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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{4} + πn

+1

度数

x = 27 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

乘开

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

乘开

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

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

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = sin (x), c = 1, d = - cos (x)

= 1 \cdot 1 + 1 \cdot (- cos (x)) + sin (x) \cdot 1 + sin (x) (- cos (x))

使用加减运算法则

+ (- a) = - a

= 1 \cdot 1 - 1 \cdot cos (x) + 1 \cdot sin (x) - sin (x) cos (x)

化简

1 \cdot 1 - 1 \cdot cos (x) + 1 \cdot sin (x) - sin (x) cos (x) : 1 - cos (x) + sin (x) - sin (x) cos (x)

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

数字相乘:

1 \cdot 1 = 1

= 1 - 1 \cdot cos (x) + 1 \cdot sin (x) - sin (x) cos (x)

乘以:

1 \cdot cos (x) = cos (x)

= 1 - cos (x) + 1 \cdot sin (x) - sin (x) cos (x)

乘以:

1 \cdot sin (x) = sin (x)

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

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

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

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

化简

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

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

对同类项分组

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

1 - 1 = 0

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

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

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

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

分解

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

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

因式分解出通项

cos (x)

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

使用指数法则:

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

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

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

因式分解出通项

sin (x)

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

因式分解出通项

(1 + sin (x))

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

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

分别求解每个部分

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

1 + sin (x) = 0 : x = \frac{3 π}{2} + 2 πn

1 + sin (x) = 0

将

1

到右边

1 + sin (x) = 0

两边减去

1

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

化简

sin (x) = - 1

sin (x) = - 1

sin (x) = - 1

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 = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

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

cos (x) - sin (x) = 0

使用三角恒等式改写

cos (x) - sin (x) = 0

在两边除以

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

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

化简

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

使用基本三角恒等式:

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

1 - tan (x) = 0

1 - tan (x) = 0

将

1

到右边

1 - tan (x) = 0

两边减去

1

1 - tan (x) - 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 = \frac{3 π}{2} + 2 πn, x = \frac{π}{4} + πn

作图

流行的例子

tan(x)=1,-pi<x<= pi

tan (x) = 1, - π < x \leq π

1/(2cos^2(x-1))=(1+tan^2(x))/(2sec^2(x))

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

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

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

tan(x)=(87.092)/(86.676)

tan (x) = \frac{87.092}{86.676}

csc (x) = - \frac{3}{2}