recent
أخبار ساخنة

# Can Sin(x) Be Greater Than One ?

## Can Sin(x) Be Greater Than One ?

The simple answer for this problem is YES !

We already know that $$-1\leq sin\left ( x \right ) \leq1$$ , $$\forall x \in \mathbb{R}$$ . Therefore, any solution to

$$\left | sin \left ( x \right ) \right | > 1$$ would implies values of $$x$$ such that $$x \notin \mathbb{R}$$.

Example:

Consider the problem $$sin\left ( \theta \right )$$ = 2 . What is the value of $$\theta$$ ?

Start by utilizing the Euler's Formula:

$$e^{i\theta} = cos\left ( \theta \right ) + isin\left ( \theta \right )$$

then apply the Euler's Formula for $$-\theta$$ to have :

$$e^{i\left ( -\theta \right )} = cos\left ( \theta \right ) - isin\left ( \theta \right )$$

Now, subtract these two formulas like follows:

$$\begin{matrix} & e^{i\theta} = cos\left ( \theta \right ) + isin\left ( \theta \right ) \\ - & \\ & e^{i\left ( -\theta \right )} = cos\left ( \theta \right ) - isin\left ( \theta \right ) \\ & ----------- \\ \Rightarrow & e^{i\theta} - e^{i\left ( -\theta \right )} = 2isin\left ( \theta \right ) \\ \end{matrix}$$

Rewrite the last equation in term of $$sin\left (\theta \right )$$

$$\Rightarrow sin\left (\theta \right ) = \frac{e^{i\theta}-e^{-i\theta}}{2i}$$

Now, Since the original problem is $$sin\left ( \theta \right )$$ = 2, we may substitute the Left Hand Side with

$$\frac{e^{i\theta}-e^{-i\theta}}{2i}$$ to have:

$$\frac{e^{i\theta}-e^{-i\theta}}{2i} =2$$ $$\Rightarrow e^{i\theta}-e^{-i\theta} = 4i$$

To find the value of $$e^{i\theta}$$ ; multiply the both sides with $$e^{i\theta}$$ like follows:

$$e^{i\theta} \left [e^{i\theta}-e^{-i\theta} \right ]$$ $$\Rightarrow e^{2i\theta} - 1 = 4ie^{i\theta}$$

$$\Rightarrow e^{2i\theta} - 4ie^{i\theta} - 1$$

Apply the Quadratic Formula $$\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$$ on the last equation to evaluate $$e^{i\theta}$$

$$e^{i\theta} = \frac{4i\pm \sqrt{-16-4\left (1 \right )\left (-1 \right )}}{2}$$

$$e^{i\theta} = \frac{4i\pm 2i\sqrt{3}}{2}$$

$$e^{i\theta} = 2i\pm i\sqrt{3}$$ $$\Rightarrow e^{i\theta} = i(2 \pm \sqrt{3} )$$

Now, multiply both sides with $$ln$$ to exclude $$\theta$$ like follows:

$$ln\left ( e^{i\theta} \right ) = ln\left ( i(2 \pm \sqrt{3} ) \right )$$

$$i \theta = ln \left ( i \right ) \pm ln \left ( 2 \pm \sqrt{3} \right )$$

$$i \theta = i\frac{\pi}{2} + ln \left ( 2 \pm \sqrt{3} \right )$$

$$\theta = \frac{\pi}{2} - iln \left ( 2 \pm \sqrt{3} \right )$$

Hence $$\theta = \frac{\pi}{2} \pm iln \left ( 2 + \sqrt{3} \right ) + 2 \pi n$$ where, $$n \in \mathbb{N}$$

العلم للجميع