{sin ^{2} ( a ) + 1}{( 1 + tan ^{2} ( a ) )} = 1

两边减去

1

\frac{sin ^{2} ( a ) + 1}{1 + tan ^{2} ( a )} - 1 = 0

化简

\frac{sin ^{2} ( a ) + 1}{1 + tan ^{2} ( a )} - 1 : \frac{sin ^{2} ( a ) - tan ^{2} ( a )}{1 + tan ^{2} ( a )}

\frac{sin ^{2} ( a ) + 1}{1 + tan ^{2} ( a )} - 1

将项转换为分式:

1 = \frac{1 ( 1 + tan ^{2} ( a ) )}{1 + tan ^{2} ( a )}

= \frac{sin ^{2} ( a ) + 1}{1 + tan ^{2} ( a )} - \frac{1 \cdot ( 1 + tan ^{2} ( a ) )}{1 + tan ^{2} ( a )}

因为分母相等，所以合并分式:

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

= \frac{sin ^{2} ( a ) + 1 - 1 \cdot ( 1 + tan ^{2} ( a ) )}{1 + tan ^{2} ( a )}

乘以:

1 \cdot (1 + tan^{2} (a)) = (1 + tan^{2} (a))

= \frac{sin ^{2} ( a ) + 1 - ( tan ^{2} ( a ) + 1 )}{1 + tan ^{2} ( a )}

乘开

sin^{2} (a) + 1 - (1 + tan^{2} (a)) : sin^{2} (a) - tan^{2} (a)

sin^{2} (a) + 1 - (1 + tan^{2} (a))

- (1 + tan^{2} (a)) : - 1 - tan^{2} (a)

- (1 + tan^{2} (a))

打开括号

= - (1) - (tan^{2} (a))

使用加减运算法则

+ (- a) = - a

= - 1 - tan^{2} (a)

= sin^{2} (a) + 1 - 1 - tan^{2} (a)

1 - 1 = 0

= sin^{2} (a) - tan^{2} (a)

= \frac{sin ^{2} ( a ) - tan ^{2} ( a )}{1 + tan ^{2} ( a )}

\frac{sin ^{2} ( a ) - tan ^{2} ( a )}{1 + tan ^{2} ( a )} = 0

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

sin^{2} (a) - tan^{2} (a) = 0

分解

sin^{2} (a) - tan^{2} (a) : (sin (a) + tan (a)) (sin (a) - tan (a))

sin^{2} (a) - tan^{2} (a)

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

sin^{2} (a) - tan^{2} (a) = (sin (a) + tan (a)) (sin (a) - tan (a))

= (sin (a) + tan (a)) (sin (a) - tan (a))

(sin (a) + tan (a)) (sin (a) - tan (a)) = 0

分别求解每个部分

sin (a) + tan (a) = 0 or sin (a) - tan (a) = 0

sin (a) + tan (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) + tan (a) = 0

用 sin, cos 表示

sin (a) + tan (a)

使用基本三角恒等式:

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

= sin (a) + \frac{sin ( a )}{cos ( a )}

化简

sin (a) + \frac{sin ( a )}{cos ( a )} : \frac{sin ( a ) cos ( a ) + sin ( a )}{cos ( a )}

sin (a) + \frac{sin ( a )}{cos ( a )}

将项转换为分式:

sin (a) = \frac{sin ( a ) cos ( a )}{cos ( a )}

= \frac{sin ( a ) cos ( a )}{cos ( a )} + \frac{sin ( a )}{cos ( a )}

因为分母相等，所以合并分式:

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

= \frac{sin ( a ) cos ( a ) + sin ( a )}{cos ( a )}

= \frac{sin ( a ) cos ( a ) + sin ( a )}{cos ( a )}

\frac{sin ( a ) + cos ( a ) sin ( a )}{cos ( a )} = 0

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

sin (a) + cos (a) sin (a) = 0

分解

sin (a) + cos (a) sin (a) : sin (a) (cos (a) + 1)

sin (a) + cos (a) sin (a)

因式分解出通项

sin (a)

= sin (a) (1 + cos (a))

sin (a) (cos (a) + 1) = 0

分别求解每个部分

sin (a) = 0 or cos (a) + 1 = 0

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

sin (a) = 0

的通解

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}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

解

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

cos (a) + 1 = 0 : a = π + 2 πn

cos (a) + 1 = 0

将

1

到右边

cos (a) + 1 = 0

两边减去

1

cos (a) + 1 - 1 = 0 - 1

化简

cos (a) = - 1

cos (a) = - 1

cos (a) = - 1

的通解

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}

a = π + 2 πn

a = π + 2 πn

合并所有解

a = 2 πn, a = π + 2 πn

sin (a) - tan (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) - tan (a) = 0

用 sin, cos 表示

sin (a) - tan (a)

使用基本三角恒等式:

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

= sin (a) - \frac{sin ( a )}{cos ( a )}

化简

sin (a) - \frac{sin ( a )}{cos ( a )} : \frac{sin ( a ) cos ( a ) - sin ( a )}{cos ( a )}

sin (a) - \frac{sin ( a )}{cos ( a )}

将项转换为分式:

sin (a) = \frac{sin ( a ) cos ( a )}{cos ( a )}

= \frac{sin ( a ) cos ( a )}{cos ( a )} - \frac{sin ( a )}{cos ( a )}

因为分母相等，所以合并分式:

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

= \frac{sin ( a ) cos ( a ) - sin ( a )}{cos ( a )}

= \frac{sin ( a ) cos ( a ) - sin ( a )}{cos ( a )}

\frac{- sin ( a ) + cos ( a ) sin ( a )}{cos ( a )} = 0

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

- sin (a) + cos (a) sin (a) = 0

分解

- sin (a) + cos (a) sin (a) : sin (a) (cos (a) - 1)

- sin (a) + cos (a) sin (a)

因式分解出通项

sin (a)

= sin (a) (- 1 + cos (a))

sin (a) (cos (a) - 1) = 0

分别求解每个部分

sin (a) = 0 or cos (a) - 1 = 0

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

sin (a) = 0

的通解

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}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

解

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

cos (a) - 1 = 0 : a = 2 πn

cos (a) - 1 = 0

将

1

到右边

cos (a) - 1 = 0

两边加上

1

cos (a) - 1 + 1 = 0 + 1

化简

cos (a) = 1

cos (a) = 1

cos (a) = 1

的通解

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}

a = 0 + 2 πn

a = 0 + 2 πn

解

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

合并所有解

a = 2 πn, a = π + 2 πn

合并所有解

a = 2 πn, a = π + 2 πn

作图

流行的例子

tan(a)= 5/9 ,b=6

tan (a) = \frac{5}{9}, b = 6

sin^3(x)sin^2(x)sin(x)=0

sin^{3} (x) sin^{2} (x) sin (x) = 0

2cos^2(x)+1=cos^2(x)

2 cos^{2} (x) + 1 = cos^{2} (x)

sin^2(x)cos^2(x)=((1-cos^4(x)))/8

sin^{2} (x) cos^{2} (x) = \frac{( 1 - cos ^{4} ( x ) )}{8}

sin^3(x)+sin(x)=2sin^{22}(x)

sin^{3} (x) + sin (x) = 2 sin^{22} (x)