Question Number 13401 by Tinkutara last updated on 19/May/17
$$\mathrm{If}\:\mathrm{sin}\:\left(\pi\:\mathrm{cos}\:\theta\right)\:=\:\mathrm{cos}\:\left(\pi\:\mathrm{sin}\:\theta\right),\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{sin}\:\mathrm{2}\theta\:=\:\pm\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Answered by ajfour last updated on 19/May/17
$$\mathrm{sin}\:{x}=\mathrm{cos}\:{y} \\ $$$${x}=\frac{\pi}{\mathrm{2}}\pm{y}\:\:\:\:\:\left({at}\:{least}\right) \\ $$$${here}\:\:\:\:\mathrm{sin}\:\left(\pi\mathrm{cos}\:\theta\right)=\mathrm{cos}\:\left(\pi\mathrm{sin}\:\theta\right) \\ $$$$\pi\mathrm{cos}\:\theta=\frac{\pi}{\mathrm{2}}+\pi\mathrm{sin}\:\theta\:\:\:\:……\left({i}\right) \\ $$$$\pi\mathrm{cos}\:\theta=\frac{\pi}{\mathrm{2}}−\pi\mathrm{sin}\:\theta\:\:\:……\left({ii}\right) \\ $$$${from}\:\left({i}\right): \\ $$$$\mathrm{cos}\:\theta−\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:,\:{squaring} \\ $$$$\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta=\frac{\mathrm{1}}{\mathrm{4}}\:\:\Rightarrow\:\:\:\mathrm{sin}\:\mathrm{2}\theta=\frac{\mathrm{3}}{\mathrm{4}}\:; \\ $$$${from}\:\left({ii}\right): \\ $$$$\mathrm{cos}\:\theta+\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\:\:,\:{squaring} \\ $$$$\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta=\frac{\mathrm{1}}{\mathrm{4}}\:\:\Rightarrow\:\:\mathrm{sin}\:\mathrm{2}\theta=−\frac{\mathrm{3}}{\mathrm{4}}\:\:. \\ $$$$ \\ $$$$ \\ $$
Commented by Tinkutara last updated on 20/May/17
$$\mathrm{In}\:\left(\mathrm{i}\right)\:\mathrm{shouldn}'\mathrm{t}\:\mathrm{it}\:\mathrm{be}\:\mathrm{something}\:\mathrm{else} \\ $$$$\mathrm{because}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}\:+\:\theta\right)\:=\:−\:\mathrm{sin}\:\theta? \\ $$
Commented by ajfour last updated on 20/May/17
$${i}\:{have}\:{obtained}\:+\frac{\mathrm{3}}{\mathrm{4}}\:{as}\:{well}. \\ $$
Commented by ajfour last updated on 20/May/17
$${if}\:\mathrm{sin}\:{A}=\mathrm{cos}\:{B} \\ $$$${we}\:{conclude}\:\:{A}=\frac{\pi}{\mathrm{2}}+{B}\:,\:{here}\:{in}\:\left({i}\right) \\ $$
Commented by Tinkutara last updated on 20/May/17
$$\mathrm{Thanks}! \\ $$