Simplify $(-2\sin(t)-2\sin(2t))^2+(2\cos(t)-2\cos(2t))^2$? In calculating the length of a deltoid one gets the following string of trigonometric functions:
$$(-2\sin(t)-2\sin(2t))^2+(2\cos(t)-2\cos(2t))^2$$
$$=4\big(\sin^2(2t)+2\sin(t)\sin(2t)+\cos^2(2t)-2\cos(t)\cos(2t)+\sin^2(t)+\cos^2(t)\big)$$
$$(\sin^2x+\cos^2x=1)$$
$$=4\big(1+2\sin(t)\sin(2t)-2\cos(t)\cos(2t)+1\big)$$
Question is, how to simplify this further?

Notice, 
$$4(1+2\sin t\sin 2t-2\cos t\cos 2t+1)$$
$$=8(1+\sin t\sin 2t-\cos t\cos 2t)$$
$$=8(1-(\cos t\cos 2t-\sin t\sin 2t)$$
using $\color{red}{\cos A\cos B-\sin A\sin B=\cos (A+B)}$, 
$$=8(1-\cos(t+2t))$$ $$=8(1-\cos 3t)$$
using $\color{red}{\cos A=1-2\sin^2\frac{A}{2}}$,
$$=8\left(1-\left(1-2\sin^2 \frac{3t}{2}\right)\right)$$
$$=8\left(1-1+2\sin^2 \frac{3t}{2}\right)$$
$$=8\left(2\sin^2 \frac{3t}{2}\right)=\color{blue}{16\sin^2 \frac{3t}{2}}$$
