Hello, zivku!
i'm having difficulty solving: \(\displaystyle \tan(x)\:=\:\frac{\sin(x+2\pi)}{\cos(x - 2\pi)}\)
I'm not surprised that you can't
solve it . . . it's an <u>identity</u>!
The compound-angle formulas:
. . \(\displaystyle \sin(A\,\pm\,B)\:=\:\sin(A)\cos(B)\,\pm\,\sin(B)\cos(A)\)
. . \(\displaystyle \cos(A\,\pm\,B)\:=\:\cos(A)\cos(B)\,\mp\,\sin(A)\sin(B)\)
Then: \(\displaystyle \,\sin(x+2\pi)\:=\:\sin(x)\cos(2\pi}\,+\,\sin(2\pi)\cos(x)\)
. . . . . . . . . . . . . . . \(\displaystyle =\:\sin(x)\cdot1\,+\,0\cdot\cos(x)\:=\:\sin(x)\)
And: \(\displaystyle \;\cos(x+2\pi)\:=\:\cos(x)\cos(2\pi)\,-\,\sin(x)\sin(2\pi)\)
. . . . . . . . . . . . . . . \(\displaystyle =\:\cos(x)\cdot1\,-\,\sin(x)\cdot0\:=\:\cos(x)\)
So the right side is: \(\displaystyle \;\frac{\sin(x+2\pi)}{\cos(x+2\pi)}\:=\:\frac{\sin(x)}{\cos(x)}\:=\:\tan(x)\)
