Solution and Deduction

dhiab · #1 $Old$ July 3rd, 2009, 09:20 AM

1 ) Solve in R this equation: $\cos 2x + 2\sin x\cos x = 0$ ( let:tan x= t)
2 )Deduct the values of :
$\cos \left( {\frac{\pi }{8}} \right),\cos \left( {\frac{{3\pi }}{8}} \right),\sin \left( {\frac{\pi }{8}} \right),\sin \left( {\frac{{3\pi }}{8}} \right)$

malaygoel · #2 $Old$ July 3rd, 2009, 09:27 AM

$cos^2\frac{\pi}{8}=\frac{cos\frac{\pi}{4}+1}{2}$

Similarly, other values can be found.

Prove It · #3 $Old$ July 3rd, 2009, 09:50 AM

Quote:

Originally Posted by dhiab $View Post$

1 ) Solve in R this equation: $\cos 2x + 2\sin x\cos x = 0$ ( let:tan x= t)
2 )Deduct the values of :
$\cos \left( {\frac{\pi }{8}} \right),\cos \left( {\frac{{3\pi }}{8}} \right),\sin \left( {\frac{\pi }{8}} \right),\sin \left( {\frac{{3\pi }}{8}} \right)$

$2\sin{x}\cos{x} = \sin{(2x)}$ .

So if $\cos{(2x)} + 2\sin{x}\cos{x} = 0$

$\cos{(2x)} + \sin{(2x)} = 0$

$\cos{(2x)} = -\sin{(2x)}$

$1 = -\frac{\sin{(2x)}}{\cos{(2x)}}$

$-\tan{(2x)} = 1$

$2x = \left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\} + 2\pi n$ , where $n \in \mathbf{Z}$

$x = \frac{1}{2}\left[\left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\} + 2\pi n\right]$

$x = \left\{\frac{\pi}{8}, \frac{5\pi}{8}\right\} + \pi n$ , where $n \in \mathbf{Z}$ .

HallsofIvy · #4 $Old$ July 3rd, 2009, 11:36 AM

Blast you, Prove It! You got in just ahead of me!

dhiab · #5 $Old$ July 3rd, 2009, 11:47 AM

Quote:

Originally Posted by Prove It $View Post$

$2\sin{x}\cos{x} = \sin{(2x)}$ .

So if $\cos{(2x)} + 2\sin{x}\cos{x} = 0$

$\cos{(2x)} + \sin{(2x)} = 0$

$\cos{(2x)} = -\sin{(2x)}$

$1 = -\frac{\sin{(2x)}}{\cos{(2x)}}$

$-\tan{(2x)} = 1$

$2x = \left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\} + 2\pi n$ , where $n \in \mathbf{Z}$

$x = \frac{1}{2}\left[\left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\} + 2\pi n\right]$

$x = \left\{\frac{\pi}{8}, \frac{5\pi}{8}\right\} + \pi n$ , where $n \in \mathbf{Z}$ .

Hello THANK YOU
You have:

Subst :

Now solve the quadratic equation.....

Prove It · #6 $Old$ July 3rd, 2009, 11:55 AM

Quote:

Originally Posted by dhiab $View Post$

Hello THANK YOU
You have:

Subst :

Now solve the quadratic equation.....

I've just realised I made a mistake...

I got to the step

$-\tan{(2x)} = 1$

$\tan{(2x)} = -1$

$2x = \left\{\pi - \frac{\pi}{4}, 2\pi - \frac{\pi}{4}\right\} + 2\pi n, n \in \mathbf{Z}$

$2x = \left\{\frac{3\pi}{4}, \frac{7\pi}{4}\right\} + 2\pi n$

$x = \frac{1}{2}\left[\left\{\frac{3\pi}{4}, \frac{7\pi}{4}\right\} + 2\pi n\right]$

$x = \left\{\frac{3\pi}{8}, \frac{7\pi}{8}\right\} + \pi n, n \in \mathbf{Z}$ .

There was nothing wrong with my solution process, I just forgot about the negative sign.

You don't need to use the other method...