If $\sin(x)+\cos(x)=y$, for what value(s) of $y$ can $\sin(x)\cos(x)=1$?

Question

If $\sin(x)+\cos(x)=y$, for what value(s) of $y$ can $\sin(x)\cos(x)=1$?

Squaring Both sides

$$(1+2\cos x\sin x)=y^2$$

I am stuck here. Can you square both sides, as $\sin(x)\cos(x)$ cannot be equal to $1$ in the real domain?

Answer 1

Note that $\sin x \cos x = 1$ iff $\sin^2 x + 2 \sin x \cos x + \cos^2 x = (\sin x + \cos x)^2 = 3$.

Note that $\sin x + \cos x = \sqrt{2} \sin (x+ {\pi \over 4})$ hence you are looking for solutions to $\sin^2 (x+ {\pi \over 4}) = {3 \over 2}$.

One can find solutions to $\sin^2 y = {3 \over 2}$ by expanding $\sin$ in terms of exponentials to get $(e^{iy}-e^{-iy})^2 = -6$, or $(e^{i2y})^2 + 4 (e^{i2y}) +1 = 0$.

Solving this gives $y = {1 \over 2} (\pi -i \log (2 \pm \sqrt{3}))+m\pi$ ($m$ an arbitrary integer), and hence $x = ({1 \over 4}+m) \pi - i {1 \over 2} \log (2 \pm \sqrt{3})$ ($m$ just flips the sign, the $\pm$ has no effect on the result).

In particular, the answer to the question is that if $\sin x \cos x =1$ then $\sin x + \cos x = \pm \sqrt{3}$.

Answer 2

$$(1+2\cos x\sin x)^2=y^2\implies \cos x\sin x=\frac{\pm y-1}2$$

And thus you need to solve

$$\frac{\pm y-1}2=1\implies \pm y=3...\text{what can you conclude?}$$

Of course, much easy, as you say at the end of your question, to observe that

$$1=\cos x\sin x=\frac{\sin 2x}2$$

and this has no solution at all in $\;\Bbb R\;$ .