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受欢迎的三角函数 >

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

解答

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

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

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

用替代法求解

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

令

: cos (a) = u

1 + u^{2} = 2 u^{2}

1 + u^{2} = 2 u^{2} : u = 1, u = - 1

1 + u^{2} = 2 u^{2}

将

1

到右边

1 + u^{2} = 2 u^{2}

两边减去

1

1 + u^{2} - 1 = 2 u^{2} - 1

化简

u^{2} = 2 u^{2} - 1

u^{2} = 2 u^{2} - 1

将

2 u^{2}

para o lado esquerdo

u^{2} = 2 u^{2} - 1

两边减去

2 u^{2}

u^{2} - 2 u^{2} = 2 u^{2} - 1 - 2 u^{2}

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 1}

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = cos (a)

代回

cos (a) = 1, cos (a) = - 1

cos (a) = 1, cos (a) = - 1

cos (a) = 1 : a = 2 πn

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

cos (a) = - 1 : a = π + 2 πn

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

作图

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