If sin4xa + cos4xb= 1a+b,prove that sin8xa3 +cos8xb3 = 1(a+b)3.
1can we proceed like this ???
sin x = z + 1/z ??? where |z| = 1 ??
cos x = (z - 1/z) / 2 i
1taking LCM on LHS of given expression we get ,
bsin4x + acos4x = aba+b
now → bsin4x + a(1 - sin2x)2 = aba+b
→bsin4x + asin4x - 2asin2x + a = aba+b
→(a+b)2sin4x - 2a(a+b)sin2x + a2 =0
it's a quadratic in sin2x so solving it we get sin2x = aa+b so cos2x = ba+b
now substitue the values in sin8xa3 + cos8xb3
u will get RHS ans 1(a+b)3
it's done[1]
1a more elegant solution i cn think is to writ it as sin2 x = a/b cos2x -------->sin2x=a/(a+b) and cos2x=b/(a+b)
and the problem is finished[1]
1mind mein calculate kar ke dekh na [1]....it is reducible to a perfect square[3][3]
111=(sin^2\alpha+cos^2\alpha )^2
From which we have asin^2\alpha=bcos^2\alpha, and the result follows.
341You can use Cauchy Schwarz as has been done in
http://www.goiit.com/posts/list/trignometry-sin-4-x-sin-2-y-cos-4-x-cos-2-y-1-then-sin-4-y-sin-78900.htm
(This exact problem too was solved in this way, but unable to find that post)