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?
On
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 $\color{red}{\cos A=1-2\sin^2\frac{A}{2}}$, $$=8\left(1-\left(1-2\sin^2 \frac{3t}{2}\right)\right)$$ $$=8\left(1-1+2\sin^2 \frac{3t}{2}\right)$$ $$=8\left(2\sin^2 \frac{3t}{2}\right)=\color{blue}{16\sin^2 \frac{3t}{2}}$$
\begin{align} \ldots &= 4\big(1 +2\sin(t)\sin(2t) -2\cos(t)\cos(2t) +1\big) \\ &=4(2 -2\cos 3t) \\ &=8(1 -\cos 3t) \\ &=16 (\sin ( 3 t/2) )^2 \\ &=\big( 4 \sin ( 3 t/2) \big)^2. \end{align}