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Question Number 20434 by pp75718276@gmail.com last updated on 26/Aug/17
If  cos x=tan y, cos y=tan z, cos z=tan x,  then the value of  sin x  is
$$\mathrm{If}\:\:\mathrm{cos}\:{x}=\mathrm{tan}\:{y},\:\mathrm{cos}\:{y}=\mathrm{tan}\:{z},\:\mathrm{cos}\:{z}=\mathrm{tan}\:{x}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{sin}\:{x}\:\:\mathrm{is} \\ $$
Answered by ajfour last updated on 27/Aug/17
1+tan^2 z=(1/(cos^2 z))  ⇒  1+cos^2 y =(1/(tan^2 x))  ⇒ 1+(1/(1+tan^2 y))=(1/(tan^2 x))  ⇒  1+(1/(1+cos^2 x))=((1−sin^2 x)/(sin^2 x))  let sin^2 x=t  then        1+(1/(2−t))=((1−t)/t)       ⇒   ((3−t)/(2−t)) = ((1−t)/t)  ⇒  3t−t^2 =2+t^2 −3t    2t^2 −6t+2=0     2(t−(3/2))^2 =(5/2)  t=(3/2)−((√5)/2)  sin x = ±(√(((3−(√5))/2) )) .
$$\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {z}=\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {z}} \\ $$$$\Rightarrow\:\:\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {y}\:=\frac{\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}} {x}} \\ $$$$\Rightarrow\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {y}}=\frac{\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}} {x}} \\ $$$$\Rightarrow\:\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}}=\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:^{\mathrm{2}} {x}} \\ $$$${let}\:\mathrm{sin}\:^{\mathrm{2}} {x}={t} \\ $$$${then}\:\:\:\:\:\:\:\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}−{t}}=\frac{\mathrm{1}−{t}}{{t}} \\ $$$$\:\:\:\:\:\Rightarrow\:\:\:\frac{\mathrm{3}−{t}}{\mathrm{2}−{t}}\:=\:\frac{\mathrm{1}−{t}}{{t}} \\ $$$$\Rightarrow\:\:\mathrm{3}{t}−{t}^{\mathrm{2}} =\mathrm{2}+{t}^{\mathrm{2}} −\mathrm{3}{t} \\ $$$$\:\:\mathrm{2}{t}^{\mathrm{2}} −\mathrm{6}{t}+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\mathrm{2}\left({t}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${t}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:{x}\:=\:\pm\sqrt{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:}\:. \\ $$
Commented by myintkhaing last updated on 27/Aug/17
Check (1/(tan^2 x))
$$\mathrm{Check}\:\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \mathrm{x}} \\ $$
Answered by Tinkutara last updated on 27/Aug/17
cos^2  x = tan^2  y = sec^2  y − 1 = cot^2  z − 1  1+cos^2  x=cot^2  z=((cos^2  z)/(1−cos^2  z))=((tan^2  x)/(1−tan^2  x))  2−sin^2  x=((sin^2  x)/(cos^2  x−sin^2  x))=((sin^2  x)/(1−2sin^2  x))  (2−sin^2  x)(1−2sin^2  x)=sin^2  x  sin^4  x−3sin^2  x+1=0  sin^2  x=((3±(√5))/2)  But sin^2  x≤1 as sin x∈[−1,1]  So sin^2  x∈[0,1]  sin^2  x=(1/2)(3−(√5))=(1/2)[(1/2)(1+5−2(√5))]  sin^2  x=((((√5)−1)/2))^2 ⇒∣sin x∣=(((√5)−1)/2)  Similarly, ∣sin y∣=∣sin z∣=(((√5)−1)/2)
$$\mathrm{cos}^{\mathrm{2}} \:{x}\:=\:\mathrm{tan}^{\mathrm{2}} \:{y}\:=\:\mathrm{sec}^{\mathrm{2}} \:{y}\:−\:\mathrm{1}\:=\:\mathrm{cot}^{\mathrm{2}} \:{z}\:−\:\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}=\mathrm{cot}^{\mathrm{2}} \:{z}=\frac{\mathrm{cos}^{\mathrm{2}} \:{z}}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{z}}=\frac{\mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}} \\ $$$$\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \:{x}=\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{sin}^{\mathrm{2}} \:{x}}=\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{x}} \\ $$$$\left(\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \:{x}\right)\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{x}\right)=\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{sin}^{\mathrm{4}} \:{x}−\mathrm{3sin}^{\mathrm{2}} \:{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${But}\:\mathrm{sin}^{\mathrm{2}} \:{x}\leqslant\mathrm{1}\:{as}\:\mathrm{sin}\:{x}\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$$${So}\:\mathrm{sin}^{\mathrm{2}} \:{x}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\right)\right] \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}=\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \Rightarrow\mid\mathrm{sin}\:{x}\mid=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$${Similarly},\:\mid\mathrm{sin}\:{y}\mid=\mid\mathrm{sin}\:{z}\mid=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$

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