Some Decic Sin and Cos Sum to Rational Numbers (Posted on 2023-07-12)

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
sin^2 x+ cos^2 x)^5 = 1^5 =1 => (sin^2 x)^5+ 5*(sin^2 x)(cos^2 x)+10*(sin^2 x)^3 cos^2 x)^2+ x)+10*(sin^2 x)^ 2 cos^2 x)^3+ 5 * sin^2 x (cos ^2 x)^4+ (cos^2x)^5=1 => sin^10 x+ cos ^10x+ 5*sin^2x*(cos ^2x)(sin^6x +cos ^6x)+ 10* sin^4x* cos^4x(sin^ 2x + cos ^2 x)=1 (sin^2x+ cos^2 x)^3=1 1= Sin ^6x + cos^6x +3sin^2x*cos^2x (sin^2x+cos ^2x) sin^6x + cos^6 x= 1 - 3*sin^2x* cos ^2x) Let sin x*cos x =p Then, sin^6x *cos^6x = 1- 3p^2 Now, 61/256 - 5*p^2(1-3p^2)+10*p^4=1 => 61/256 +5p^2(1-3*p^2)+10*p^4=1 => 5*p^2-5*p^4 = 195/256 => p^2-p^4 =39/256 Let p^2=u(say) Then, u -u^2 = 39/256 256u^2 -256u+39=0 Or, u= 3/16 and u = 13/16 Or, u= p^2 =3/16 --> p = v3/4 or, p= -v3/4 u = p^2=13/16 --> p = v13/4 or, p = -v13/4 Now, p= v3/4= sinx*cos x => 2* sinx* cos x = v3/2 => sin (2x) = v3/2 => 2x = pi/3 + n *(2pi) => x = pi/6 + n*pi and, 2x= 2*pi/3+ 2n*pi => x = pi/3 + n*pi Now, p= -v3/4 = sin x* cos x => 2* sinx* cos x = -v3/2 => sin (2x) = -v3/2 => 2x = (2*pi)/3 + 2*n*2pi => x = pi/3 + n*pi Also, x= 5*pi/3+2*n*pi => x = 5pi/6 + npi Thus, x= pi/6 + npi, pi/3+npi, 2pi/3+n*pi, 5pi/6+ n*pi

	Subject	Author	Date
	No Subject	laxmigame	2023-07-19 04:45:14
	Solution	Brian Smith	2023-07-14 10:56:18
	Solution hints	Brian Smith	2023-07-12 23:18:46
	re: what I find Overlooked	Jer	2023-07-12 16:15:33
	what I find	Charlie	2023-07-12 14:37:51