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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)

Grade:11

1 Answers

Imrul Kayes
20 Points
6 years ago
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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