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受欢迎的三角函数 >

cos(2t)-sin(t)=0.5,0<= t<= 2pi

解答

cos (2 t) - sin (t) = 0.5, 0 \leq t \leq 2 π

解答

t = π + 0.94247 \dots, t = - 0.94247 \dots + 2 π, t = 0.31415 \dots, t = π - 0.31415 \dots

+1

度数

t = 23 4^{\circ}, t = 30 6^{\circ}, t = 1 8^{\circ}, t = 16 2^{\circ}

求解步骤

cos (2 t) - sin (t) = 0.5, 0 \leq t \leq 2 π

两边减去

0.5

cos (2 t) - sin (t) - 0.5 = 0

使用三角恒等式改写

- 0.5 + cos (2 t) - sin (t)

使用倍角公式:

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

= - 0.5 + 1 - 2 sin^{2} (t) - sin (t)

化简

= - 2 sin^{2} (t) - sin (t) + 0.5

0.5 - sin (t) - 2 sin^{2} (t) = 0

用替代法求解

0.5 - sin (t) - 2 sin^{2} (t) = 0

令

: sin (t) = u

0.5 - u - 2 u^{2} = 0

0.5 - u - 2 u^{2} = 0 : u = - \frac{1 + 5}{4}, u = \frac{5 - 1}{4}

0.5 - u - 2 u^{2} = 0

在两边乘以

10

0.5 - u - 2 u^{2} = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

0.5 \cdot 10 - u \cdot 10 - 2 u^{2} \cdot 10 = 0 \cdot 10

整理后得

5 - 10 u - 20 u^{2} = 0

5 - 10 u - 20 u^{2} = 0

改写成标准形式

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

- 20 u^{2} - 10 u + 5 = 0

使用求根公式求解

- 20 u^{2} - 10 u + 5 = 0

二次方程求根公式:

若

a = - 20, b = - 10, c = 5

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 20 ) \cdot 5}{2 ( - 20 )}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 20 ) \cdot 5}{2 ( - 20 )}

(- 10)^{2} - 4 (- 20) \cdot 5 = 105

(- 10)^{2} - 4 (- 20) \cdot 5

使用法则

- (- a) = a

= (- 10)^{2} + 4 \cdot 20 \cdot 5

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 10)^{2} = 1 0^{2}

= 1 0^{2} + 4 \cdot 20 \cdot 5

数字相乘:

4 \cdot 20 \cdot 5 = 400

= 1 0^{2} + 400

1 0^{2} = 100

= 100 + 400

数字相加:

100 + 400 = 500

= 500

500

质因数分解:

2^{2} \cdot 5^{3}

500

500

除以

2500 = 250 \cdot 2

= 2 \cdot 250

250

除以

2250 = 125 \cdot 2

= 2 \cdot 2 \cdot 125

125

除以

5125 = 25 \cdot 5

= 2 \cdot 2 \cdot 5 \cdot 25

25

除以

525 = 5 \cdot 5

= 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

= 2^{2} \cdot 5^{3}

= 5^{3} \cdot 2^{2}

使用指数法则:

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

= 2^{2} \cdot 5^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2} 5^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 2 \cdot 55

整理后得

= 105

u_{1, 2} = \frac{- ( - 10 ) \pm 10 5}{2 ( - 20 )}

将解分隔开

u_{1} = \frac{- ( - 10 ) + 10 5}{2 ( - 20 )}, u_{2} = \frac{- ( - 10 ) - 10 5}{2 ( - 20 )}

u = \frac{- ( - 10 ) + 10 5}{2 ( - 20 )} : - \frac{1 + 5}{4}

\frac{- ( - 10 ) + 10 5}{2 ( - 20 )}

去除括号:

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

= \frac{10 + 10 5}{- 2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{10 + 10 5}{- 40}

使用分式法则:

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

= - \frac{10 + 10 5}{40}

消掉

\frac{10 + 10 5}{40} : \frac{1 + 5}{4}

\frac{10 + 10 5}{40}

分解

10 + 105 : 10 (1 + 5)

10 + 105

改写为

= 10 \cdot 1 + 105

因式分解出通项

10

= 10 (1 + 5)

= \frac{10 ( 1 + 5 )}{40}

约分:

10

= \frac{1 + 5}{4}

= - \frac{1 + 5}{4}

u = \frac{- ( - 10 ) - 10 5}{2 ( - 20 )} : \frac{5 - 1}{4}

\frac{- ( - 10 ) - 10 5}{2 ( - 20 )}

去除括号:

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

= \frac{10 - 10 5}{- 2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{10 - 10 5}{- 40}

使用分式法则:

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

10 - 105 = - (105 - 10)

= \frac{10 5 - 10}{40}

分解

105 - 10 : 10 (5 - 1)

105 - 10

改写为

= 105 - 10 \cdot 1

因式分解出通项

10

= 10 (5 - 1)

= \frac{10 ( 5 - 1 )}{40}

约分:

10

= \frac{5 - 1}{4}

二次方程组的解是:

u = - \frac{1 + 5}{4}, u = \frac{5 - 1}{4}

u = sin (t)

代回

sin (t) = - \frac{1 + 5}{4}, sin (t) = \frac{5 - 1}{4}

sin (t) = - \frac{1 + 5}{4}, sin (t) = \frac{5 - 1}{4}

sin (t) = - \frac{1 + 5}{4}, 0 \leq t \leq 2 π : t = π + arcsin (\frac{1 + 5}{4}), t = - arcsin (\frac{1 + 5}{4}) + 2 π

sin (t) = - \frac{1 + 5}{4}, 0 \leq t \leq 2 π

使用反三角函数性质

sin (t) = - \frac{1 + 5}{4}

sin (t) = - \frac{1 + 5}{4}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

t = arcsin (- \frac{1 + 5}{4}) + 2 πn, t = π + arcsin (\frac{1 + 5}{4}) + 2 πn

t = arcsin (- \frac{1 + 5}{4}) + 2 πn, t = π + arcsin (\frac{1 + 5}{4}) + 2 πn

在

0 \leq t \leq 2 π

范围内的解

t = π + arcsin (\frac{1 + 5}{4}), t = - arcsin (\frac{1 + 5}{4}) + 2 π

sin (t) = \frac{5 - 1}{4}, 0 \leq t \leq 2 π : t = arcsin (\frac{5 - 1}{4}), t = π - arcsin (\frac{5 - 1}{4})

sin (t) = \frac{5 - 1}{4}, 0 \leq t \leq 2 π

使用反三角函数性质

sin (t) = \frac{5 - 1}{4}

sin (t) = \frac{5 - 1}{4}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

t = arcsin (\frac{5 - 1}{4}) + 2 πn, t = π - arcsin (\frac{5 - 1}{4}) + 2 πn

t = arcsin (\frac{5 - 1}{4}) + 2 πn, t = π - arcsin (\frac{5 - 1}{4}) + 2 πn

在

0 \leq t \leq 2 π

范围内的解

t = arcsin (\frac{5 - 1}{4}), t = π - arcsin (\frac{5 - 1}{4})

合并所有解

t = π + arcsin (\frac{1 + 5}{4}), t = - arcsin (\frac{1 + 5}{4}) + 2 π, t = arcsin (\frac{5 - 1}{4}), t = π - arcsin (\frac{5 - 1}{4})

以小数形式表示解

t = π + 0.94247 \dots, t = - 0.94247 \dots + 2 π, t = 0.31415 \dots, t = π - 0.31415 \dots

作图

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