Let $f : [0;1] \to \mathbb{R}$ be a continuous function such that $f(0) = 0$. Which of the following statements are true? Let $f : [0;1] \to \mathbb{R}$ be a continuous function such that $f(0) = 0$. Which of the following statements are true?
a. If $\int_ 0^{\pi} f(t) \cos nt\, dt = 0,$ for all $n \in {0} \cup \mathbb{N}$, then $f= 0$.
b. If $\int_ 0^{\pi} f(t) \sin nt\, dt = 0,$ for all $n \in  \mathbb{N}$, then $f= 0$.
c. If $\int_ 0^{\pi} f(t)\,t^n\, dt = 0,$ for all $n \in \{0\} \cup \mathbb{N}$, then $f= 0$.

My thoughts:-
(a) it is not true as we can take $f(t)=\sin{nt}$.
(b)no idea.
(c)it is true.   
Does my answers correct?
and what about (b)
 A: All the three are correct. 
c. is true due to Stone-Weierstrass Theorem.
For a. extend $f$ to $[-\pi,\pi]$ as an even function. Then, if a. holds for $\cos nx$ in $[-\pi,\pi]$, by symmetry, and for $\sin nx$ since $f$ is even.
For b. extend $f$ to $[-\pi,\pi]$ as an odd function.
For a. and b. we use the fact that: If $f:[-\pi,\pi]\to\mathbb R$ is continuous and
$$
\int_{\pi}^\pi f(x)\,\cos nx\,dx=\int_{\pi}^\pi f(x)\,\sin nx\,dx=0, \quad\text{for every}\,\,
n\in\mathbb N,
$$ 
then $f$ is constant.
A: Parseval's identity shows that a) and b) are true: For a) extend $f$ as an even function; For b), extend $f$ as an odd function.
c) is correct 
we don not need $f(0)=0$
there exist $M\gt0$ such that $|f(x)|\lt M$
By Stone-Weierstrass Theorem,  $\forall  \epsilon \gt0$, there exist a polynomial $P(x)$ such that 
$$\int_0^\pi|f(x)-P(x)| dx\lt  \epsilon.$$ 
Since $\int_ 0^{\pi} f(t)\,t^n\, dt = 0 $, we have $\int_0^\pi f(x)P(x) dx=0$, so
$$\int_0^\pi f^2(x) dx = \int_0^\pi f^2(x) dx - \int_0^\pi f(x)P(x) dx = \\
\int_0^\pi f(x) (f(x) -  P(x) )dx\leq \int_0^\pi |f(x) (f(x) -  P(x) )|dx \\
\leq M \int_0^\pi |f(x) -  P(x) |dx \lt M\epsilon$$
Hence, 
$$\int_0^\pi f^2(x) dx  =0$$
so $f\equiv 0$
