# Newton-Raphson - How many solutions?

#### Skykai

Hi, I was asked to find a solution to the following problem using the Newton-Raphson process:

x=(sin(x)+cos(x))^1/2

which I found to be 1.14955 (5dp)

But how to I determine how many solutions there are in total?

I plotted the graph and looked into tangents but think I'm barking up the wrong tree.. I can't quite get my head around this so any help would be greatly appreciated..! Thank you

#### v8archie

Math Team
Think about the range of possible values of $\sin{(x)} + \cos{(x)}$. This then gives you a range of $x$ within which all possible solutions must occur.

Note that $\sin{(x)} + \cos{(x)} = \sqrt{2}\left( \frac{1}{\sqrt2}\sin{(x)} + \frac{1}{\sqrt2}\cos{(x)} \right)$ can be expressed as a single sinusoidal function using $\sin{(A+B)} = \sin{(A)}\cos{(B)} + \cos{(A)}\sin{(B)}$

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#### Skykai

Thank you v8archie but I have I admit that I'm still very lost. I can see that the range is between â€“âˆš2 and âˆš2 but I can't see how you managed to manipulate that formula into that trig identity, nor do I understand why?

Sorry if i'm being stupid, I think I've just been staring at this problem for too many hours so nothing is making sense anymore.

Would you mind explaining a little bit more please?

#### JeffM1

Before we get to your question

$-\ 1 \le sin(x) \le 1\ and\ -\ 1 \le cos(x) \le 1 \implies$

$-\ 2 \le sin(x) + cos(x) \le 2 \implies$

$\sqrt{sin(x) + cos(x)} \in \mathbb R\ only\ if\ 0 \le sin(x) + cos(x) \le 2 \implies$

$0 \le x \le \sqrt{2}\ if\ x \in \mathbb R\ and\ x = \sqrt{sin(x) + cos(x)}.$

But your answer implies that $x > \sqrt{2}.$

So I am confused from the get-go.

Now consider the potential x-intercepts in $[0,\ \sqrt{2}]$ for the function

$f(x) = \sqrt{sin(x) + cos(x)} - x \implies f(0) = 1.$

Moreover, $f'(x) = -\ \dfrac{1}{2\sqrt{sin(x) + cos(x)}} - 1 < 0.$

So what is the maximum number of possible x-intercepts in $[0,\ \sqrt{2}]$

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#### skeeter

Math Team
x=(sin(x)+cos(x))^1/2

which I found to be 1.14955 (5dp)
But your answer implies that $x > \sqrt{2}$
$1.14955 < \sqrt{2} \approx 1.414$

$\sin{x}+\cos{x} = \sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right) \implies 0 < \sqrt{\sin{x}+\cos{x}} = \sqrt{\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)} < \sqrt{2} \approx 1.1892...$

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#### JeffM1

$1.14955 < \sqrt{2} \approx 1.414$
Yes indeed. Somehow I misread 1.14955 as 1.4955.

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#### skeeter

Math Team
Yes indeed. Somehow I misread 1.14955 as 1.4955.
... "it" happens.