Image of $\gamma(t)=(\sin(t+k)\cos(t), \sin(t+k)\sin(t))$ is a circle. I would like to show the image of $\gamma(t)=(\sin(t+k)\cos(t), \sin(t+k)\sin(t))$ is a circle. 
I was hinted that I should appeal to the Isoperimetric Inequality. Hence I calculate the area that the simply closed positively oriented curve $\gamma(t)$ encloses: $$A(\gamma(t))=\frac{1}{2}\int_0^T(x\dot y-\dot xy )dt$$ where $T=2\pi$ in our case. $$\dot x=\cos(t+k)\cos(t)-\sin(t+k)\sin(t), \dot y=\cos(t+k)\sin(t)+\sin(t+k)\cos(t))\implies x\dot y-\dot xy= \sin^2(t+k)\implies A(\gamma(t))=\frac{1}{2}\int_0^{2\pi}\sin^2(t+k)dt=\frac{\pi}{2}$$
Similarly, I calculate the closed curve's length $l(\gamma(t))$: $$l(\gamma(t))=\int_0^{2\pi}|\dot \gamma(t)|dt=\int_0^{2\pi}1dt=2\pi$$
According to the isoperimetric Equality, $$A(\gamma(t))=\frac{1}{4\pi}l(\gamma(t))^2$$ if and only if $\gamma(t)$ is a circle. Clearly, $\frac{\pi}{2}\ne\frac{1}{4\pi}(2\pi)^2$. $$$$What am I missing here? 

The problem is in setting $T = 2\pi$, because your function is actually $\pi$ periodic. Hence the area enclosed is
$$A = \frac 1 2 \int_0^{\pi} x\dot y - y \dot x = \frac 1 2 \int_0^{\pi} \sin^2(t + k) \, dt = \frac{\pi}{4}$$
while the corresponding length is 
$$\ell(\gamma(t)) = \int_0^{\pi} \, dt = \pi.$$
Now the equality
$$A = \frac{\ell^2}{4\pi}$$
holds.

For a purely algebraic calculation that also reveals the true periodicity of the curve, note that
$$x(t) = \frac 1 2 \big(\sin(2t + k) + \sin(k)\big)$$
and
$$y(t) = -\frac 1 2 \big(\cos(2t + k) - \cos(k)\big).$$
