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受欢迎的三角函数 >

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

解答

sin (x^{2} + 1) - cos (x^{2} + 1) = 0

解答

x = \frac{π + 4 πn - 4}{2}, x = - \frac{π + 4 πn - 4}{2}

+1

度数

x = 0^{\circ} + 98.02420 \dots^{\circ} n, x = 0^{\circ} - 98.02420 \dots^{\circ} n

求解步骤

sin (x^{2} + 1) - cos (x^{2} + 1) = 0

使用三角恒等式改写

sin (x^{2} + 1) - cos (x^{2} + 1) = 0

在两边除以

cos (x^{2} + 1), cos (x^{2} + 1) \neq = 0

\frac{sin ( x ^{2} + 1 ) - cos ( x ^{2} + 1 )}{cos ( x ^{2} + 1 )} = \frac{0}{cos ( x ^{2} + 1 )}

化简

\frac{sin ( x ^{2} + 1 )}{cos ( x ^{2} + 1 )} - 1 = 0

使用基本三角恒等式:

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

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

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

将

1

到右边

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

两边加上

1

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

化简

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

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

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

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x^{2} + 1 = \frac{π}{4} + πn

x^{2} + 1 = \frac{π}{4} + πn

解

x^{2} + 1 = \frac{π}{4} + πn : x = \frac{π + 4 πn - 4}{2}, x = - \frac{π + 4 πn - 4}{2}

x^{2} + 1 = \frac{π}{4} + πn

将

1

到右边

x^{2} + 1 = \frac{π}{4} + πn

两边减去

1

x^{2} + 1 - 1 = \frac{π}{4} + πn - 1

化简

x^{2} = \frac{π}{4} + πn - 1

x^{2} = \frac{π}{4} + πn - 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

x = \frac{π}{4} + πn - 1, x = - \frac{π}{4} + πn - 1

化简

\frac{π}{4} + πn - 1 : \frac{π + 4 πn - 4}{2}

\frac{π}{4} + πn - 1

化简

\frac{π}{4} + πn - 1 : \frac{π + 4 πn - 4}{4}

\frac{π}{4} + πn - 1

将项转换为分式:

πn = \frac{πn 4}{4}, 1 = \frac{1 \cdot 4}{4}

= \frac{π}{4} + \frac{πn \cdot 4}{4} - \frac{1 \cdot 4}{4}

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{π + πn \cdot 4 - 1 \cdot 4}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{π + 4 πn - 4}{4}

= \frac{π + πn \cdot 4 - 4}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{π + 4 πn - 4}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{π + 4 πn - 4}{2}

化简

- \frac{π}{4} + πn - 1 : - \frac{π + 4 πn - 4}{2}

- \frac{π}{4} + πn - 1

化简

\frac{π}{4} + πn - 1 : \frac{π + 4 πn - 4}{4}

\frac{π}{4} + πn - 1

将项转换为分式:

πn = \frac{πn 4}{4}, 1 = \frac{1 \cdot 4}{4}

= \frac{π}{4} + \frac{πn \cdot 4}{4} - \frac{1 \cdot 4}{4}

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{π + πn \cdot 4 - 1 \cdot 4}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{π + 4 πn - 4}{4}

= - \frac{π + 4 πn - 4}{4}

化简

\frac{π + πn \cdot 4 - 4}{4} : \frac{π + 4 πn - 4}{2}

\frac{π + πn \cdot 4 - 4}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{π + 4 πn - 4}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{π + 4 πn - 4}{2}

= - \frac{π + 4 πn - 4}{2}

x = \frac{π + 4 πn - 4}{2}, x = - \frac{π + 4 πn - 4}{2}

x = \frac{π + 4 πn - 4}{2}, x = - \frac{π + 4 πn - 4}{2}

作图

流行的例子

sin (8 x) = 0

sin(θ/2)=(sqrt(2))/2

sin (\frac{θ}{2}) = \frac{2}{2}

9 tan (3 x) = 9

sin (θ) = \frac{40}{41}

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

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