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Question Number 149309 by ArielVyny last updated on 04/Aug/21

if t=tanx what is the value of sinx and cosx ?

$${if}\:\:\:\:{t}={tanx}\:{what}\:{is}\:{the}\:{value}\:{of} \\ $$$${sinx}\:{and}\:{cosx}\:? \\ $$

Answered by nimnim last updated on 04/Aug/21

sinx=(t/( (√(1+t^2 )))), cosx=(1/( (√(1+t^2 ))))

$${sinx}=\frac{{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }},\:\:{cosx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$

Answered by puissant last updated on 04/Aug/21

sin(x)=((∣t∣)/( (√(1+t^2 )))) ; cos(x)=(1/( (√(1+t^2 ))))..

$$\mathrm{sin}\left(\mathrm{x}\right)=\frac{\mid\mathrm{t}\mid}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\:\:;\:\:\mathrm{cos}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}.. \\ $$

Answered by mathmax by abdo last updated on 04/Aug/21

cos^2 x =(1/(1+tan^2 x))=(1/(1+t^2 )) ⇒cos(x)=+^− (1/( (√(1+t^2 )))) sin^2 x=1−cos^2 x =1−(1/(1+t^2 ))=(t^2 /(1+t^2 )) ⇒sinx =((∣t∣)/( (√(1+t^2 ))))

$$\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\Rightarrow\mathrm{cos}\left(\mathrm{x}\right)=\overset{−} {+}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }} \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{x}=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }=\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\Rightarrow\mathrm{sinx}\:=\frac{\mid\mathrm{t}\mid}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }} \\ $$

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