Question Number 102635 by ketto255 last updated on 10/Jul/20
Answered by bobhans last updated on 10/Jul/20
$$\mathrm{sin}\:\left(\mathrm{90}^{{o}} −\theta\right)\:=\:\mathrm{cos}\:\theta\:=\pm\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta}\:= \\ $$$$\pm\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{25}}}\:=\:\pm\:\frac{\sqrt{\mathrm{23}}}{\mathrm{5}} \\ $$
Answered by Rasheed.Sindhi last updated on 10/Jul/20
$$\mathrm{sin}\left(\mathrm{90}−\theta\right)=\mathrm{cos}\:\theta\: \\ $$$$\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta\:=\mathrm{1}\: \\ $$$$\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} \theta\:=\mathrm{1} \\ $$$$\mathrm{cos}\:\theta=\pm\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{25}}}=\pm\frac{\sqrt{\mathrm{23}}}{\mathrm{5}} \\ $$$$\mathrm{sin}\left(\mathrm{90}−\theta\right)=\pm\frac{\sqrt{\mathrm{23}}}{\mathrm{5}}\:\:\bigstar \\ $$