Question Number 81003 by mathocean1 last updated on 08/Feb/20
$$\left(\mathrm{E}\right):\:\mathrm{sin2}{x}={cosx}+{sinx}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}.\:{show}\:{that}\:\left(\mathrm{E}\right)\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\: \\ $$$$\left(\mathrm{E}'\right):\:\mathrm{2cos}^{\mathrm{2}} \mathrm{X}−\sqrt{\mathrm{2}}\mathrm{cosX}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{with}\:\mathrm{X}={x}−\frac{\pi}{\mathrm{4}}. \\ $$
Commented by MJS last updated on 08/Feb/20
$$\mathrm{wrong} \\ $$
Commented by mathocean1 last updated on 09/Feb/20
$$\mathrm{how}\:\mathrm{sir}? \\ $$
Commented by jagoll last updated on 09/Feb/20
$$\mathrm{sin}\:{x}+\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\:=\:\sqrt{\mathrm{2}\:}\:\mathrm{cos}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:{X}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${now}\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{1}+\mathrm{2sin}\:{x}\mathrm{cos}\:{x}\:−\mathrm{1} \\ $$$$\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} −\mathrm{1}\:=\:\left(\sqrt{\mathrm{2}\:}\:\mathrm{cos}\:{X}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$${so}\:{E}\:\Rightarrow\:\mathrm{2cos}\:^{\mathrm{2}} {X}−\mathrm{1}\:=\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:{X}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2cos}\:^{\mathrm{2}} {X}−\sqrt{\mathrm{2}}\:\mathrm{cos}\:{X}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$
Commented by MJS last updated on 09/Feb/20
$$\mathrm{haha}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{subtract}\:\mathrm{correctly}\:\mathrm{anymore}... \\ $$$$\mathrm{time}\:\mathrm{for}\:\mathrm{the}\:\mathrm{primary}\:\mathrm{school}\:\mathrm{again} \\ $$