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受欢迎的三角函数 >

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

解答

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

解答

a = 2 πn, a = π + 2 πn, a = 1.10714 \dots + πn

+1

度数

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n, a = 63.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

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

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

化简

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

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

将项转换为分式:

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

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

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

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

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

1 = \frac{1 cos ^{2} ( a )}{cos ^{2} ( a )}

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

乘以:

1 \cdot cos^{2} (a) = cos^{2} (a)

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

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

2 \cdot \frac{sin ( a )}{cos ( a )} = \frac{2 sin ( a )}{cos ( a )}

2 \cdot \frac{sin ( a )}{cos ( a )}

分式相乘:

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

= \frac{sin ( a ) \cdot 2}{cos ( a )}

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

乘

\frac{cos ^{2} ( a ) + sin ^{2} ( a )}{cos ^{2} ( a )} sin^{2} (a) : \frac{sin ^{2} ( a ) ( cos ^{2} ( a ) + sin ^{2} ( a ) )}{cos ^{2} ( a )}

\frac{cos ^{2} ( a ) + sin ^{2} ( a )}{cos ^{2} ( a )} sin^{2} (a)

分式相乘:

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

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

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

cos^{2} (a), cos (a)

的最小公倍数:

cos^{2} (a)

cos^{2} (a), cos (a)

最小公倍数 (LCM)

计算出由出现在

cos^{2} (a)

或

cos (a)

中的因子组成的表达式

= cos^{2} (a)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos^{2} (a)

对于

\frac{sin ( a ) \cdot 2}{cos ( a )} :

将分母和分子乘以

cos (a)

\frac{sin ( a ) \cdot 2}{cos ( a )} = \frac{sin ( a ) \cdot 2 cos ( a )}{cos ( a ) cos ( a )} = \frac{sin ( a ) \cdot 2 cos ( a )}{cos ^{2} ( a )}

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

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

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

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

(cos^{2} (a) + sin^{2} (a)) sin^{2} (a) - 2 cos (a) sin (a) = 0

分解

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

(cos^{2} (a) + sin^{2} (a)) sin^{2} (a) - 2 cos (a) sin (a)

使用指数法则:

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

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

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

因式分解出通项

sin (a)

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

sin (a) (sin (a) (cos^{2} (a) + sin^{2} (a)) - 2 cos (a)) = 0

使用三角恒等式改写

sin (a) (sin (a) (cos^{2} (a) + sin^{2} (a)) - 2 cos (a))

使用毕达哥拉斯恒等式:

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

= sin (a) (- 2 cos (a) + sin (a) \cdot 1)

化简

sin (a) (- 2 cos (a) + sin (a) \cdot 1) : sin (a) (- 2 cos (a) + sin (a))

sin (a) (- 2 cos (a) + sin (a) \cdot 1)

乘以:

sin (a) \cdot 1 = sin (a)

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

= sin (a) (- 2 cos (a) + sin (a))

sin (a) (- 2 cos (a) + sin (a)) = 0

分别求解每个部分

sin (a) = 0 or - 2 cos (a) + sin (a) = 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

- 2 cos (a) + sin (a) = 0 : a = arctan (2) + πn

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

使用三角恒等式改写

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

在两边除以

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

\frac{- 2 cos ( a ) + sin ( a )}{cos ( a )} = \frac{0}{cos ( a )}

化简

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

使用基本三角恒等式:

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

- 2 + tan (a) = 0

- 2 + tan (a) = 0

将

2

到右边

- 2 + tan (a) = 0

两边加上

2

- 2 + tan (a) + 2 = 0 + 2

化简

tan (a) = 2

tan (a) = 2

使用反三角函数性质

tan (a) = 2

tan (a) = 2

的通解

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

a = arctan (2) + πn

a = arctan (2) + πn

合并所有解

a = 2 πn, a = π + 2 πn, a = arctan (2) + πn

以小数形式表示解

a = 2 πn, a = π + 2 πn, a = 1.10714 \dots + πn

作图

流行的例子

(tan(a))/2 = 2/11

\frac{tan ( a )}{2} = \frac{2}{11}

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

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

cos(x)[3sin(x)-2]=0

cos (x) [3 sin (x) - 2] = 0

sin^3(x)+sin(x)-4=0

sin^{3} (x) + sin (x) - 4 = 0

solvefor a,sin^2(a)+cos^2(b)=1

so l v e f or a, sin^{2} (a) + cos^{2} (b) = 1