$$\sqrt2\sin\left(x+\frac\pi4\right)=\sqrt2\left(\sin(x)\cos\left(\frac\pi4\right)+\cos(x)\sin\left(\frac\pi4\right)\right)=\sin(x)+\cos(x)$$
Could you solve it from opposite? $$\sin(x)+\cos (x)=\sqrt2\sin\left(x+\frac\pi4\right)$$
Second line is completely looking like magic to me. I was trying using Double angle formula. But, I couldn't relate it to anyone.
$$\sin (x)+\cos (x)=\sin (x)+\sin (\frac{\pi}{2}+x)$$
But, I know that
$$\sin (x+y) != \sin (x) +\sin (y)$$
$$\sin (48+30)=0.978147$$$$\sin (48) +\sin (30)=1.243144825$$
So, they aren't equal. That's why I can't use formula of $$\sin (a+b)$$
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$\begingroup$$$ \begin{aligned} \sin x + \cos x&= \sqrt{2} \cdot \dfrac{1}{\sqrt{2}} \, \left( \sin x + \cos x \right) \\ &= \sqrt{2} \, \left( \dfrac{1}{\sqrt{2}}\sin x + \dfrac{1}{\sqrt{2}}\cos x \right) \\ &= \sqrt{2} \big( \sin x \cos(\pi/4) + \sin(\pi/4) \cos x \big) \\ &= \sqrt{2} \sin(x + \pi/4) \end{aligned} $$
$\endgroup$ 4 $\begingroup$From$$\tag1\sin(x+y)=\sin x\cos y+\cos x\sin y$$and$$\tag2\sin(-y)=-\sin y $$(both of which identities I assume as known and will not prove here), we obtain$$\tag3\sin(x-y)=\sin x\cos y-\cos x\sin y$$and by adding $(1)+(3)$,$$\tag4\sin(x+y)+ \sin(x-y)=2\sin x\cos y. $$Substitue $x+y\leftarrow u$ and $x-y\leftarrow v$, i.e., $x\leftarrow\frac{u+v}2$ and $y\leftarrow\frac{u-v}2$ to arrive at$$ \sin u+\sin v=2\sin\frac{u+v}2\cos\frac{u-v}2$$
$\endgroup$ $\begingroup$Using the formula $\sin (x+y)$ as mentioned here .
See for first line we know $\cos\left(\dfrac\pi4\right)$ and $\sin\left(\dfrac\pi4\right)$ is equal to $\dfrac{1}{\sqrt{2}}$
So the expression reduces to $$\sqrt2\sin\left(x+\frac\pi4\right)=\sqrt{2}(\frac{\sin x}{\sqrt{2}} + \frac{\cos x}{\sqrt{2}}) $$
Now I hope you can reduce it to second line.
For the second line use the formula I gave in starting of answer. Do tell if you can't solve yet in comment
$\endgroup$ $\begingroup$Note :
$$ a\sin(\theta)+ b\cos(\theta) = R\sin(\theta+\beta)$$
Using Double angle identity :
$$a\sin(\theta)+ b\cos(\theta) = \underbrace{R\cos(\beta)}_{\text{relates to a }}\sin(\theta)+ \underbrace{R\sin(\beta)}_{\text{relates to b}}\sin(\theta) $$
$$\tag{1} a \sin(\theta) = R\cos(\beta)\sin(\theta)$$$$ a = R\cos(\beta)$$$$ \tag{2}b \cos(\theta) = R\sin(\beta)\cos(\theta) $$
$$b = R\sin(\beta) $$Then ,
$$ \tag{3}\frac{b}{a} = \frac{R\sin(\beta)}{R\cos(\beta)}=\tan(\beta)$$
Given that :$$ 1\sin(\theta)+1\cos(\theta)$$
$$ ~a,b = 1 $$
Using Pythagorean theorem :
$$R = \sqrt{a^2 + b ^2}$$
$$R = \sqrt{1+1}= \sqrt{2}$$
$$ \tan(\beta) = 1 ; ~~\arctan(1) = \frac{\pi}{4} ; ~~ \beta = \frac{\pi}{4} $$
Thus,
$$R\sin(\theta + \beta) = \sqrt{2}\sin \left(\theta + \frac{\pi}{4} \right) $$
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