Find-tan-142-5-without-tables-

Question Number 173447 by ajfour last updated on 11/Jul/22

$${Find}\:\mathrm{tan}\:\left(\mathrm{142}.\mathrm{5}°\right)\:{without} \\ $$$${tables}. \\ $$

Commented by mr W last updated on 12/Jul/22

welcome back sir! let t=tan 15° tan 30°=((2t)/(1−t^2 ))=(1/( (√3))) t^2 +2(√3)t−1=0 ⇒t=2−(√3) let s=tan 142.5° (1/(tan 15°))=tan 75°=−tan 285°=−tan (2×142.5°) (1/(2−(√3)))=−((2s)/(1−s^2 )) s^2 −2(2−(√3))s−1=0 ⇒s=tan 142.5°=2−(√3)−2(√(2−(√3))) =((√2)−(√3))((√2)+1)

$$\boldsymbol{{welcome}}\:\boldsymbol{{back}}\:\boldsymbol{{sir}}! \\ $$$$ \\ $$$${let}\:{t}=\mathrm{tan}\:\mathrm{15}° \\ $$$$\mathrm{tan}\:\mathrm{30}°=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${t}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${let}\:{s}=\mathrm{tan}\:\mathrm{142}.\mathrm{5}° \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{15}°}=\mathrm{tan}\:\mathrm{75}°=−\mathrm{tan}\:\mathrm{285}°=−\mathrm{tan}\:\left(\mathrm{2}×\mathrm{142}.\mathrm{5}°\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}=−\frac{\mathrm{2}{s}}{\mathrm{1}−{s}^{\mathrm{2}} } \\ $$$${s}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){s}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{s}=\mathrm{tan}\:\mathrm{142}.\mathrm{5}°=\mathrm{2}−\sqrt{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$

Commented by ajfour last updated on 11/Jul/22

(√(2−(√3)))=p 2−(√3)=p^2 =(1/(2+(√3))) p^2 +(1/p^2 )=4 (p+(1/p))^2 =6 p+(1/p)=(√6) p^2 −(√6)p+1=0 p=((√6)/2)−((√2)/2) hence rewriting another way s=2−(√3)−2p =2−(√3)−(√6)+(√2) thank you sir.

$$\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}={p} \\ $$$$\mathrm{2}−\sqrt{\mathrm{3}}={p}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${p}^{\mathrm{2}} +\frac{\mathrm{1}}{{p}^{\mathrm{2}} }=\mathrm{4} \\ $$$$\left({p}+\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}} =\mathrm{6} \\ $$$${p}+\frac{\mathrm{1}}{{p}}=\sqrt{\mathrm{6}} \\ $$$${p}^{\mathrm{2}} −\sqrt{\mathrm{6}}{p}+\mathrm{1}=\mathrm{0} \\ $$$${p}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${hence}\:{rewriting}\:{another}\:{way} \\ $$$${s}=\mathrm{2}−\sqrt{\mathrm{3}}−\mathrm{2}{p} \\ $$$$\:\:\:=\mathrm{2}−\sqrt{\mathrm{3}}−\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}} \\ $$$${thank}\:{you}\:{sir}. \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa11 last updated on 11/Jul/22

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

Commented by Tawa11 last updated on 13/Jul/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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