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受欢迎的三角函数 >

sin(x)+cos(x)=sec(x)

解答

sin (x) + cos (x) = sec (x)

解答

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

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

sin (x) + cos (x) = sec (x)

两边减去

sec (x)

sin (x) + cos (x) - sec (x) = 0

用 sin, cos 表示

cos (x) - sec (x) + sin (x)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

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

化简

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

cos (x) - \frac{1}{cos ( x )} + sin (x)

将项转换为分式:

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

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

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 ( x ) cos ( x ) - 1 + sin ( x ) cos ( x )}{cos ( x )}

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

cos (x) cos (x) - 1 + sin (x) cos (x)

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

cos (x) cos (x)

使用指数法则:

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

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

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= cos (x) sin (x) - sin^{2} (x)

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

分解

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

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

使用指数法则:

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

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

= - sin (x) sin (x) + sin (x) cos (x)

因式分解出通项

sin (x)

= sin (x) (- sin (x) + cos (x))

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

分别求解每个部分

sin (x) = 0 or - sin (x) + cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

- sin (x) + cos (x) = 0 : x = \frac{π}{4} + πn

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

使用三角恒等式改写

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

在两边除以

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

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

化简

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

使用基本三角恒等式:

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

- tan (x) + 1 = 0

- tan (x) + 1 = 0

将

1

到右边

- tan (x) + 1 = 0

两边减去

1

- tan (x) + 1 - 1 = 0 - 1

化简

- tan (x) = - 1

- tan (x) = - 1

两边除以

- 1

- tan (x) = - 1

两边除以

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

化简

tan (x) = 1

tan (x) = 1

tan (x) = 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 = \frac{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

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

作图

流行的例子

sin (x) = - 0.2761

sin (2 y) = 0

7sin^2(θ)-15sin(θ)+8=0

7 sin^{2} (θ) - 15 sin (θ) + 8 = 0

sin (5 t) = 0

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

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