Does this apply: $\sin^2(\pi t-\pi) + \cos^2(\pi t-\pi) = 1$ So I'm solving for the length of the following parametric description:

$$
\Bigg[x(t)=3\cos(\pi t-\pi)\,\,\,\,\,\,\,y(t)=3\sin(\pi t - \pi)$$
I applied the formula for solving for length, namely:
$$\int_{0}^{2} \sqrt{(-3\pi \sin(\pi t - \pi))^2+(3\pi \cos(\pi t - \pi))^2 }\,\,\, dt$$
$$\int_{0}^{2} \sqrt{9\pi^2\times1}\,\,\, dt$$

My question is:
Is it correct that [ $(\sin(\pi t - \pi))^2+ (\cos(\pi t - \pi))^2 = 1$ ] because of the rule: $\cos^2(x) + \sin^2(x) = 1$ ? Can I always apply this convention as long as I have the same on the inside of the sin and cos?

Highly appreciated,

-Bowser
Yes it is, the relation $\sin^2(x)+\cos^2(x)=1$ holds for any $x$ value
Hint you can verify using formulae $$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)$$ and $$cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$$