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Question Number 144829 by mathlove last updated on 29/Jun/21
sin^3 x+cos^3 x=1−(1/2)sin^2 x x=?
sin3x+cos3x=1−12sin2xx=?
Answered by ajfour last updated on 29/Jun/21
(1−s^2 )^3 =(1−(s^2 /2)−s^3 )^2 1−s^6 +3s^4 −3s^2 =1+(s^4 /4)+s^6 −s^2 +s^5 −2s^3 ⇒ s^6 +(s^5 /2)−((11s^4 )/8)−s^3 +s^2 =0 ⇒ s=0 or s^4 +(s^3 /2)−((11s^2 )/8)−s+1=0 there are two real positive roots to this quartic; say s=α,β now sin x=0,α,β x=nπ+(−1)^n θ where θ=0,α,β
(1−s2)3=(1−s22−s3)21−s6+3s4−3s2=1+s44+s6−s2+s5−2s3⇒s6+s52−11s48−s3+s2=0⇒s=0ors4+s32−11s28−s+1=0therearetworealpositiverootstothisquartic;says=α,βnowsinx=0,α,βx=nπ+(−1)nθwhereθ=0,α,β
Answered by liberty last updated on 29/Jun/21
sin^3 x+cos^3 x=sin^2 x+cos^2 x−(1/2)sin^2 x sin^3 x+cos^3 x=(1/2)sin^2 x+cos^2 x cos^3 x−cos^2 x=(1/2)sin^2 x−sin^3 x cos^2 x(cos x−1)=sin^2 x((1/2)−sin x) cos^2 x(−2sin^2 (1/2)x)=sin^2 x((1/2)−2sin (1/2)xcos (1/2)x) let tan (x/2)=t⇒tan x=((2t)/(1−t^2 )) ⇒−2((t^2 /(1+t^2 )))=(((2t)/(1−t^2 )))((1/2)−((2t)/(1+t^2 ))) ⇒((−2t^2 )/(1+t^2 ))= (((2t)/(1−t^2 )))(((1+t^2 −4t)/(2(1+t^2 )))) ⇒(((2t)/(1+t^2 )))[((t^2 −4t+1)/(2−2t^2 )) +t ]=0 ⇒(((2t)/(1+t^2 )))[((t^2 −4t+1+2t−2t^3 )/(2−2t^2 ))]=0 ⇒(((2t)/(1+t^2 )))(((2t^3 −t^2 +2t−1)/(2−2t^2 )))=0 we get t=0 ⇒tan (x/2) = 0 →(x/2) = 0+kπ ⇒x= 2kπ ; k∈Z for 2t^3 −t^2 +2t−1=0 test t=(1/2)→(1/4)−(1/4)+1−1=0 (valid) ⇒factorise (2t−1)(t^2 +1)=0 ⇒tan (x/2)=(1/2) ; (x/2)= arctan ((1/2))+kπ ⇒x=2arctan ((1/2))+2kπ
sin3x+cos3x=sin2x+cos2x−12sin2xsin3x+cos3x=12sin2x+cos2xcos3x−cos2x=12sin2x−sin3xcos2x(cosx−1)=sin2x(12−sinx)cos2x(−2sin212x)=sin2x(12−2sin12xcos12x)lettanx2=t⇒tanx=2t1−t2⇒−2(t21+t2)=(2t1−t2)(12−2t1+t2)⇒−2t21+t2=(2t1−t2)(1+t2−4t2(1+t2))⇒(2t1+t2)[t2−4t+12−2t2+t]=0⇒(2t1+t2)[t2−4t+1+2t−2t32−2t2]=0⇒(2t1+t2)(2t3−t2+2t−12−2t2)=0wegett=0⇒tanx2=0→x2=0+kπ⇒x=2kπ;k∈Zfor2t3−t2+2t−1=0testt=12→14−14+1−1=0(valid)⇒factorise(2t−1)(t2+1)=0⇒tanx2=12;x2=arctan(12)+kπ⇒x=2arctan(12)+2kπ

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