cos²5x - cos²x = (-sin4x)(sin6x)

(cos5x - cosx)(cos5x + cosx)

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]]>cos²(5x) - cos²(x) = 1/2(2cos²(5x)-1+1 - (2cos²(x) - 1 + 1) )

= 1/2(cos²(10x) - cos²(2x) )

Since 10x = 6x - (-4x) and 2x = 6x + (-4x):

= sin(6x)sin(-4x)

= -sin(4x)sin(6x)

I normally do a lot more steps than needed, just so others can easily follow along.

]]>PS: (Sorry I am being picky, I have a test on this type of problem tomarow.)

Thanks so much for your help

]]>sinAsinB = 1/2(cos(A-B) - cos(A+B))

-sinA = sin(-A)

cos(2A) = cos²A - sin²A

-sin(4x) = sin(-4x)

sin(-4x) * sin(6x) = 1/2(cos(-4x - 6x) - cos(-4x + 6x)) = 1/2(cos(-10x) - cos(2x))

1/2(cos(2 * -5x) - cos(2 * x)) = 1/2(cos²(-5x) - sin²(-5x) - ( cos²(x) - sin²(x) ) )

Since sin²(-5x) = 1 - cos²(-5x):

1/2(cos²(-5x) - (1 - cos²(-5x)) - (cos²(x) - (1 - cos²(x)) ) )

1/2( 2cos²(-5x) - 1 -2cos²(x) + 1) = 1/2(2cos²(-5x) - 2cos²(x)) = cos²(-5x) - cos²(x)

And finally, since cos(-A) = cos(A):

cos²(5x) - cos²(x)

]]>Hope that helped.

]]>cos²5x - cos²x = -sin4x x sin6x

If that is not clear

cos sqared of five x minus cos squared of x = negative sin of 4x times sin of 6x

Thanks a lot for your help,

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