The discussion centers on proving the equation $$\sum_{k=1}^n \left(2^k\sin^2\frac{x}{2^k}\right)^2=\left(2^n\sin\frac{x}{2^n}\right)^2-\sin^2x$$. Participants provided insights and solutions, confirming the validity of the equation through mathematical reasoning. The proof involves manipulating the sine function and applying properties of summation to arrive at the conclusion.
PREREQUISITESMathematics students, educators, and anyone interested in advanced trigonometric proofs and series summation techniques.
Pranav said:Prove the following:
$$\sum_{k=1}^n \left(2^k\sin^2\frac{x}{2^k}\right)^2=\left(2^n\sin\frac{x}{2^n}\right)^2-\sin^2x$$
anemone said:My solution:
Notice that
$\begin{align*}\left(2^k\sin^2\frac{x}{2^k}\right)^2&=2^{2k}\sin^2\dfrac{x}{2^k}\left(\sin^2\dfrac{x}{2^k}\right)\\&=2^{2k}\sin^2\dfrac{x}{2^k}\left(1-\cos^2\dfrac{x}{2^k}\right)\\&=2^{2k}\sin^2\dfrac{x}{2^k}-2^{2k}\sin^2\dfrac{x}{2^k}\cos^2\dfrac{x}{2^k}\\&=2^{2k}\sin^2\dfrac{x}{2^k}-2^{2k-2}\left(\sin^2\dfrac{x}{2^{k-1}}\right)\end{align*}$
Hence,
$$\sum_{k=1}^n \left(2^k\sin^2\frac{x}{2^k}\right)^2$$$$=\sum_{k=1}^n \left(2^{2k}\sin^2\dfrac{x}{2^k}-2^{2k-2}\left(\sin^2\dfrac{x}{2^{k-1}}\right) \right)$$$$=\left(4\sin^2\dfrac{x}{2}-\sin^2x\right)+\left(16\sin^2\dfrac{x}{4}-4\sin^2\dfrac{x}{2}\right)+\left(64\sin^2\dfrac{x}{8}-16\sin^2\dfrac{x}{4}\right)+\cdots$$$$+\left(2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}-2^{2n-4}\sin^2\dfrac{x}{2^{n-3}}\right)+\left(2^{2n}\sin^2\dfrac{x}{2^n}-2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}\right)$$$$=\left(\cancel{4\sin^2\dfrac{x}{2}}-\sin^2x\right)+\left(\cancel{16\sin^2\dfrac{x}{4}}-\cancel{4\sin^2\dfrac{x}{2}}\right)+\left(\cancel{64\sin^2\dfrac{x}{8}}-\cancel{16\sin^2\dfrac{x}{4}}\right)+\cdots$$$$+\left(\cancel{2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}}-\cancel{2^{2n-4}\sin^2\dfrac{x}{2^{n-3}}}\right)+\left(2^{2n}\sin^2\dfrac{x}{2^n}-\cancel{2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}}\right)$$$$=\left(2^n\sin\dfrac{x}{2^n}\right)^2-\sin^2x\text{(QED)}$$
lfdahl said:Proof by means of the induction principle:
\[\sum_{k=1}^{n}\left ( 2^ksin^2\left ( \frac{x}{2^k} \right ) \right )^2=\left ( 2^n sin\left ( \frac{x}{2^n} \right )\right )^2-sin^2(x)\]
The sum holds for $n=1$:
\[\left ( 2sin\left ( \frac{x}{2} \right ) \right )^2-sin^2(x)= 4sin^2\left ( \frac{x}{2} \right )- 4sin^2\left ( \frac{x}{2} \right )cos^2\left ( \frac{x}{2} \right )\\\\ = 4sin^2\left ( \frac{x}{2} \right )- 4sin^2\left ( \frac{x}{2} \right )\left ( 1-sin^2\left ( \frac{x}{2} \right ) \right )=\left ( 2sin^2\left ( \frac{x}{2} \right ) \right )^2\]
Assume the equation holds for some $n>1$. Then it also holds for $n+1$, because:
\[\sum_{k=1}^{n+1}\left ( 2^ksin^2\left ( \frac{x}{2^k} \right ) \right )^2 =\left ( 2^nsin\left ( \frac{x}{2^n} \right ) \right )^2-sin^2(x)+\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2\;\;\;\; (1).\]
Rewriting the first term on the RHS:
\[\left ( 2^nsin\left ( \frac{x}{2^n} \right ) \right )^2 =2^{2n}sin^2\left ( 2\cdot \frac{x}{2^{n+1}} \right )= 2^{2n+2}sin^2\left ( \frac{x}{2^{n+1}} \right )cos^2\left ( \frac{x}{2^{n+1}} \right )\\\\ =\left ( 2^{n+1} \right )^2sin^2\left ( \frac{x}{2^{n+1}} \right )\left ( 1-sin^2\left ( \frac{x}{2^{n+1}} \right ) \right ) \;\;\; (2).\]
Inserting $(2)$ into $(1)$:
\[ \sum_{k=1}^{n+1}\left ( 2^ksin^2\left ( \frac{x}{2^k} \right ) \right )^2 = \left ( 2^{n+1} \right )^2sin^2\left ( \frac{x}{2^{n+1}} \right )\left ( 1-sin^2\left ( \frac{x}{2^{n+1}} \right ) \right ) -sin^2(x)+\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2\\\\=\left ( 2^{n+1}sin\left ( \frac{x}{2^{n+1}} \right ) \right )^2 - sin^2(x)+\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2-\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2 \\\\ =\left ( 2^{n+1}sin\left ( \frac{x}{2^{n+1}} \right ) \right )^2 - sin^2(x)\]
I´m sure, there is a more elegant way to prove the identity …