Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 113769 by deleteduser12 last updated on 15/Sep/20

prove that, tan (7(1/2))°=(√6)−(√3)+(√2)−2

$${prove}\:{that},\:\mathrm{tan}\:\left(\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}}\right)°=\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}−\mathrm{2} \\ $$

Answered by Dwaipayan Shikari last updated on 15/Sep/20

tan((π/(24)))=((sin(π/(24)))/(cos(π/(24))))=((2sin^2 (π/(24)))/(2sin(π/(24))cos(π/(24))))=((1−cos(π/(12)))/(sin(π/(12)))) cos(π/(12))=cos((π/3)−(π/4))=(((√3)+1)/(2(√2))) sin(π/(12))=(((√3)−1)/(2(√2))) So ((1−cos(π/(12)))/(sin(π/(12))))=((2(√2)−(√3)+1)/( (√3)−1))=((2(√2)((√3)+1))/2)−(2+(√3))=(√6)−(√3)+(√2)−2

$${tan}\left(\frac{\pi}{\mathrm{24}}\right)=\frac{{sin}\frac{\pi}{\mathrm{24}}}{{cos}\frac{\pi}{\mathrm{24}}}=\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{24}}}{\mathrm{2}{sin}\frac{\pi}{\mathrm{24}}{cos}\frac{\pi}{\mathrm{24}}}=\frac{\mathrm{1}−{cos}\frac{\pi}{\mathrm{12}}}{{sin}\frac{\pi}{\mathrm{12}}} \\ $$$${cos}\frac{\pi}{\mathrm{12}}={cos}\left(\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${sin}\frac{\pi}{\mathrm{12}}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${So}\:\frac{\mathrm{1}−{cos}\frac{\pi}{\mathrm{12}}}{{sin}\frac{\pi}{\mathrm{12}}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{1}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}}−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)=\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}−\mathrm{2} \\ $$$$ \\ $$$$ \\ $$

Answered by $@y@m last updated on 15/Sep/20

tan 30^o =((2tan 15^o )/(1−tan^2 15^o )) (1/( (√3)))=((2x)/(1−x^2 )) where x=tan 15^o 1−x^2 =2(√3)x x^2 +2(√3)x−1=0 x=((−2(√3)±(√(12+4)))/2) x=((−2(√3)±4)/2)=−(√3)±2 But tan 15^o >0 ∴tan 15^o =2−(√3) ...(1) Similarly, tan 15^o =((2tan (7(1/2))^o )/(1−tan^2 (7(1/2))^o )) 2−(√3)=((2y)/(1−y^2 )) where y=tan (7(1/2))^o (2−(√3))y^2 +2y−(2−(√3))=0 y=((−2±(√(4+4(4+3−4(√3)))))/(2(2+(√3)))) y=((−2±2(√(1+(7−4(√3)))))/(2(2−(√3)))) y=((−1±(√(8−4(√3))))/(2+(√3))) y=((−1±((√6)−(√2)))/(2+(√3))) But y>0 ∴ y=(((√6)−(√2)−1)/(2−(√3))) y=(((√6)−(√2)−1)/(2−(√3))) ×((2+(√3))/(2+(√3))) y=((2(√6)−2(√2)−2−(√6)+(√(18))−(√3))/(2^2 −(√3^2 ))) y=(√6)−2(√2)−2+3(√2)−(√3) y=(√6)+(√2)−(√3)−2

$$\mathrm{tan}\:\mathrm{30}^{\mathrm{o}} =\frac{\mathrm{2tan}\:\mathrm{15}^{\mathrm{o}} }{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{15}^{\mathrm{o}} } \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:{where}\:{x}=\mathrm{tan}\:\mathrm{15}^{{o}} \\ $$$$\mathrm{1}−{x}^{\mathrm{2}} =\mathrm{2}\sqrt{\mathrm{3}}{x} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{2}\sqrt{\mathrm{3}}\pm\sqrt{\mathrm{12}+\mathrm{4}}}{\mathrm{2}} \\ $$$${x}=\frac{−\mathrm{2}\sqrt{\mathrm{3}}\pm\mathrm{4}}{\mathrm{2}}=−\sqrt{\mathrm{3}}\pm\mathrm{2} \\ $$$${But}\:\mathrm{tan}\:\mathrm{15}^{{o}} >\mathrm{0} \\ $$$$\therefore\mathrm{tan}\:\mathrm{15}^{{o}} =\mathrm{2}−\sqrt{\mathrm{3}}\:...\left(\mathrm{1}\right) \\ $$$${Similarly}, \\ $$$$\mathrm{tan}\:\mathrm{15}^{\mathrm{o}} =\frac{\mathrm{2tan}\:\left(\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{o}} }{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{o}} } \\ $$$$\mathrm{2}−\sqrt{\mathrm{3}}=\frac{\mathrm{2}{y}}{\mathrm{1}−{y}^{\mathrm{2}} }\:{where}\:{y}=\mathrm{tan}\:\left(\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{o}} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){y}^{\mathrm{2}} +\mathrm{2}{y}−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)=\mathrm{0} \\ $$$${y}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{4}\left(\mathrm{4}+\mathrm{3}−\mathrm{4}\sqrt{\mathrm{3}}\right)}}{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \\ $$$${y}=\frac{−\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{1}+\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)}}{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)} \\ $$$${y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{8}−\mathrm{4}\sqrt{\mathrm{3}}}}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${y}=\frac{−\mathrm{1}\pm\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${But}\:{y}>\mathrm{0} \\ $$$$\therefore\:{y}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}\: \\ $$$$\:{y}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}\:×\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${y}=\frac{\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}−\sqrt{\mathrm{6}}+\sqrt{\mathrm{18}}−\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{2}} −\sqrt{\mathrm{3}^{\mathrm{2}} }} \\ $$$${y}=\sqrt{\mathrm{6}}−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}+\mathrm{3}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}} \\ $$$${y}=\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{2} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com