# Thread: smallest "p" value for sin

1. ## smallest "p" value for sin

If p is a positive real number and 2 sin x = 2 sin(x + p) for every real value of x, what is the smallest possible value for p, in degrees?

How can I show that it really is 360?

2. Hello donnagirl
Originally Posted by donnagirl
If p is a positive real number and 2 sin x = 2 sin(x + p) for every real value of x, what is the smallest possible value for p, in degrees?

How can I show that it really is 360?
Divide by $2$; then expand $\sin (x+p)$, and say:
$\sin x \equiv \sin(x+p)$

$\Rightarrow \sin x \equiv \sin x \cos p + \cos x \sin p$

Then compare the coefficients of $\sin x$ and $\cos x$:
$\Rightarrow \left\{ \begin{array}{l} \cos p =1\\ \sin p = 0\end{array}\right .$
And the smallest positive value of $p$ that satisfies these equations is $p = 360^o$

3. Hmmm, I still don't see it Grandad--how did you obtain the coefficients values?

4. Hello donnagirl
Originally Posted by donnagirl
Hmmm, I still don't see it Grandad--how did you obtain the coefficients values?
You'll see that I've used a $\equiv$ sign, rather than a $=$ sign, to show that the equation is true for all values of $x$. This means that we can say that the coefficients of $\sin x$ and $\cos x$ on both sides will be the same. So:
$\sin x \equiv \sin x \cos p + \cos x \sin p$

$\Rightarrow \color{red}1\color{black}\sin x+\color{red}0\color{black}\cos x \equiv \color{red}\cos p\color{black}\sin x + \color{red}\sin p\color{black}\cos x$

$\Rightarrow \left\{\begin{array}{l}\cos p = 1\\ \sin p = 0\end{array}\right .$
OK now?