Question Number 128130 by physicstutes last updated on 04/Jan/21
$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\:\mathrm{sin}\left(\mathrm{2}{x}\:+\:\mathrm{2}{h}\right)−\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{2cos}\left(\mathrm{2}{x}\:+{h}\right)\mathrm{sin}\:{h} \\ $$
Answered by Olaf last updated on 04/Jan/21
$$\mathrm{sin}{a}−\mathrm{sin}{b}\:=\:\mathrm{2sin}\left(\frac{{a}−{b}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right)\:\left(\mathrm{1}\right) \\ $$$${a}\:=\:\mathrm{2}{x}+\mathrm{2}{h}\:\mathrm{and}\:{b}\:=\:\mathrm{2}{x} \\ $$$$\left(\mathrm{1}\right)\::\:\:\mathrm{sin}\left(\mathrm{2}{x}+\mathrm{2}{h}\right)−\mathrm{sin}\left(\mathrm{2}{x}\right)\:=\: \\ $$$$\mathrm{2sin}{h}\mathrm{cos}\left(\mathrm{2}{x}+{h}\right) \\ $$
Answered by Olaf last updated on 04/Jan/21
![You can use Moivre : sin(2x+2h)−sin2x = ((e^(i(2x+2h)) −e^(−i(2x+2h)) )/(2i))−((e^(2ix) −e^(−2ix) )/(2i)) = (1/(2i))[e^(ih) (e^(i(2x+h)) +e^(−i(2x+h)) )−e^(−ih) (e^(i(2x+h)) +e^(−i(2x+h)) )] = 2((e^(ih) −e^(−ih) )/(2i))(((e^(i(2x+h)) +e^(−i(2x+h)) )/2)) = 2sinhcos(2x+h)](Q128141.png)
$$\mathrm{You}\:\mathrm{can}\:\mathrm{use}\:\mathrm{Moivre}\:: \\ $$$$\mathrm{sin}\left(\mathrm{2}{x}+\mathrm{2}{h}\right)−\mathrm{sin2}{x}\:=\: \\ $$$$\frac{{e}^{{i}\left(\mathrm{2}{x}+\mathrm{2}{h}\right)} −{e}^{−{i}\left(\mathrm{2}{x}+\mathrm{2}{h}\right)} }{\mathrm{2}{i}}−\frac{{e}^{\mathrm{2}{ix}} −{e}^{−\mathrm{2}{ix}} }{\mathrm{2}{i}}\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{i}}\left[{e}^{{ih}} \left({e}^{{i}\left(\mathrm{2}{x}+{h}\right)} +{e}^{−{i}\left(\mathrm{2}{x}+{h}\right)} \right)−{e}^{−{ih}} \left({e}^{{i}\left(\mathrm{2}{x}+{h}\right)} +{e}^{−{i}\left(\mathrm{2}{x}+{h}\right)} \right)\right]\:= \\ $$$$\mathrm{2}\frac{{e}^{{ih}} −{e}^{−{ih}} }{\mathrm{2}{i}}\left(\frac{{e}^{{i}\left(\mathrm{2}{x}+{h}\right)} +{e}^{−{i}\left(\mathrm{2}{x}+{h}\right)} }{\mathrm{2}}\right)\:=\: \\ $$$$\mathrm{2sin}{h}\mathrm{cos}\left(\mathrm{2}{x}+{h}\right) \\ $$