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Question Number 187330 by Humble last updated on 16/Feb/23
show that ((1−cosθ)/(1+cosθ))=tan^2 ((1/2)θ)
$$ \\ $$$${show}\:{that}\:\frac{\mathrm{1}−{cos}\theta}{\mathrm{1}+{cos}\theta}={tan}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\:\:\: \\ $$
Answered by Frix last updated on 16/Feb/23
What are we allowed to use? tan^2 x =((sin^2 x)/(cos^2 x))=(((1−cos 2x)/2)/((1+cos 2x)/2))=((1−cos 2x)/(1+cos 2x)) ⇒ ((1−cos θ)/(1+cos θ))=(((1−cos θ)/2)/((1+cos θ)/2))=((sin^2 (θ/2))/(cos^2 (θ/2)))=tan^2 (θ/2) or cos θ =((e^(iθ) +e^(−iθ) )/2)=((1+e^(2iθ) )/(2e^(iθ) )) sin θ =((1−e^(2iθ) )/(2e^(iθ) ))i ((1−cos θ)/(1+cos θ))=((−(((1−e^(iθ) )^2 )/(2e^(iθ) )))/(((1+e^(iθ) )^2 )/(2e^(iθ) )))=−(((1−e^(iθ) )/(1+e^(iθ) )))^2 = =(((1−e^(iθ) )/(1+e^(iθ) ))i)^2 =(((1−e^(2i(θ/2)) )/(1+e^(2i(θ/2)) ))i)^2 =(((((1−e^(2i(θ/2)) )/(2e^(i(θ/2)) ))i)/((1+e^(2i(θ/2)) )/(2e^(i(θ/2)) ))))^2 = =(((sin (θ/2))/(cos (θ/2))))^2 =tan^2 (θ/2)
$$\mathrm{What}\:\mathrm{are}\:\mathrm{we}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{use}? \\ $$$$\mathrm{tan}^{\mathrm{2}} \:{x}\:=\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}=\frac{\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}}{\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}}=\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}} \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{cos}\:\theta}=\frac{\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{2}}}{\frac{\mathrm{1}+\mathrm{cos}\:\theta}{\mathrm{2}}}=\frac{\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}{\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}=\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{or} \\ $$$$\mathrm{cos}\:\theta\:=\frac{\mathrm{e}^{\mathrm{i}\theta} +\mathrm{e}^{−\mathrm{i}\theta} }{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{e}^{\mathrm{2i}\theta} }{\mathrm{2e}^{\mathrm{i}\theta} } \\ $$$$\mathrm{sin}\:\theta\:=\frac{\mathrm{1}−\mathrm{e}^{\mathrm{2i}\theta} }{\mathrm{2e}^{\mathrm{i}\theta} }\mathrm{i} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{cos}\:\theta}=\frac{−\frac{\left(\mathrm{1}−\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{2}} }{\mathrm{2e}^{\mathrm{i}\theta} }}{\frac{\left(\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{2}} }{\mathrm{2e}^{\mathrm{i}\theta} }}=−\left(\frac{\mathrm{1}−\mathrm{e}^{\mathrm{i}\theta} }{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }\right)^{\mathrm{2}} = \\ $$$$=\left(\frac{\mathrm{1}−\mathrm{e}^{\mathrm{i}\theta} }{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }\mathrm{i}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}−\mathrm{e}^{\mathrm{2i}\frac{\theta}{\mathrm{2}}} }{\mathrm{1}+\mathrm{e}^{\mathrm{2i}\frac{\theta}{\mathrm{2}}} }\mathrm{i}\right)^{\mathrm{2}} =\left(\frac{\frac{\mathrm{1}−\mathrm{e}^{\mathrm{2i}\frac{\theta}{\mathrm{2}}} }{\mathrm{2e}^{{i}\frac{\theta}{\mathrm{2}}} }\mathrm{i}}{\frac{\mathrm{1}+\mathrm{e}^{\mathrm{2i}\frac{\theta}{\mathrm{2}}} }{\mathrm{2e}^{{i}\frac{\theta}{\mathrm{2}}} }}\right)^{\mathrm{2}} = \\ $$$$=\left(\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}\right)^{\mathrm{2}} =\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$
Commented by Humble last updated on 16/Feb/23
awesome!!. thank you, sir
$${awesome}!!.\:{thank}\:{you},\:{sir} \\ $$
Commented by Frix last updated on 16/Feb/23
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