Show that $-\sin⁡(3x - \pi /2)$ is the same as $\sin(3x + \pi /2)$ I am trying to prove that 
$$-\sin⁡(3x  - \frac{\pi}2)=\sin(3x + \frac{\pi}2)$$
I know both terms are the same because their graphs are, so I want to proof  this equality using trigonometric identities.
$$ \sin( A - B) = \sin A\cos B - \cos A\sin B $$
$$ \sin( A + B) = \sin A\cos B + \cos A\sin B $$
$$- \sin( A - B) = - \sin A\cos B + \cos A\sin B $$
This is what I have got so far. I am not sure how I can get rid of the "-" sign on the third line.
Thanks in advance!
 A: Using your second identity and choosing $B=\pi$,
$$\sin(A+\pi)=\sin(A)\cos(\pi)+\cos(A)\sin(\pi)=-1\times\sin(A)+\cos(A)\times0=-\sin(A)$$
or simply $$\boxed{\sin(A+\pi)=-\sin(A)}$$
Now with $A=3x-\pi/2$ we have
$$\sin(3x-\pi/2+\pi)=-\sin(3x-\pi/2)$$
where the LHS is $\sin(3x+\pi/2)$.
A: Note that $$-\sin⁡(3x-\pi /2)=-\left( \sin { (3x)\cos { (\pi /2) } -\cos { (3x) } \sin { (\pi /2) }  }  \right) =-\left( -\cos { (3x) }  \right) =\cos { (3x) } \\$$
$$ \sin  (3x+\pi /2)=\sin { (3x)\cos { (\pi /2) } +\cos { (3x) } \sin { (\pi /2) }  } =\cos { (3x) } \\ $$
Second Method :
Let assume $-⁡\sin { \left( 3x-\pi /2 \right)  } \neq \sin  \left( 3x+\pi /2 \right) $ then
$$\sin  \left( 3x+\pi /2 \right) +\sin  \left( 3x-\pi /2 \right) \neq 0\\ 2\sin { \frac { 3x+\pi /2+3x-\pi /2 }{ 2 }  } \cos { \frac { 3x+\pi /2-3x+\pi /2 }{ 2 }  } =0\\ 2\sin { (3x) } \cos { (\pi /2) } \neq 0\\ 0\neq 0\\ $$
which shows contradiction
A: Hint: use complementary angle formulas:
$$A=\sin(3x+\pi/2)=\sin(\pi/2-(-3x))=\cos(-3x)=\cos(3x).$$
$$B=-\sin(3x-\pi/2)=\sin(\pi/2-3x)=\cos(3x).$$
Hence $A=B$.
A: The two arguments differ by $\pi$, which correspond to antipodal points, hence sines (and cosines) of opposite signs.
