Question

cos2X-cos6X Если cosX=1/sqrt(3)

Answer 1

$\cos2x-\cos6x=2\cos^2x -1-2\cos^23x+1=2\cos^2x-2(4\cos^3x-3\cos x)=$
$=2(( \frac{1}{ \sqrt{3} } )^2-4\cdot (\frac{1}{ \sqrt{3} })^3-3\cdot \frac{1}{ \sqrt{3} })=2\cdot \frac{3-13\sqrt{3}}{9}= \frac{6-26\sqrt{3}}{9}$

Answer 2

cos2x-cos6x=-2sin(-2x)sin4x=2sin2xsin4x=2sin2x*2sin2xcos2x=4sin²2xcos2x=
=4*8/9*(-1/3)=-32/27
cosx=1/√3
cos2x=2cos²x-1=2*1/3 -1=2/3-1=-1/3
sin²2x=1-cos²2x=1-1/9=8/9