Question

Sin2x/cosx-sin(pi+2)

Answer 1

$\frac{sin(2x)}{cos(x)} -sin(\pi+2)=\frac{2sin(x)cos(x)}{cos(x)} -(sin(\pi)cos(2)+cos(\pi)sin(2))=\\=2sin(x)-(-sin(2))=2sin(x)+sin(2)$

Answer 2

To simplify the expression $\frac{\sin(2x)}{\cos(x)}-\sin(\pi+2)$, we first simplify $\sin(\pi+2)$:

$\sin(\pi+2)=\sin(\pi)\cos(2)+\cos(\pi)\sin(2)=-\sin(2)$

Now, we substitute this back into the expression:

$\frac{\sin(2x)}{\cos(x)}-\sin(\pi+2)=\frac{\sin(2x)}{\cos(x)}-(-\sin(2))=\frac{\sin(2x)}{\cos(x)}+\sin(2)$

We can simplify this further using the trigonometric identity $\sin(2x)=2\sin(x)\cos(x)$:

$\frac{2\sin(x)\cos(x)}{\cos(x)}+\sin(2)=2\sin(x)+\sin(2)$