Question Number 206899 by MATHEMATICSAM last updated on 29/Apr/24
$$\mathrm{If}\:\mathrm{tan}\theta\:=\:\frac{\mathrm{2}{x}\left({x}\:+\:\mathrm{1}\right)}{\mathrm{2}{x}\:+\:\mathrm{1}}\:\mathrm{then}\:\mathrm{find}\:\mathrm{sin}\theta\:\mathrm{and} \\ $$$$\mathrm{cos}\theta. \\ $$
Answered by mathzup last updated on 29/Apr/24
$$\mathrm{1}+{tan}^{\mathrm{2}} \theta\:=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}\:\Rightarrow{cos}^{\mathrm{2}} \theta\:=\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{4}{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }}=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\:+\mathrm{4}{x}^{\mathrm{4}} +\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{\mathrm{4}{x}^{\mathrm{4}} +\mathrm{8}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}\:\Rightarrow \\ $$$$\Rightarrow{cos}\theta\:=\overset{−} {+}\sqrt{\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{\mathrm{4}{x}^{\mathrm{4}} +\mathrm{8}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}} \\ $$$${sin}\theta=\overset{−} {+}\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} \theta} \\ $$
Answered by Frix last updated on 29/Apr/24
$${c}=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\wedge{s}=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\Rightarrow \\ $$$$\mathrm{cos}\:\theta\:=\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}} \\ $$$$\mathrm{sin}\:\theta\:=\frac{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}} \\ $$
Answered by lepuissantcedricjunior last updated on 03/May/24
![tan(𝛉)=((2x(x+1))/(2x+1)) or calculons sin𝛉 et cos𝛉 on a tan𝛉=((sin𝛉)/(cos𝛉))=((2x(1+x))/(2x+1)) =>((sin^2 𝛉)/(1−sin^2 𝛉))=[((2x(1+x))/(2x+1))]^2 =>sin^2 𝛉[4x^2 +8x^2 +4x^4 +4x^2 +4x+1]=(([2x(x+1)]^2 )/) sin^2 𝛉=(([2x(1+x)]^2 )/((2x+1)^2 +[2x(1+x)^2 )) =>sin𝛉=∓((2x(1+x))/( (√((2x+1)^2 +[2x(x+1)]^2 )))) =>cos𝛉=∓((2x+1)/( (√((2x+1)^2 +[2x(1+x)]^2 )))) ........le puissant Dr...................](https://www.tinkutara.com/question/Q207026.png)
$$\boldsymbol{{tan}}\left(\boldsymbol{\theta}\right)=\frac{\mathrm{2}\boldsymbol{{x}}\left(\boldsymbol{{x}}+\mathrm{1}\right)}{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}} \\ $$$$\boldsymbol{{or}}\:\boldsymbol{{calculons}}\:\boldsymbol{{sin}\theta}\:\boldsymbol{{et}}\:\boldsymbol{{cos}\theta} \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{a}} \\ $$$$\boldsymbol{{tan}\theta}=\frac{\boldsymbol{{sin}\theta}}{\boldsymbol{{cos}\theta}}=\frac{\mathrm{2}\boldsymbol{{x}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}} \\ $$$$=>\frac{\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\theta}}{\mathrm{1}−\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\theta}}=\left[\frac{\mathrm{2}\boldsymbol{{x}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}}\right]^{\mathrm{2}} \\ $$$$=>\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\theta}\left[\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{x}}+\mathrm{1}\right]=\frac{\left[\mathrm{2}\boldsymbol{{x}}\left(\boldsymbol{{x}}+\mathrm{1}\right)\right]^{\mathrm{2}} }{} \\ $$$$\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\theta}=\frac{\left[\mathrm{2}\boldsymbol{{x}}\left(\mathrm{1}+\boldsymbol{{x}}\right)\right]^{\mathrm{2}} }{\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{2}} +\left[\mathrm{2}\boldsymbol{{x}}\left(\mathrm{1}+\boldsymbol{{x}}\right)^{\mathrm{2}} \right.} \\ $$$$=>\boldsymbol{{sin}\theta}=\mp\frac{\mathrm{2}\boldsymbol{{x}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}{\:\sqrt{\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{2}} +\left[\mathrm{2}\boldsymbol{{x}}\left(\boldsymbol{{x}}+\mathrm{1}\right)\right]^{\mathrm{2}} }} \\ $$$$=>\boldsymbol{{cos}\theta}=\mp\frac{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}}{\:\sqrt{\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{2}} +\left[\mathrm{2}\boldsymbol{{x}}\left(\mathrm{1}+\boldsymbol{{x}}\right)\right]^{\mathrm{2}} }} \\ $$$$……..\boldsymbol{{le}}\:\boldsymbol{{puissant}}\:\boldsymbol{{Dr}}………………. \\ $$$$ \\ $$