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受欢迎的三角函数 >

solvefor t,cos(2t)-sin(t)=sqrt(o)

解答

求解

t, cos (2 t) - sin (t) = o

解答

t = arcsin (- \frac{1 + - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 + - 8 o + 9}{4}) + 2 πn, t = arcsin (- \frac{1 - - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 - - 8 o + 9}{4}) + 2 πn

求解步骤

cos (2 t) - sin (t) = o

两边减去

o

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

使用三角恒等式改写

cos (2 t) - sin (t) - o

使用倍角公式:

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

= 1 - 2 sin^{2} (t) - sin (t) - o

1 - sin (t) - o - 2 sin^{2} (t) = 0

用替代法求解

1 - sin (t) - o - 2 sin^{2} (t) = 0

令

: sin (t) = u

1 - u - o - 2 u^{2} = 0

1 - u - o - 2 u^{2} = 0 : u = - \frac{1 + - 8 o + 9}{4}, u = - \frac{1 - - 8 o + 9}{4}

1 - u - o - 2 u^{2} = 0

改写成标准形式

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

- 2 u^{2} - u + 1 - o = 0

使用求根公式求解

- 2 u^{2} - u + 1 - o = 0

二次方程求根公式:

若

a = - 2, b = - 1, c = 1 - o

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) ( 1 - o )}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) ( 1 - o )}{2 ( - 2 )}

化简

(- 1)^{2} - 4 (- 2) (1 - o) : - 8 o + 9

(- 1)^{2} - 4 (- 2) (1 - o)

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 (1 - o)

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 (1 - o) = 8 (1 - o)

4 \cdot 2 (1 - o)

数字相乘:

4 \cdot 2 = 8

= 8 (- o + 1)

= 1 + 8 (- o + 1)

乘开

1 + 8 (1 - o) : - 8 o + 9

1 + 8 (1 - o)

乘开

8 (1 - o) : 8 - 8 o

8 (1 - o)

使用分配律:

a (b - c) = ab - a c

a = 8, b = 1, c = o

= 8 \cdot 1 - 8 o

数字相乘:

8 \cdot 1 = 8

= 8 - 8 o

= 1 + 8 - 8 o

数字相加:

1 + 8 = 9

= - 8 o + 9

= - 8 o + 9

u_{1, 2} = \frac{- ( - 1 ) \pm - 8 o + 9}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + - 8 o + 9}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - - 8 o + 9}{2 ( - 2 )}

u = \frac{- ( - 1 ) + - 8 o + 9}{2 ( - 2 )} : - \frac{1 + - 8 o + 9}{4}

\frac{- ( - 1 ) + - 8 o + 9}{2 ( - 2 )}

去除括号:

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

= \frac{1 + - 8 o + 9}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1 + - 8 o + 9}{- 4}

使用分式法则:

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

= - \frac{1 + - 8 o + 9}{4}

u = \frac{- ( - 1 ) - - 8 o + 9}{2 ( - 2 )} : - \frac{1 - - 8 o + 9}{4}

\frac{- ( - 1 ) - - 8 o + 9}{2 ( - 2 )}

去除括号:

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

= \frac{1 - - 8 o + 9}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1 - - 8 o + 9}{- 4}

使用分式法则:

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

= - \frac{1 - - 8 o + 9}{4}

二次方程组的解是:

u = - \frac{1 + - 8 o + 9}{4}, u = - \frac{1 - - 8 o + 9}{4}

u = sin (t)

代回

sin (t) = - \frac{1 + - 8 o + 9}{4}, sin (t) = - \frac{1 - - 8 o + 9}{4}

sin (t) = - \frac{1 + - 8 o + 9}{4}, sin (t) = - \frac{1 - - 8 o + 9}{4}

sin (t) = - \frac{1 + - 8 o + 9}{4} : t = arcsin (- \frac{1 + - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 + - 8 o + 9}{4}) + 2 πn

sin (t) = - \frac{1 + - 8 o + 9}{4}

使用反三角函数性质

sin (t) = - \frac{1 + - 8 o + 9}{4}

sin (t) = - \frac{1 + - 8 o + 9}{4}

的通解

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

t = arcsin (- \frac{1 + - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 + - 8 o + 9}{4}) + 2 πn

t = arcsin (- \frac{1 + - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 + - 8 o + 9}{4}) + 2 πn

sin (t) = - \frac{1 - - 8 o + 9}{4} : t = arcsin (- \frac{1 - - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 - - 8 o + 9}{4}) + 2 πn

sin (t) = - \frac{1 - - 8 o + 9}{4}

使用反三角函数性质

sin (t) = - \frac{1 - - 8 o + 9}{4}

sin (t) = - \frac{1 - - 8 o + 9}{4}

的通解

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

t = arcsin (- \frac{1 - - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 - - 8 o + 9}{4}) + 2 πn

t = arcsin (- \frac{1 - - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 - - 8 o + 9}{4}) + 2 πn

合并所有解

t = arcsin (- \frac{1 + - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 + - 8 o + 9}{4}) + 2 πn, t = arcsin (- \frac{1 - - 8 o + 9}{4}) + 2 πn, t = π + arcsin (\frac{1 - - 8 o + 9}{4}) + 2 πn

作图

流行的例子

cot (x) = - \frac{15}{8}

tan (x^{\circ}) = \frac{30}{73}

cos (x) = (\frac{1}{2})

-2cos(θ)+2sqrt(3)=sqrt(3)

- 2 cos (θ) + 23 = 3

solvefor x,4cos^3(x)=3cos(x)

so l v e f or x, 4 cos^{3} (x) = 3 cos (x)