Trigonometry (11th Edition) Clone

Write $P$ in terms of $\sin(2\pi t)$: $$P=16k-16k\sin^2(2\pi t)$$
From part a), we have the expression of $P$ $$P=16k\cos^2(2\pi t)$$ From Pythagorean Identity, we have $$\cos^2\theta=1-\sin^2\theta$$ Therefore, similarly in $P$, $$P=16k[1-\sin^2(2\pi t)]$$ $$P=16k-16k\sin^2(2\pi t)$$