Question Number 149309 by ArielVyny last updated on 04/Aug/21
$${if}\:\:\:\:{t}={tanx}\:{what}\:{is}\:{the}\:{value}\:{of} \\ $$$${sinx}\:{and}\:{cosx}\:? \\ $$
Answered by nimnim last updated on 04/Aug/21
$${sinx}=\frac{{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }},\:\:{cosx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$
Answered by puissant last updated on 04/Aug/21
$$\mathrm{sin}\left(\mathrm{x}\right)=\frac{\mid\mathrm{t}\mid}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\:\:;\:\:\mathrm{cos}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}.. \\ $$
Answered by mathmax by abdo last updated on 04/Aug/21
$$\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\Rightarrow\mathrm{cos}\left(\mathrm{x}\right)=\overset{−} {+}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }} \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{x}=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }=\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\Rightarrow\mathrm{sinx}\:=\frac{\mid\mathrm{t}\mid}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }} \\ $$