y = (\sin x+a)(\cos x+a)

Let t = \sin x+\cos x

\sin x\cos x = \dfrac{1}{2} t^2-\dfrac{1}{2}

Then (1) is converted to

On the other hand,

t = \sin x+\cos x

=\sqrt{2} (\dfrac{\sqrt{2} }{2} \sin x+ \dfrac{\sqrt{2} }{2} \cos x)

= \sqrt{2} \sin (x+\dfrac{\pi}{4} )

Therefore, the domain of t in function (2) is ,

Evaluate the quadratic function (2)

y = \dfrac{1}{2} t^2 +at+a^2-\dfrac{1}{2}

= \dfrac{1}{2}(t^2+2at+a^2)-\dfrac{1}{2} a^2+a^2-\dfrac{1}{2}

= \dfrac{1}{2}(t+a)^2+\dfrac{1}{2}(a^2-1)

When

-\sqrt{2} \leq a \leq \sqrt{2}

y_{min} = \dfrac{1}{2}(a^2-1)

if a\geq \sqrt{2}, the minimum value happens when t = -\sqrt{2}

Substitute to (2)

y_{min} = \dfrac{1}{2} ( -\sqrt{2})^2 -a\sqrt{2}+a^2-\dfrac{1}{2}

= a^2-a\sqrt{2}+\dfrac{1}{2}

if a\leq - \sqrt{2} , the minimum value happens when t = \sqrt{2}

Substitute to (2)

y_{min} = \dfrac{1}{2} ( \sqrt{2})^2 +a\sqrt{2}+a^2-\dfrac{1}{2}

= a^2+a\sqrt{2}+\dfrac{1}{2}

In summary, the minimum value of the function in terms a is

y_{min} = \begin{cases} \dfrac{1}{2}(a^2-1) &\text{if } -\sqrt{2} \leq a \leq \sqrt{2} \\ \\ a^2-a\sqrt{2}+\dfrac{1}{2} & \text{if } a\geq \sqrt{2}\\ \\ a^2+a\sqrt{2}+\dfrac{1}{2} &\text{if } a\leq - \sqrt{2} \end{cases}