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
 A: Yes it is, the relation $\sin^2(x)+\cos^2(x)=1$ holds for any $x$ value
A: Notice:


*

*$$3\cos(\pi t-\pi)=-3\cos(\pi t)$$

*$$3\sin(\pi t-\pi)=-3\sin(\pi t)$$


Now:


*

*For the integrand part:


$$\sqrt{\frac{\text{d}}{\text{d}t}\left(-3\cos(\pi t)\right)^2+\frac{\text{d}}{\text{d}t}\left(-3\sin(\pi t)\right)^2}=\sqrt{9\pi^2\sin^2(\pi t)+9\pi^2\cos^2(\pi t)}=$$
$$\sqrt{9\pi^2\left[\sin^2(\pi t)+\cos^2(\pi t)\right]}=\sqrt{9\pi^2}\sqrt{\sin^2(\pi t)+\cos^2(\pi t)}=$$
$$3\pi\sqrt{\sin^2(\pi t)+\cos^2(\pi t)}$$
Now, know that for any value of $x$, we have the identity $\cos^2(x)+\sin^2(x)=1$, and notice $b>a$:
$$3\pi\int_{a}^{b}\sqrt{\sin^2(\pi t)+\cos^2(\pi t)}\space\text{d}t=3\pi\int_{a}^{b}\sqrt{1}\space\text{d}t=3\pi\int_{a}^{b}1\space\text{d}t=3\pi\left[t\right]_{a}^{b}=3\pi\left(b-a\right)$$
And in your example $a=0$ and $b=2$:
$$\text{I}=3\pi\int_{0}^{2}\sqrt{\sin^2(\pi t)+\cos^2(\pi t)}\space\text{d}t=3\pi\left(2-0\right)=6\pi$$
A: Hint you can verify using formulae $$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)$$ and $$cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$$
