Integral of $-4\sin(2t - (pi/2)) $ weird behavior on wolfram alpha

Question

I'm confused by what Wolfram Alpha is doing with my function: $$-4\sin{(2t - (\pi/2))}$$ on why the it gets replaced by $$4\cos{(2t)}$$.

Is it equal? Link: See behavior here

Answer 1

Yes. Using the angle-addition formula, $$ \sin{(2t-\pi/2)} = \sin{2t}\cos{(\pi/2)}-\cos{2t}\sin{(\pi/2)} = -\cos{2t}. $$

Answer 2

Yes, through two identities:

$$-4\sin(2t-\pi/2)=-4\sin(-(\pi/2-2t)) \\ =4\sin(\pi/2-2t) \\ =4\cos(2t).$$

The first identity is in the second equation, and is the fact that $\sin$ is odd. The second one is in the third equation, and is the basic cofunction identity (easily seen by drawing a right triangle).