Is $\frac{2}{π}(\sin t+\cos t)$ the same as $\frac{4}{π\sqrt2}\sin\left(t+\frac{\pi}{4}\right)$?

Question

I've been solving some problems on signals for practice and I solved a particular one in two ways. The first way gave me this result : $$V(t)=\frac{2}{π}(\sin t+\cos t)$$

The second way gave: $$V(t)=\frac{4}{π\sqrt2}\sin\left(t+\frac{\pi}{4}\right)$$

Plotting this two online they seem to be the same. How can I prove this mathematically? I haven't used trigonometric formulas much so I don't have many ideas. I tried turning the sint to a cosine and then applying the addition formula of cosa+cosb but that doesn't help.

Answer 1

HINT

Recall that

$$\sin (t+\pi/4)=\sin t\cos (\pi/4)+\cos t\sin (\pi/4)$$

Answer 2

Hint 1:$$$$$2$ functions $f(x)$ and $g(x)$ are considered equal if and only if they have the same domain $D$, and the same value of $f(x)$ and $g(x)$ for any $x\in D$. $$$$ Hint 2:$$$$ $V_1(t)=\frac{2}{π}(\sin t+\cos t)$ and $V_2(t)=\frac{4}{π\sqrt2}\sin(t+π/4)$ have the same domain ($R$ - the set of Real Numbers). $$$$ Hint 3: $$$$ $$V_1(t)=\frac{2}{π}(\sin t+\cos t)=\dfrac{2\sqrt 2}{\pi}\left(\dfrac{\sin t}{\sqrt 2}+\dfrac{\cos t}{\sqrt 2}\right)=\frac{4}{π\sqrt2}sin(t+π/4)=V_2(t)$$ $$$$ Note that $\sin (t+\pi/4)=\sin t\cos(\pi/4)+\sin(\pi/4)\cos t$