Q.Derive an expression for equation of stationary wave on a stretched string.

[!TLDR] Adding two oppositely-travelling equal waves $y_1,y_2$ via the sum-to-product identity gives $y=(2a\cos\frac{2\pi x}{\lambda})\sin(2\pi nt)$ -- x and t are separated, not combined, so it's stationary, with nodes at $x=(2p+1)\lambda/4$ and antinodes at $x=p\lambda/2$. [!ANSWER] $y=A\sin(2\pi nt)$, with $A=2a\cos(2\pi x/\lambda)$; this is the equation of a stationary wave.

Consider two simple harmonic progressive waves of equal amplitude a and wavelength $\lambda$, travelling in opposite directions along the x-axis: $y_1=a\sin\left\{2\pi\left(nt-\dfrac{x}{\lambda}\right)\right\}$ and $y_2=a\sin\left\{2\pi\left(nt+\dfrac{x}{\lambda}\right)\right\}$. By the principle of superposition, the resultant displacement is $y=y_1+y_2$. Using the sum-to-product identity $\sin C+\sin D=2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$, with $C=2\pi(nt-x/\lambda)$ and $D=2\pi(nt+x/\lambda)$, so that $\dfrac{C+D}{2}=2\pi nt$ and $\dfrac{C-D}{2}=-\dfrac{2\pi x}{\lambda}$ (and cosine is even, so the sign does not matter): $y=2a\sin(2\pi nt)\cos\left(\dfrac{2\pi x}{\lambda}\right)$, i.e. $y=\left(2a\cos\dfrac{2\pi x}{\lambda}\right)\sin(2\pi nt)$. Writing the bracketed, x-dependent factor as $A=2a\cos\left(\dfrac{2\pi x}{\lambda}\right)$, the result is $y=A\sin(2\pi nt)$. Since x and t appear SEPARATELY here (x only inside the amplitude A, never combined with t into one travelling argument), this is NOT a progressive wave -- it is a STATIONARY wave: every particle oscillates with the same frequency n, but with an amplitude A that varies periodically with position. NODES (zero amplitude) occur where $\cos(2\pi x/\lambda)=0$, i.e. $x=(2p+1)\dfrac{\lambda}{4}$ for $p=0,1,2,\ldots$; ANTINODES (maximum amplitude $\pm2a$) occur where $\cos(2\pi x/\lambda)=\pm1$, i.e. $x=p\dfrac{\lambda}{2}$ for $p=0,1,2,\ldots$. [!ANSWER] $y=A\sin(2\pi nt)$, with $A=2a\cos(2\pi x/\lambda)$; this is the equation of a stationary wave.

Add the two equal, oppositely-travelling wave equations using the sum-to-product trigonometric identity, then identify the x-dependent factor as a stationary amplitude and locate nodes/antinodes from where it is zero/extremal.

Forgetting to use the sum-to-product identity correctly (getting the arguments $\frac{C+D}{2}$ and $\frac{C-D}{2}$ mixed up), which flips which variable (x or t) ends up inside the amplitude term versus the oscillating sine term.