Just double the angle

Calculus Level 3

( ∑ j = 0 ∞ ( − 1 ) j x 2 j ( 2 j ) ! ) ( ∑ k = 0 ∞ ( − 1 ) k x 2 k + 1 ( 2 k + 1 ) ! ) \large \left( \sum_{j=0}^\infty \dfrac{(-1)^{j} x^{2j}}{(2j)!} \right) \left( \sum_{k=0}^\infty \dfrac{(-1)^{k} x^{2k+1}}{(2k+1)!} \right)

Which of the following must be equal to the expression above?

∑ q = 0 ∞ ( − 1 ) q ( 2 x ) 2 q 2 ( 2 q ) ! \sum_{q=0}^\infty \dfrac{(-1)^{q} (2x)^{2q}}{2(2q)!} ∑ s = 0 ∞ ( − 1 ) s + 1 ( 2 x ) 2 s 2 ( 2 s ) ! \sum_{s=0}^\infty \dfrac{(-1)^{s+1} (2x)^{2s}}{2(2s)!} ∑ p = 0 ∞ ( − 1 ) p ( 2 x ) 2 p + 1 2 ( 2 p + 1 ) ! \sum_{p=0}^\infty \dfrac{(-1)^{p} (2x)^{2p+1}}{2(2p+1)!} ∑ r = 0 ∞ ( − 1 ) r + 1 ( 2 x ) 2 r + 1 2 ( 2 r + 1 ) ! \sum_{r=0}^\infty \dfrac{(-1)^{r+1} (2x)^{2r+1}}{2(2r+1)!}

Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

Michael Mendrin
Aug 15, 2016

The title gives away the solution. We're looking at the series expansion of

C o s ( x ) S i n ( x ) = 1 2 S i n ( 2 x ) Cos(x)Sin(x)=\dfrac{1}{2}Sin(2x)

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