# Trigonometric Function

1. Feb 1, 2006

### Psi-String

I have stuck on this problem for long time

$$sin^2 \alpha = \frac{\alpha}{2}$$

I never meet this kind of problem before, and I have no idea about this. Could someone tell me how to solve this kind of problem?

(Ans: $$\alpha = 1.39 rad$$)

2. Feb 1, 2006

### VietDao29

This kind of problem can be done with Newton's method.
$$\sin ^ 2 \alpha = \frac{\alpha}{2}$$
$$\Leftrightarrow 2 \sin ^ 2 \alpha = \alpha$$
$$\Leftrightarrow - 2 \sin ^ 2 \alpha + \alpha = 0$$
$$\Leftrightarrow \cos (2 \alpha) - 1 + \alpha = 0$$
Now let $$f(x) = \cos (2x) + x - 1$$
Since f(0) = 0, that means x = 0 is 1 solution to the question.
Now let's choose x0.
$f(1) = \cos 2$. Since $$2 \in ]\frac{\pi}{2} ; \ \pi[$$, we have: $f(1) = \cos 2 < 0$
$f(2) = \cos 4 + 1 > 0$.
So we have 1 more solution on the interval ]1; 2[.
Pick any x0 on the interval ]1; 2[.
Then apply the Newton's method:
$$x_{n + 1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}$$. And let n increase without bond. The solution will be $$x = \lim_{n \rightarrow \infty} x_n$$.
If you get x = 0 (just try another x0) (we already know this solution, we want to find another on the interval ]1; 2[).
You may want to try some x0 that's closed to 2.
Just try it and see what you get.
So there are 2 solutions to the problem.
Can you go from here?
By the way, I believe you cannot get any solution that reads:
$x \approx 1.39 \mbox{ rad}$. That may be a typo though.

Last edited: Feb 1, 2006
3. Feb 1, 2006

### Psi-String

hmmm.... though a little bit vague, I think I can solve it.

Thanks

4. Feb 1, 2006

### HallsofIvy

Staff Emeritus
You do understand that $\alpha= 0$ is an obvious solution and are asking for non-zero solutions?

The point is that there is no "algebraic method" to solve an equation in which the unknown occurs both in and outside of a trascendental function- such as $sin^2 \alpha= \frac{\alpha}{2}$.

As VietDao29 said, a good way to solve such equation is "Newton's method". A solution to f(x)= 0 is given by xn+1= xn- \frac{f(x_0}{f'(x_0}[/tex]

In this problem $f(x)= sin^2(x)- \frac{x}{2}. f'(x)= 2 sin(x)cos(x)-\frac{1}{2}$ so $x_{n+1}= x_n- \frac{sin^(x)- \frac{x}{2}}{2sin(x)cos(x)-\frac{1}{2}}$

Pick a starting value for x (perhaps $x= \frac{\pi}{2}$ would be keep you away from 0) and do the arithmetic.

5. Feb 2, 2006

### Integral

Staff Emeritus
A good way to start this type of problem is by plotting the functions $\sin^2( \alpha )$ and $\frac { \alpha } 2$.

Roots are at the points of intersection.

6. Feb 2, 2006

### Cyrus

For small angle, sin^2(alpha) ~= (alpha^2). So a place to look might be when alpha = 1/2

My calculator converges on .5545

7. Feb 2, 2006

### VietDao29

Then, there's probably something wrong with your calculator. :tongue2:
You can always put x back to the original equation, and find out that it is not a solution... :tongue:
By the way, the problem reads:
$$\sin ^ 2 \alpha = \frac{\alpha}{2}$$.

8. Feb 2, 2006

### Cyrus

No, that ans is flat out wrong. Plug it in, you will get .9676672 = .695, I dont think so. The two solutions are .5545 and I think something like 1.83ish.

9. Feb 2, 2006

### Integral

Staff Emeritus
I get 1.849076836 and 0.554572091

10. Feb 2, 2006

### VietDao29

Whooops, I almost missed one solution x = 0.554572091... My bad.
I must be careful next time... :grumpy:
But don't forget the obvious solution x = 0.