Menu Close

sin-pi-7-cos-pi-14-tan-3pi-14-2cos-pi-7-1-

Tinku Tara 0 Comments
Pin




Question Number 135604 by liberty last updated on 14/Mar/21
((sin (π/7).cos (π/14))/(tan (3π/14)(2cos (π/7)−1))) =?
$$\:\frac{\mathrm{sin}\:\left(\pi/\mathrm{7}\right).\mathrm{cos}\:\left(\pi/\mathrm{14}\right)}{\mathrm{tan}\:\left(\mathrm{3}\pi/\mathrm{14}\right)\left(\mathrm{2cos}\:\left(\pi/\mathrm{7}\right)−\mathrm{1}\right)}\:=? \\ $$
Answered by EDWIN88 last updated on 14/Mar/21
Remark { ((sin 3x=3sin x−4sin^3 x)),((cos 3x=4cos^3 x−3cos x)) :} let x = (π/(14)) ⇒((sin (π/7)cos (π/14))/(tan (3π/14)(2cos (π/7)−1))) = ((sin 2x cos x)/(tan 3x (2cos 2x−1))) = ((2sin x cos^2 x cos 3x)/(sin 3x(4cos^2 x−3) )) = ((2sin x cos^2 x (4cos^3 x−3cos x))/((3sin x−4sin^3 x)(4cos^2 x−3))) = ((2cos^3 x (4cos^2 x−3))/((3−4sin^2 x)(4cos^2 x−3))) = ((2cos^3 x)/(4cos^2 x−1)) according cos 2x = cos (π/7) = Q is one of the roots of 8Q^3 −4Q^2 −4Q+1 = 0 therefore 8(1+Q)^3 = 7(1+2Q)^2 (7/8) = (((1+cos 2x)^3 )/((1+2cos 2x)^2 )) = (((2cos^2 x)^3 )/((4cos^2 x−1)^2 )) (7/8)= ((8cos^6 x)/((4cos^2 x−1)^2 )) ; (7/(16)) = (((2cos^3 x)^2 )/((4cos^2 x−1)^2 )) ∴ ((2cos^3 x)/(4cos^2 x−1)) = ((√7)/4)
$$\:\mathrm{Remark}\:\begin{cases}{\mathrm{sin}\:\mathrm{3x}=\mathrm{3sin}\:\mathrm{x}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{x}}\\{\mathrm{cos}\:\mathrm{3x}=\mathrm{4cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{3cos}\:\mathrm{x}}\end{cases} \\ $$$$\mathrm{let}\:\mathrm{x}\:=\:\frac{\pi}{\mathrm{14}}\:\Rightarrow\frac{\mathrm{sin}\:\left(\pi/\mathrm{7}\right)\mathrm{cos}\:\left(\pi/\mathrm{14}\right)}{\mathrm{tan}\:\left(\mathrm{3}\pi/\mathrm{14}\right)\left(\mathrm{2cos}\:\left(\pi/\mathrm{7}\right)−\mathrm{1}\right)}\: \\ $$$$=\:\frac{\mathrm{sin}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{x}}{\mathrm{tan}\:\mathrm{3x}\:\left(\mathrm{2cos}\:\mathrm{2x}−\mathrm{1}\right)}\:=\:\frac{\mathrm{2sin}\:\mathrm{x}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{cos}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{3x}\left(\mathrm{4cos}^{\mathrm{2}} \:\mathrm{x}−\mathrm{3}\right)\:} \\ $$$$=\:\frac{\mathrm{2sin}\:\mathrm{x}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:\left(\mathrm{4cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{3cos}\:\mathrm{x}\right)}{\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{x}\right)\left(\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{3}\right)} \\ $$$$=\:\frac{\mathrm{2cos}\:^{\mathrm{3}} \mathrm{x}\:\left(\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{3}\right)}{\left(\mathrm{3}−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{3}\right)} \\ $$$$=\:\frac{\mathrm{2cos}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}} \\ $$$$\mathrm{according}\:\mathrm{cos}\:\mathrm{2x}\:=\:\mathrm{cos}\:\left(\pi/\mathrm{7}\right)\:=\:\mathrm{Q}\:\mathrm{is}\:\mathrm{one} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{8Q}^{\mathrm{3}} −\mathrm{4Q}^{\mathrm{2}} −\mathrm{4Q}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{therefore}\:\mathrm{8}\left(\mathrm{1}+\mathrm{Q}\right)^{\mathrm{3}} \:=\:\mathrm{7}\left(\mathrm{1}+\mathrm{2Q}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{7}}{\mathrm{8}}\:=\:\frac{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2x}\right)^{\mathrm{3}} }{\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{2x}\right)^{\mathrm{2}} }\:=\:\frac{\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{3}} }{\left(\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{7}}{\mathrm{8}}=\:\frac{\mathrm{8cos}\:^{\mathrm{6}} \mathrm{x}}{\left(\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:;\:\frac{\mathrm{7}}{\mathrm{16}}\:=\:\frac{\left(\mathrm{2cos}\:^{\mathrm{3}} \mathrm{x}\right)^{\mathrm{2}} }{\left(\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\therefore\:\frac{\mathrm{2cos}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}}\:=\:\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}\: \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *