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受欢迎的三角函数 >

solvefor t,x=acos^2(t)+bsin^2(t)

解答

求解

t, x = a cos^{2} (t) + b sin^{2} (t)

解答

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

求解步骤

x = a cos^{2} (t) + b sin^{2} (t)

交换两边

a cos^{2} (t) + b sin^{2} (t) = x

两边减去

x

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

使用三角恒等式改写

- x + cos^{2} (t) a + sin^{2} (t) b

使用毕达哥拉斯恒等式:

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

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

= - x + (1 - sin^{2} (t)) a + sin^{2} (t) b

- x + (1 - sin^{2} (t)) a + sin^{2} (t) b = 0

用替代法求解

- x + (1 - sin^{2} (t)) a + sin^{2} (t) b = 0

令

: sin (t) = u

- x + (1 - u^{2}) a + u^{2} b = 0

- x + (1 - u^{2}) a + u^{2} b = 0 : u = \frac{x - a}{- a + b}, u = - \frac{x - a}{- a + b}; - a + b \neq = 0

- x + (1 - u^{2}) a + u^{2} b = 0

将

x

到右边

- x + (1 - u^{2}) a + u^{2} b = 0

两边加上

x

- x + (1 - u^{2}) a + u^{2} b + x = 0 + x

化简

(1 - u^{2}) a + u^{2} b = x

(1 - u^{2}) a + u^{2} b = x

展开

(1 - u^{2}) a + u^{2} b : a - a u^{2} + b u^{2}

(1 - u^{2}) a + u^{2} b

= a (1 - u^{2}) + b u^{2}

乘开

a (1 - u^{2}) : a - a u^{2}

a (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = a, b = 1, c = u^{2}

= a \cdot 1 - a u^{2}

= 1 \cdot a - a u^{2}

乘以:

1 \cdot a = a

= a - a u^{2}

= a - a u^{2} + u^{2} b

= a - a u^{2} + b u^{2}

a - a u^{2} + b u^{2} = x

将

a

到右边

a - a u^{2} + b u^{2} = x

两边减去

a

a - a u^{2} + b u^{2} - a = x - a

化简

- a u^{2} + b u^{2} = x - a

- a u^{2} + b u^{2} = x - a

分解

- a u^{2} + b u^{2} : u^{2} (- a + b)

- a u^{2} + b u^{2}

因式分解出通项

u^{2}

= u^{2} (- a + b)

u^{2} (- a + b) = x - a

两边除以

- a + b; - a + b \neq = 0

u^{2} (- a + b) = x - a

两边除以

- a + b; - a + b \neq = 0

\frac{u ^{2} ( - a + b )}{- a + b} = \frac{x}{- a + b} - \frac{a}{- a + b}; - a + b \neq = 0

化简

\frac{u ^{2} ( - a + b )}{- a + b} = \frac{x}{- a + b} - \frac{a}{- a + b}

化简

\frac{u ^{2} ( - a + b )}{- a + b} : u^{2}

\frac{u ^{2} ( - a + b )}{- a + b}

约分:

- a + b

= u^{2}

化简

\frac{x}{- a + b} - \frac{a}{- a + b} : \frac{x - a}{- a + b}

\frac{x}{- a + b} - \frac{a}{- a + b}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

u^{2} = \frac{x - a}{- a + b}; - a + b \neq = 0

u^{2} = \frac{x - a}{- a + b}; - a + b \neq = 0

u^{2} = \frac{x - a}{- a + b}; - a + b \neq = 0

对于

x^{2} = f (a)

解为

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

u = \frac{x - a}{- a + b}, u = - \frac{x - a}{- a + b}; - a + b \neq = 0

u = sin (t)

代回

sin (t) = \frac{x - a}{- a + b}, sin (t) = - \frac{x - a}{- a + b}; - a + b \neq = 0

sin (t) = \frac{x - a}{- a + b}, sin (t) = - \frac{x - a}{- a + b}; - a + b \neq = 0

sin (t) = \frac{x - a}{- a + b} : t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn

sin (t) = \frac{x - a}{- a + b}

使用反三角函数性质

sin (t) = \frac{x - a}{- a + b}

sin (t) = \frac{x - a}{- a + b}

的通解

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

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn

sin (t) = - \frac{x - a}{- a + b} : t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

sin (t) = - \frac{x - a}{- a + b}

使用反三角函数性质

sin (t) = - \frac{x - a}{- a + b}

sin (t) = - \frac{x - a}{- a + b}

的通解

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

t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

合并所有解

t = arcsin (\frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = arcsin (- \frac{x - a}{- a + b}) + 2 πn, t = π + arcsin (\frac{x - a}{- a + b}) + 2 πn

作图

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