Answer
$y_1+y_2=2A\cos (\dfrac{2\pi t}{T}) \cos (\dfrac{ 2\pi x}{\lambda})$
Work Step by Step
By using the Sum and Difference formulas: $\cos (a+b)=\cos a \cos b -\sin a \sin b$ and $\cos (a-b)=\cos a \cos b +\sin a \sin b$
$y_1+y_2=A[\cos 2\pi (\dfrac{t}{T} -\dfrac{x}{\lambda})+ \cos 2\pi(\dfrac{t}{T} -\dfrac{x}{\lambda})]$
Now, $y_1+y_2=A[\cos 2\pi (\dfrac{t}{T} -\dfrac{x}{\lambda})+ \cos 2\pi(\dfrac{t}{T} -\dfrac{x}{\lambda})]$
or, $y_1+y_2=2A\cos (\dfrac{2\pi t}{T}) \cos (\dfrac{ 2\pi x}{\lambda})$
