1. Solution and Deduction

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

Similarly, other values can be found.

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

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

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

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

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

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

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

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

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

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

5. Originally Posted by Prove It
$\displaystyle 2\sin{x}\cos{x} = \sin{(2x)}$.

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

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

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

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

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

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

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

$\displaystyle x = \left\{\frac{\pi}{8}, \frac{5\pi}{8}\right\} + \pi n$, where $\displaystyle n \in \mathbf{Z}$.
Hello THANK YOU
You have:

Subst :

6. Originally Posted by dhiab
Hello THANK YOU
You have:

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

I got to the step

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

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

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

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

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

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

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

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

7. Prove it said:

.
but :

8. Originally Posted by dhiab
Prove it said:

.
but :

9. Originally Posted by Prove It