Question Number 226401 by fantastic2 last updated on 27/Nov/25
$${most}\:{hated}\:{trigonometric} \\ $$$${problem}: \\ $$$$\mathrm{sin}\left(\:\frac{\pi}{\mathrm{7}}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)=? \\ $$
Commented by AgniMath last updated on 27/Nov/25
$${bhannat}\:{maths}\:{op}. \\ $$
Commented by fantastic2 last updated on 27/Nov/25
$${yea}\:{brooo} \\ $$
Commented by fantastic2 last updated on 27/Nov/25
$${are}\:{you}\:{a}\:{jee}\:{aspirant}? \\ $$
Commented by AgniMath last updated on 27/Nov/25
$${yes}.\:{But}\:{main}\:{focus}\:{is}\:{clearing}\:{ISI} \\ $$
Commented by fantastic2 last updated on 27/Nov/25
$${ISI}? \\ $$
Commented by AgniMath last updated on 27/Nov/25
$${Indian}\:{Statistical}\:{Institute}\:\left({B}\:{Math}\right) \\ $$
Answered by peace2 last updated on 27/Nov/25
![sin(π−x)=sin(x)⇒ P^2 =Π_(k=1) ^6 sin(((kπ)/7)) Z^7 −1=0⇒z=e^((2ik(π)/7) ;k∈[0,6]∩N z^7 −1=Π(z−e^((2ikπ)/7) ) ((z^7 −1)/(z−1))=Π_(k=1) ^6 (z−e^((i2kπ)/7) ) ⇒7=Π_(k=1) ^6 e^((ikπ)/7) [e^(−((ikπ)/7)) −e^((ikπ)/7) ] 7=e^(i3π) Π(−2isin(((kπ)/7))) ⇒p^2 =(7/(64))⇒p=((√7)/8)](https://www.tinkutara.com/question/Q226402.png)
$${sin}\left(\pi−{x}\right)={sin}\left({x}\right)\Rightarrow \\ $$$${P}^{\mathrm{2}} =\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{7}}\right) \\ $$$${Z}^{\mathrm{7}} −\mathrm{1}=\mathrm{0}\Rightarrow{z}={e}^{\frac{\mathrm{2}{ik}\left(\pi\right.}{\mathrm{7}}} ;{k}\in\left[\mathrm{0},\mathrm{6}\right]\cap\mathbb{N} \\ $$$${z}^{\mathrm{7}} −\mathrm{1}=\Pi\left({z}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{7}}} \right) \\ $$$$\frac{{z}^{\mathrm{7}} −\mathrm{1}}{{z}−\mathrm{1}}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}\left({z}−{e}^{\frac{{i}\mathrm{2}{k}\pi}{\mathrm{7}}} \right) \\ $$$$\Rightarrow\mathrm{7}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{e}^{\frac{{ik}\pi}{\mathrm{7}}} \left[{e}^{−\frac{{ik}\pi}{\mathrm{7}}} −{e}^{\frac{{ik}\pi}{\mathrm{7}}} \right] \\ $$$$\mathrm{7}={e}^{{i}\mathrm{3}\pi} \Pi\left(−\mathrm{2}{isin}\left(\frac{{k}\pi}{\mathrm{7}}\right)\right) \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{64}}\Rightarrow{p}=\frac{\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$
Answered by AgniMath last updated on 27/Nov/25
$$\underset{{k}=\mathrm{1}} {\overset{\frac{{p}−\mathrm{1}}{\mathrm{2}}} {\prod}}\mathrm{sin}\left(\frac{{k}\pi}{{p}}\right)\:=\:\frac{\sqrt{{p}}}{\mathrm{2}^{\frac{{p}−\mathrm{1}}{\mathrm{2}}} }\:\left({Key}\:{formula}\:{when}\:{p}\right. \\ $$$$\left.{is}\:{prime}.\right) \\ $$$${Put}\:{p}\:=\:\mathrm{7} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\mathrm{sin}\left(\frac{{k}\pi}{\mathrm{7}}\right)\:=\:\frac{\sqrt{\mathrm{7}}}{\mathrm{2}^{\mathrm{3}} }\:=\:\frac{\sqrt{\mathrm{7}}}{\mathrm{8}}\:\left({Ans}\right) \\ $$
Answered by Frix last updated on 28/Nov/25
$$\mathrm{Using}\:\mathrm{trigonometric}\:\mathrm{formulas}: \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:={x}>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)={x} \\ $$$$\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:=\mathrm{4}{x} \\ $$$$\left(\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}=\mathrm{4}{x}\right)^{\mathrm{3}} \\ $$$$… \\ $$$$\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:=\frac{\mathrm{256}{x}^{\mathrm{3}} }{\mathrm{7}} \\ $$$$\mathrm{4}{x}=\frac{\mathrm{256}{x}^{\mathrm{3}} }{\mathrm{7}}\wedge{x}>\mathrm{0}\:\Rightarrow\:{x}=\frac{\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$
Answered by fantastic2 last updated on 28/Nov/25
![let x=sin( (π/7))sin (((2π)/7))sin (((3π)/7)) y=cos ( (π/7)) cos (((2π)/7))cos (((3π)/7)) xy=(sin( (π/7))cos ((π/7)))(sin (((2π)/7))cos (((2π)/7)))(sin (((3π)/7))cos (((3π)/7))) 8xy=2(sin( (π/7))cos ((π/7)))2(sin (((2π)/7))cos (((2π)/7)))2(sin (((3π)/7))cos (((3π)/7))) 8xy=sin( ((2π)/7))sin (((4π)/7))sin (((6π)/7)) 8xy=sin( ((2π)/7))sin (π−((3π)/7))sin (π−(π/7)) 8xy=sin( (π/7))sin (((2π)/7))sin (((3π)/7)) 8xy=x ⇒y=(1/8) x^2 =sin^2 ( (π/7))sin^2 (((2π)/7))sin^2 (((3π)/7)) x^2 =(((1−cos (((2π)/7))))/2)×(((1−cos (((4π)/7))))/2)×(((1−cos (((6π)/7))))/2) 8x^2 =(1−cos (((2π)/7)))(1−(cos (π−((3π)/7)))(1−(cos (π−(π/7))) 8x^2 =(1−cos (((2π)/7)))(1+cos (((3π)/7)))(1+cos ((π/7))) let cos (((3π)/7))=a ,cos (((2π)/7))=b,cos ((π/7))=c ⇒8x^2 =(1+a)(1−b)(1+c) 8x^2 =1+a−b+c−ab−bc+ca−abc[i.e (1/8)] 8x^2 =(7/8)+(a−b+c)−(ab+bc−ca) (a−b+c)−(ab+bc−ca) =(a−b+c)−(1/2)[2cos (((3π)/7))cos (((2π)/7))+2cos (((2π)/7))cos ((π/7))−2cos ((π/7))cos (((3π)/7))] =(a−b+c)−(1/2)[cos (((5π)/7))+cos ((π/7))+cos (((3π)/7))+cos ((π/7))−cos (((4π)/7))−cos (((2π)/7))] =(a−b+c)−(1/2)[cos (π−((2π)/7))+2c+a−cos (π−((3π)/7))−b] =(a−b+c)−(1/2)[2a−2b+2c] =(a−b+c)−(a−b+c)=0 So 8x^2 =(7/8)+0 ⇒x=((√7)/8)✓](https://www.tinkutara.com/question/Q226443.png)
$${let} \\ $$$${x}=\mathrm{sin}\left(\:\frac{\pi}{\mathrm{7}}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\ $$$${y}=\mathrm{cos}\:\left(\:\frac{\pi}{\mathrm{7}}\right)\:\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{cos}\:\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\ $$$${xy}=\left(\mathrm{sin}\left(\:\frac{\pi}{\mathrm{7}}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)\right)\left(\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\right)\left(\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\right) \\ $$$$\mathrm{8}{xy}=\mathrm{2}\left(\mathrm{sin}\left(\:\frac{\pi}{\mathrm{7}}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)\right)\mathrm{2}\left(\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\right)\mathrm{2}\left(\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\right) \\ $$$$\mathrm{8}{xy}=\mathrm{sin}\left(\:\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{sin}\:\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)\mathrm{sin}\:\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right) \\ $$$$\mathrm{8}{xy}=\mathrm{sin}\left(\:\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{sin}\:\left(\pi−\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\mathrm{sin}\:\left(\pi−\frac{\pi}{\mathrm{7}}\right) \\ $$$$\mathrm{8}{xy}=\mathrm{sin}\left(\:\frac{\pi}{\mathrm{7}}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\ $$$$\mathrm{8}{xy}={x} \\ $$$$\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${x}^{\mathrm{2}} =\mathrm{sin}^{\mathrm{2}} \left(\:\frac{\pi}{\mathrm{7}}\right)\mathrm{sin}^{\mathrm{2}} \:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{sin}^{\mathrm{2}} \:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\ $$$${x}^{\mathrm{2}} =\frac{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\right)}{\mathrm{2}}×\frac{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)\right)}{\mathrm{2}}×\frac{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)\right)}{\mathrm{2}} \\ $$$$\mathrm{8}{x}^{\mathrm{2}} =\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\right)\left(\mathrm{1}−\left(\mathrm{cos}\:\left(\pi−\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\right)\left(\mathrm{1}−\left(\mathrm{cos}\:\left(\pi−\frac{\pi}{\mathrm{7}}\right)\right)\right.\right. \\ $$$$\mathrm{8}{x}^{\mathrm{2}} =\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\right)\left(\mathrm{1}+\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\right)\left(\mathrm{1}+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)\right) \\ $$$${let}\:\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)={a}\:,\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)={b},\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)={c} \\ $$$$\Rightarrow\mathrm{8}{x}^{\mathrm{2}} =\left(\mathrm{1}+{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}+{c}\right) \\ $$$$\mathrm{8}{x}^{\mathrm{2}} =\mathrm{1}+{a}−{b}+{c}−{ab}−{bc}+{ca}−{abc}\left[{i}.{e}\:\frac{\mathrm{1}}{\mathrm{8}}\right] \\ $$$$\mathrm{8}{x}^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{8}}+\left({a}−{b}+{c}\right)−\left({ab}+{bc}−{ca}\right) \\ $$$$\left({a}−{b}+{c}\right)−\left({ab}+{bc}−{ca}\right) \\ $$$$=\left({a}−{b}+{c}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{2cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{2cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)−\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{7}}\right)\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\right] \\ $$$$=\left({a}−{b}+{c}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{7}}\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)−\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\right] \\ $$$$=\left({a}−{b}+{c}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\:\left(\pi−\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{2}{c}+{a}−\mathrm{cos}\:\left(\pi−\frac{\mathrm{3}\pi}{\mathrm{7}}\right)−{b}\right] \\ $$$$=\left({a}−{b}+{c}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}{a}−\mathrm{2}{b}+\mathrm{2}{c}\right] \\ $$$$=\left({a}−{b}+{c}\right)−\left({a}−{b}+{c}\right)=\mathrm{0} \\ $$$${So} \\ $$$$\mathrm{8}{x}^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{8}}+\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{7}}}{\mathrm{8}}\checkmark \\ $$