What is sin⁡(x)+cos⁡(x) in terms of sine?

Explanation:
Using Pythagorean Identity
${\displaystyle {\sin }^{2}x+{\cos }^{2}x=1,\text{ so }{\cos }^{2}x=1-{\sin }^{2}x}$
${\displaystyle \cos x=\pm \sqrt{1-{\sin }^{2}x}}$
${\displaystyle \sin x+\cos x=\sin x\pm \sqrt{1-{\sin }^{2}x}}$
Using complement / cofunction identity
${\displaystyle \cos x=\sin (\frac{\pi }{2}-x)}$
${\displaystyle \sin x+\cos x=\sin x+\sin (\frac{\pi }{2}-2)}$

Explanation
Suppose that ${\displaystyle \sin x+\cos x=R\sin (x+\alpha )}$
Then
${\displaystyle \sin x+\cos x=R\sin x\cos \alpha +R\cos x\sin \alpha }$
${\displaystyle =\left(R\cos \alpha \right)\sin x+\left(R\sin \alpha \right)\cos x}$
The coefficients of ${\displaystyle \sin x}$ and of ${\displaystyle \cos x}$ must be equal so
${\displaystyle R\cos \alpha =1}$
${\displaystyle R\sin \alpha =1}$
Squaring and adding, we get
${\displaystyle R^2{\cos }^2\alpha +R^2{\sin }^2\alpha =2\text{ so }{R}^2({\cos }^2\alpha +{\sin }^2\alpha )=2}$
${\displaystyle R=\sqrt{2}}$
And now
${\displaystyle \cos \alpha =\frac{1}{\sqrt{2}}}$
${\displaystyle \sin \alpha =\frac{1}{\sqrt{2}}}$
${\displaystyle \text{So }\alpha ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}}$
${\displaystyle \sin x+\cos x=\sqrt{2}\sin (x+\frac{\pi }{4})}$

Explanation:
Using pythagorean identity,
${\sin }^{2}x+{\cos }^{2}x=1$
$\text{So, }{\cos }^{2}x=1-{\sin }^{2}x$
By taking square root on both the sides,
$\cos x+\sin x=\sin \pm \sqrt{1-{\sin }^{2}x}$
Using complement or cofunction identity,
$\cos x=\sin (\frac{\pi }{2}-x)$
$\sin x+\cos x=\sin x+\sin (\frac{\pi }{2}-x)$
Thus, the expression for $\sin x+\cos x$ in terms of sine is $\sin x+\sin (\frac{\pi }{2}–x).$