-2
$\begingroup$

We just started on trig formulas in my class and this was one of the problems on our homework that stumped me. I couldn't remember any formulas for $\sin(2x)$ so any help would be appreciated.

$\endgroup$
1
  • 5
    $\begingroup$ maybe it was the other way around? $\endgroup$
    – Asinomás
    Dec 29, 2016 at 3:55

4 Answers 4

11
$\begingroup$

This isn't true.

Let $x=0$, then $\sin(2x) =\sin (0) = 0$, while $\cos(x) = \cos(0) = 1$.

If, as pointed out in the comments, it's the other way around, we have that $\cos(x) =0 \Rightarrow \sin(2x) = 2\sin(x)\cos(x) = 2\sin(x) * 0 = 0$, where recognizing $\sin(2x) = 2\sin(x)\cos(x)$ is the crux of the matter.

$\endgroup$
0
5
$\begingroup$

$$\sin2x=2\sin x\cos x$$

Now what if $\sin x=0?$


$$\sin2x=0\implies2x=n\pi$$ where $n$ is any integer

$\cos x=\cos\dfrac{n\pi}2$ which will be $=0$ if $n$ is odd

$\endgroup$
2
$\begingroup$

$$\sin(2x)=2\sin(x)\cos(x)=0\Rightarrow \sin(x)=0 \text{ or} \cos(x)=0$$

You can see that with $\sin(x)=0$ we get $x=k\pi$ where $\cos(x)=\pm1$. Hence the argument is not valid.

$\endgroup$
0
$\begingroup$

$\sin(\theta)=0$ precisely when $\theta$ is a multiple of $\pi$. So if $\sin(2x)=0$, that means that $2x$ is a multiple of $\pi$, or $x$ is a multiple of $\pi/2$.

However, it is not true that if $x$ is a multiple of $\pi/2$ then $\cos(x)=0$. This is true for odd multiples of $\pi/2$ (e.g. $\pi/2, 3\pi/2, 5\pi/2, \dots$) but false for even multiples of $\pi/2$ (e.g. $\pi, 2\pi, 3\pi, \dots$).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .