If 32 sin^6 x = 10 - 15cos2x + b cos4x - a cos6x then prove ,
tan x tan5x = (a+b-5)/(a-b+7)

Question

If 32 sin^6 x = 10 - 15cos2x + b cos4x - a cos6x then prove , tan x tan5x = (a+b-5)/(a-b+7)

Imrul Kayes · Answer

32 sin^6x = 4.(2sin^2 x)^3 = 4(1- cos2x)^3 = 4(1-cos^3 2x -3cos 2x +3cos^2 2x) = 4- 12cos 2x + 6(1+cos 4x) – (cos 6x+ 3cos2x) = 10 – 15cos 2x +6cos 4x -cos6x then, 10- 15cos 2x +6cos 4x – cos 6x =10 – 15cos2x + b cos4x – a cos6x or, (6-b)cos4x = (1-a)cos6x or,cos4x/cos6x = (1-a)/(6-b) or, (cos4x – cos 6x)/(cos 4x + cos6x) = (1-a-6+b)/(1-a+6-b) or, tan5x tanx =(-a+b-5)/(7-a-b) this isthe correct answer due to the minus sign in the question before acos6x. it should be + sign to get the mentioned answer.

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If 32 sin^6 x = 10 - 15cos2x + b cos4x - a cos6x then prove , tan x tan5x = (a+b-5)/(a-b+7)

If 32 sin^6 x = 10 - 15cos2x + b cos4x - a cos6x then prove ,
tan x tan5x = (a+b-5)/(a-b+7)

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If 32 sin^6 x = 10 - 15cos2x + b cos4x - a cos6x then prove , tan x tan5x = (a+b-5)/(a-b+7)

If 32 sin^6 x = 10 - 15cos2x + b cos4x - a cos6x then prove , tan x tan5x = (a+b-5)/(a-b+7)

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If 32 sin^6 x = 10 - 15cos2x + b cos4x - a cos6x then prove ,
tan x tan5x = (a+b-5)/(a-b+7)