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Question Number 193423 by mustafazaheen last updated on 13/Jun/23

when tan(θ/2)=(1/a) then find cosθ=? from the a

$$\mathrm{when}\:\:\:\mathrm{tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{a}} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{cos}\theta=?\:\mathrm{from}\:\mathrm{the}\:\mathrm{a} \\ $$

Answered by deleteduser1 last updated on 13/Jun/23

tan((θ/2)+(θ/2))=((2tan((θ/2)))/(1−tan^2 ((θ/2))))⇒tan(θ)=((2a)/(a^2 −1)) sec^2 (θ)=1+tan^2 θ=1+((4a^2 )/((a^2 −1)^2 ))=(((a^2 −1)^2 +4a^2 )/((a^2 −1)^2 )) ⇒cosθ=((a^2 −1)/( (√((a^2 +1)^2 ))))=((a^2 −1)/(a^2 +1))=1−(2/(a^2 +1))

$${tan}\left(\frac{\theta}{\mathrm{2}}+\frac{\theta}{\mathrm{2}}\right)=\frac{\mathrm{2}{tan}\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}\Rightarrow{tan}\left(\theta\right)=\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$${sec}^{\mathrm{2}} \left(\theta\right)=\mathrm{1}+{tan}^{\mathrm{2}} \theta=\mathrm{1}+\frac{\mathrm{4}{a}^{\mathrm{2}} }{\left({a}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=\frac{\left({a}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} }{\left({a}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{cos}\theta=\frac{{a}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }}=\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1}−\frac{\mathrm{2}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$

Commented by mustafazaheen last updated on 13/Jun/23

T

$$\:\underbrace{{T}} \\ $$

Answered by Subhi last updated on 13/Jun/23

cos(θ)=1−2sin^2 ((θ/2)) = 2cos^2 ((θ/2))−1 sin((θ/2))=(√((1−cos(θ))/2)) cos((θ/2))=(√((1+cos(θ))/2)) ((sin((θ/2)))/(cos((θ/2))))=tan((θ/2))=(1/a)=(√((1−cos(θ))/(1+cos(θ)))) a^2 (1−cos(θ))=1+cos(θ) ⇛ a^2 −1=(1+a^2 )cos(θ) cos(θ)=((a^2 −1)/(a^2 +1))

$${cos}\left(\theta\right)=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)\:=\:\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)−\mathrm{1} \\ $$$${sin}\left(\frac{\theta}{\mathrm{2}}\right)=\sqrt{\frac{\mathrm{1}−{cos}\left(\theta\right)}{\mathrm{2}}} \\ $$$${cos}\left(\frac{\theta}{\mathrm{2}}\right)=\sqrt{\frac{\mathrm{1}+{cos}\left(\theta\right)}{\mathrm{2}}} \\ $$$$\frac{{sin}\left(\frac{\theta}{\mathrm{2}}\right)}{{cos}\left(\frac{\theta}{\mathrm{2}}\right)}={tan}\left(\frac{\theta}{\mathrm{2}}\right)=\frac{\mathrm{1}}{{a}}=\sqrt{\frac{\mathrm{1}−{cos}\left(\theta\right)}{\mathrm{1}+{cos}\left(\theta\right)}} \\ $$$${a}^{\mathrm{2}} \left(\mathrm{1}−{cos}\left(\theta\right)\right)=\mathrm{1}+{cos}\left(\theta\right)\:\Rrightarrow\:{a}^{\mathrm{2}} −\mathrm{1}=\left(\mathrm{1}+{a}^{\mathrm{2}} \right){cos}\left(\theta\right) \\ $$$${cos}\left(\theta\right)=\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$

Commented by mustafazaheen last updated on 13/Jun/23

⋛

$$\:\cancel{\lesseqgtr} \\ $$

Commented by York12 last updated on 16/Jun/23

⋛

$$\:\cancel{\lesseqgtr} \\ $$

Answered by BaliramKumar last updated on 13/Jun/23

cosθ = ((1−tan^2 ((θ/2)))/(1+tan^2 ((θ/2)))) = ((1−((1/a))^2 )/(1+((1/a))^2 )) = ((1−(1/a^2 ))/(1+(1/a^2 ))) cosθ = ((a^2 −1)/(a^2 +1)) pythagorean triplet⇒ 2a_(P) , (a^2 −1)_(B) , (a^2 +1)_(H) tanθ = ((2a)/(a^2 −1))

$$\mathrm{cos}\theta\:=\:\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{a}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{a}}\right)^{\mathrm{2}} }\:=\:\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }}\: \\ $$$$\mathrm{cos}\theta\:=\:\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{pythagorean}\:\mathrm{triplet}\Rightarrow\:\underset{\mathrm{P}} {\underbrace{\mathrm{2a}}},\:\underset{\mathrm{B}} {\underbrace{\left(\mathrm{a}^{\mathrm{2}} −\mathrm{1}\right)}},\:\underset{\mathrm{H}} {\underbrace{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{1}\right)}} \\ $$$$\mathrm{tan}\theta\:=\:\frac{\mathrm{2a}}{\mathrm{a}^{\mathrm{2}} −\mathrm{1}} \\ $$

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