- #1

Werg22

- 1,427

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y=cos(x)

x=cos(y)

cos^-1(x)=cos(x)

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- Thread starter Werg22
- Start date

- #1

Werg22

- 1,427

- 1

y=cos(x)

x=cos(y)

cos^-1(x)=cos(x)

- #2

VietDao29

Homework Helper

- 1,426

- 3

You can use Newton's method to solve this:

[tex]x_n = \frac{f(x_{n - 1})}{f'(x_{n - 1})}[/tex]

And the solution x is:

[tex]x = \lim_{n \rightarrow \infty} x_n[/tex]

Newton's method

You can change the equation a bit so it's easier to take the dirivative of the function:

Since the graph of Arccos(x) is the reflection of the graph Cos(x) across the line y = x.

So the intersection of the two graph Arccos(x) and Cos(x) is right on the line y = x. So the equation can be changed to:

[tex]\cos(x) = x \Leftrightarrow \cos(x) - x = 0[/tex]

Let f(x) = cos(x) - x.

So f'(x) = -sin(x) - 1.

Using the formula, you have:

[tex]x_n = \frac{\cos (x_{n - 1}) - x_{n - 1}}{-\sin (x_{n - 1}) - 1}[/tex]

Since the two graph cos(x) and x will cut each other at some x lies between 0, and pi. So you just choose [tex]x_0 \in [0, \ \pi][/tex], eg: x_0 = 0.5,...

Viet Dao,

[tex]x_n = \frac{f(x_{n - 1})}{f'(x_{n - 1})}[/tex]

And the solution x is:

[tex]x = \lim_{n \rightarrow \infty} x_n[/tex]

Newton's method

You can change the equation a bit so it's easier to take the dirivative of the function:

Since the graph of Arccos(x) is the reflection of the graph Cos(x) across the line y = x.

So the intersection of the two graph Arccos(x) and Cos(x) is right on the line y = x. So the equation can be changed to:

[tex]\cos(x) = x \Leftrightarrow \cos(x) - x = 0[/tex]

Let f(x) = cos(x) - x.

So f'(x) = -sin(x) - 1.

Using the formula, you have:

[tex]x_n = \frac{\cos (x_{n - 1}) - x_{n - 1}}{-\sin (x_{n - 1}) - 1}[/tex]

Since the two graph cos(x) and x will cut each other at some x lies between 0, and pi. So you just choose [tex]x_0 \in [0, \ \pi][/tex], eg: x_0 = 0.5,...

Viet Dao,

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