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受欢迎的三角函数 >

8sin(t)+8cos(t)=8

解答

8 sin (t) + 8 cos (t) = 8

解答

t = \frac{π}{2} + 2 πn, t = 2 πn

+1

度数

t = 9 0^{\circ} + 36 0^{\circ} n, t = 0^{\circ} + 36 0^{\circ} n

求解步骤

8 sin (t) + 8 cos (t) = 8

两边减去

8 cos (t)

8 sin (t) = 8 - 8 cos (t)

两边进行平方

(8 sin (t))^{2} = (8 - 8 cos (t))^{2}

两边减去

(8 - 8 cos (t))^{2}

64 sin^{2} (t) - 64 + 128 cos (t) - 64 cos^{2} (t) = 0

使用三角恒等式改写

- 64 + 128 cos (t) - 64 cos^{2} (t) + 64 sin^{2} (t)

使用毕达哥拉斯恒等式:

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

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

= - 64 + 128 cos (t) - 64 cos^{2} (t) + 64 (1 - cos^{2} (t))

化简

- 64 + 128 cos (t) - 64 cos^{2} (t) + 64 (1 - cos^{2} (t)) : 128 cos (t) - 128 cos^{2} (t)

- 64 + 128 cos (t) - 64 cos^{2} (t) + 64 (1 - cos^{2} (t))

乘开

64 (1 - cos^{2} (t)) : 64 - 64 cos^{2} (t)

64 (1 - cos^{2} (t))

使用分配律:

a (b - c) = ab - a c

a = 64, b = 1, c = cos^{2} (t)

= 64 \cdot 1 - 64 cos^{2} (t)

数字相乘:

64 \cdot 1 = 64

= 64 - 64 cos^{2} (t)

= - 64 + 128 cos (t) - 64 cos^{2} (t) + 64 - 64 cos^{2} (t)

化简

- 64 + 128 cos (t) - 64 cos^{2} (t) + 64 - 64 cos^{2} (t) : 128 cos (t) - 128 cos^{2} (t)

- 64 + 128 cos (t) - 64 cos^{2} (t) + 64 - 64 cos^{2} (t)

对同类项分组

= 128 cos (t) - 64 cos^{2} (t) - 64 cos^{2} (t) - 64 + 64

同类项相加:

- 64 cos^{2} (t) - 64 cos^{2} (t) = - 128 cos^{2} (t)

= 128 cos (t) - 128 cos^{2} (t) - 64 + 64

- 64 + 64 = 0

= 128 cos (t) - 128 cos^{2} (t)

= 128 cos (t) - 128 cos^{2} (t)

= 128 cos (t) - 128 cos^{2} (t)

128 cos (t) - 128 cos^{2} (t) = 0

用替代法求解

128 cos (t) - 128 cos^{2} (t) = 0

令

: cos (t) = u

128 u - 128 u^{2} = 0

128 u - 128 u^{2} = 0 : u = 0, u = 1

128 u - 128 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

- 128 u^{2} + 128 u = 0

使用求根公式求解

- 128 u^{2} + 128 u = 0

二次方程求根公式:

若

a = - 128, b = 128, c = 0

u_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 128 ) \cdot 0}{2 ( - 128 )}

u_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 128 ) \cdot 0}{2 ( - 128 )}

12 8^{2} - 4 (- 128) \cdot 0 = 128

12 8^{2} - 4 (- 128) \cdot 0

使用法则

- (- a) = a

= 12 8^{2} + 4 \cdot 128 \cdot 0

使用法则

0 \cdot a = 0

= 12 8^{2} + 0

12 8^{2} + 0 = 12 8^{2}

= 12 8^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 128

u_{1, 2} = \frac{- 128 \pm 128}{2 ( - 128 )}

将解分隔开

u_{1} = \frac{- 128 + 128}{2 ( - 128 )}, u_{2} = \frac{- 128 - 128}{2 ( - 128 )}

u = \frac{- 128 + 128}{2 ( - 128 )} : 0

\frac{- 128 + 128}{2 ( - 128 )}

去除括号:

(- a) = - a

= \frac{- 128 + 128}{- 2 \cdot 128}

数字相加/相减:

- 128 + 128 = 0

= \frac{0}{- 2 \cdot 128}

数字相乘:

2 \cdot 128 = 256

= \frac{0}{- 256}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{256}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 128 - 128}{2 ( - 128 )} : 1

\frac{- 128 - 128}{2 ( - 128 )}

去除括号:

(- a) = - a

= \frac{- 128 - 128}{- 2 \cdot 128}

数字相减:

- 128 - 128 = - 256

= \frac{- 256}{- 2 \cdot 128}

数字相乘:

2 \cdot 128 = 256

= \frac{- 256}{- 256}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{256}{256}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 0, u = 1

u = cos (t)

代回

cos (t) = 0, cos (t) = 1

cos (t) = 0, cos (t) = 1

cos (t) = 0 : t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 0

cos (t) = 0

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}

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 1 : t = 2 πn

cos (t) = 1

cos (t) = 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}

t = 0 + 2 πn

t = 0 + 2 πn

解

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn

合并所有解

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn, t = 2 πn

将解代入原方程进行验证

将它们代入

8 sin (t) + 8 cos (t) = 8

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{2} + 2 πn

的解:

真

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

8 sin (t) + 8 cos (t) = 8

代入

t = \frac{π}{2} + 2 π 1

8 sin (\frac{π}{2} + 2 π 1) + 8 cos (\frac{π}{2} + 2 π 1) = 8

整理后得

8 = 8

\Rightarrow 真

检验

\frac{3 π}{2} + 2 πn

的解:

假

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

8 sin (t) + 8 cos (t) = 8

代入

t = \frac{3 π}{2} + 2 π 1

8 sin (\frac{3 π}{2} + 2 π 1) + 8 cos (\frac{3 π}{2} + 2 π 1) = 8

整理后得

- 8 = 8

\Rightarrow 假

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

8 sin (t) + 8 cos (t) = 8

代入

t = 2 π 1

8 sin (2 π 1) + 8 cos (2 π 1) = 8

整理后得

8 = 8

\Rightarrow 真

t = \frac{π}{2} + 2 πn, t = 2 πn

作图

流行的例子

5 sin (\frac{π}{3} x) = 2

4cos^2(x)-5cos(x)+1=0

4 cos^{2} (x) - 5 cos (x) + 1 = 0

cos(x+60)=sin(x)

cos (x + 6 0^{\circ}) = sin (x)

csc^2(x)=2cot^2(x)

csc^{2} (x) = 2 cot^{2} (x)

2sin^2(x)+sin(2x)=0

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