Is $ \sin 2x \ge 0 $ same as $\sin x\cos x\ge 0 $?

Question

Or differently phrased do $$ \sin 2x \ge 0$$ $$\sin x\cos x\ge 0 $$ have the same solution.

$\sin 2x$ is positive from $ 0\le x \le \pi/2$ , while $\sin x\cos x$ are positive in the first and third quadrant. Why is this, when $\sin 2x=2\sin x\cos x$, am i making a mistake.

Does this mean i cannot use double angle identity under the square root?

Edit: as vadim123 said the question was wrong, so i changed it

Answer 1

Since $\forall x\in \mathbb{R}$

$$\sin 2x=2\sin x\cos x$$

we have that in general

$$|\sin 2x|=|2\sin x\cos x|\neq |\sin x\cos x|$$

but

$$\sin 2x\ge 0 \iff 2\sin x\cos x\ge 0 \iff \sin x\cos x\ge 0 $$

therefore $ \sin 2x \ge 0$ and $\sin x\cos x\ge 0 $ have the same solutions.

Answer 2

Just because $\sin(2x)$ and $\sin(x)\cos (x)$ always have the same sign (which is true) doesn't mean they have the same magnitude. $|\sin(2x)|$ is twice as large as $|\sin(x)\cos(x)|$, so they are not the same (except when they are both $0$).

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(When this answer was posted, the title of the question asked about $|\sin 2x|$ and $|\sin x \cos x|$. The OP later removed the absolute-value signs without comment).

Answer 3

Anywhere you can use $\sin(2x)$ you can also use $2\sin(x)\cos(x)$. This is an identity. We can show this using the double angle formula:

$$\sin(a + b) = \sin(a)\cos(b) + \sin(b)\cos(a) \implies \sin(2x) = 2\sin(x)\cos(x)$$

Your question is asking if the solution to $\sin(2x)\ge0$ is the same as that of $\sin(x)\cos(x)\ge0$. Indeed, if we multiply the latter by $2$, we see that the solution must be the same! Why? Because $\sin(2x)=2\sin(x)\cos(x)$ is an identity: they're two representations of the same mathematical object!

Either inequality is thus solved by $x +\pi n$ such that $x\in[0,\pi/2]$ with integer $n$. This is because $\sin(2x)$ is $\pi$ periodic. Written another way, we could use the periodicity to write that the inequality is satisfied for $$x\in \dots[-\pi,-\pi/2]\cup[0,\pi/2]\cup[\pi,3\pi/2]\cup\dots$$

Answer 4

Note if $M > 0$ then $M*a*b > 0$ if $a > 0; b>0$ or $a < 0; b< 0$. And $M*a*b < 0$ if $a < 0; b > 0$ or $a > 0; b < 0$. And $M*a*b = 0$ if $a = 0$ or $b =0$. It does not matter whether $M =1$ or $M = 2$ or $M = \frac {37.95}{\pi}$.

So we would expect $\cos x \sin x \ge 0$ for the exact same values of $x$ as we would expect $\sin 2x = 2\cos x \sin x\ge 0$ or for the exact same values we would expect $\frac {37.95}{\pi}\cos x \sin x \ge 0$.

And it works out to be exactly the case. $\cos x \sin x \ge 0$ if $x$ is in the first or third quadrant (as those are the two quadrants where $\sin x$ and $\cos x$ are both the same sign.) And $\sin M \ge 0$ if $M$ is in the first or second quadrant. And $M = 2x$ is in the first quadrant if $x$ is in the first half of the first or third quadrant. ANd $M$ is in the second quadrant if $x$ is in the second half of the first or third quadrant.

Or in other words:

If $0 \le x \le \frac \pi 2$ then $\sin x \ge 0$ and $\cos x \ge 0$ so $\sin x\cos x \ge 0$. Also $0 \le 2x \le \pi$ so $\sin 2x \ge 0$.

If $\frac \pi 2 \le x \le \pi$ then $\sin x \ge 0$ and $\cos x \le 0$ so $\sin x\cos x \le 0$. Also $\pi \le 2x \le 2\pi$ so $\sin 2x \le 0$.

If $\pi \le x \le \frac {3\pi} 2$ then $\sin x \le 0$ and $\cos x \le 0$ so $\sin x\cos x \ge 0$. Also $2\pi \le 2x \le 3\pi$ so $\sin 2x \ge 0$.

If $\frac {3\pi} 2 \le x \le 2\pi$ then $\sin x \le 0$ and $\cos x \ge 0$ so $\sin x\cos x \le 0$. Also $3\pi \le 2x \le 4\pi$ so $\sin 2x \le 0$.

As well $\sin 2x = 2\cos x\sin x$ and $\cos x\sin x$ and $M*\cos x \sin x; M > 0$ should all be $> 0$ if $\cos x, \sin x$ are the same sign, and should all be $< 0$ of $\cos x, \sin y$ are opposite signs, and should $= 0$ if one or the other of $\cos x, \sin x$ equals $0$. So all turns out to work exactly as it should.