General trignometry doubt We know $\cos(-x) = \cos x$
Now my doubt is that the condition is true only when $x\in [0,\pi]$
as in fourth quadrant cosine is positive .
Or it belongs to every real number .
 A: We can think about this geometrically! 
Let $x_{\theta}$ denote a unit vector that makes the angle $\theta$ with the positive $x$-axis. The function $\cos(\theta)$ is the answer to the question

"What is the horizontal length of $x_{\theta}$?"

Now, consider the vector $x_{-\theta}$. This is just the reflection of of $x_{\theta}$ across the $x$-axis. So the answer to the question "What is the horizontal length of $x_{-\theta}$?" is the same! This is exactly the value of $\cos(\theta)$, so therefore 
$$\cos(-\theta) = \cos(\theta)$$
As for your question about values $x\in[0,\pi]$, this is simply asking questions about vectors in the upper-half plane. Since a reflection across the $x$-axis leaves this plane when $x\in(0,\pi)$, talking about $-\theta$ doesn't make sense in that context. 
Using this intuition about $\cos(\theta)$, what can you say about $\sin(-\theta)$?
A: Case 1:
$x \in [0,\pi]$.  Then $cos(x) = \cos(-x)$.  You seem to accept that.
Case 2:
$x \in [-\pi, 0]$.  Then $-x \in [0,\pi]$.  So $\cos(x) = \cos(-(-x)) = \cos(-x)$.
Case 3:
$x \in \mathbb R$.  Then there is some $k \in \mathbb Z$ so the $x = k(2\pi) + x'; -\pi < x' \le \pi$.  So $\cos (x) = \cos(x' + 2k\pi) = \cos(x') = \cos(-x') = \cos(-x - 2k\pi) = \cos (-x)$.
(Because $trig(x + 2k\pi) = trig(x)$ for all integers $k$ and any trig function.  [Because circles have circumference of $2\pi$.])
A: You have the prosthaphaeresis formula
$$ \cos{A}-\cos{B} = -2\sin{\left(\frac{A+B}{2}\right)}\sin{\left(\frac{A-B}{2}\right)}, $$
(subtract two instances of the addition formula for $\cos$) and putting $B=-A$, the first sine is $\sin{0}=0$, so $\cos{A}-\cos{(-A)}=0$.
A: If $x$ is in the fourth quadrant, $-x$ is in the first quadrant. As both cosines are positive, there is no contradiction.
A: Don't confuse the sign of $x$ with the sign of cosine. That is, sometimes $\cos(-x)$ is positive (because $x<\frac{\pi}{2}$).
As the comments point out, $\cos(x)=\cos(-x)$ for all $x$. To understand this intuitively, imagine the unit circle. Recall how the $x$-value of any coordinate on the unit circle is equal to cosine. Now randomly choose an angle $x$ on the unit circle. Note that the angle $-x$ is the reflection of $x$ across the $y$-axis, which means these two points have the same $x$-coordinates. Therefore, $\cos(x)=\cos(-x)$.
A: cosine is positive in the 1st and 4th quadrants and negative in 2nd and 3rd.
If $x$ lies in 1st quadrant $-x$ will lie in 4th.
If $x$ lies in 2nd quadrant $-x$ will lie in 3rd.
If $x$ lies in 3rd quadrant $-x$ will lie in 2nd.
If $x$ lies in 4th quadrant $-x$ will lie in 3rd.
Thus the sign of cosine will be same for $x$ lying in any quadrant or in other words, $cos(x)=cos(-x)$ for any real number $x$.
