Question:
Find the value $$f^{(2)}_{n}(x)=\sin^2{x}+\sin^2{(2x)}+\sin^2{(3x)}+\cdots+\sin^2{(nx)}$$
My solution:
since
$$\sin^2{x}=\dfrac{1}{2}(1-\cos{(2x)})$$ so $$f_{n}(x)=\dfrac{n}{2}-\dfrac{1}{2}(\cos{(2x)}+\cos{(4x)}+\cdots+\cos{(2nx)})$$ since $$2\sin{x}\cos{y}=\sin{(x+y)}-\sin{(y-x)}$$ so $$2\sin{x}\cdot[\cos{(2x)}+\cos{(4x)}+\cdots+\cos{(2nx)}]=\sin{(2n+1)x}-\sin{x}$$
Have other methods? Thank you
Question 2: Find this sum closed form $$f^{(3)}_{n}(x)=\sin^3{x}+\sin^3{(2x)}+\cdots+\sin^3{(nx)}$$ $$\cdots$$ $$f^{(m)}_{n}(x)=\sin^m{x}+\sin^m{(2x)}+\cdots+\sin^m{(nx)}=?$$
$\endgroup$ 21 Answer
$\begingroup$HINT: Using Euler's formula, $\displaystyle 2i\sin y=e^{iy}-e^{-iy}$
$$\sin^m(rx)=\left(\frac{e^{irx}-e^{-irx}}{2i}\right)^m$$
Expand and use Euler's formula to find $\sin^m(rx)$ as summation of multiples angles of $rx$ sine & cosine ratios
For example, $$\sin^3(rx)=\left(\frac{e^{irx}-e^{-irx}}{2i}\right)^3=\frac{e^{i3rx}-e^{-i3rx}+3\left(e^{irx}-e^{-irx}\right)}{-8i}=\frac{2i\sin3rx+3(2i\sin rx)}{-8i}$$
$\endgroup$ 2
