sin^6 x + cos^6 x=1 - (3/4)sin^2 2x
work on one side only!
LS looks like the sum of cubes
sin^6 x + cos^6 x
= (sin^2x)^3 + (cos^2x)^3
= (sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + sin^4x)
= (1)(sin^4x - (sin^2x)(cos^2x) + sin^4x)
Now let's do some "aside"
(sin^2x + cos^2)^2 would be
sin^4x + 2(sin^2x)(cos^2x) + cos^4x
we 'almost' have that above, differing only by the coefficient of the middle term. We can fix that by saying
(sin^4x - (sin^2x)(cos^2x) + sin^4x)
= sin^4x + 2(sin^2x)(cos^2x) + cos^4x - 4(sin^2x)(cos^2x)
= (sin^2x + cos^2x) - 3(sin^2x)(cos^2x)
= 1 - 3(sin^2x)(cos^2x)
almost there!
recall sin 2A = 2(sinA)(cosA)
so 3(sin^2x)(cos^2x)
= 3(sinxcosx)^2
= 3((1/2)sin 2x)^2
= (3/4)sin^2 2x
so
1 - 3(sin^2x)(cos^2x)
= 1 - (3/4)sin^2 2x
= RS !!!!!!
Q.E.D.
(sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + sin^4x) I DON'T AGREE WITH THIS
SHOULDNT IT BE
(sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + COS^4 X ??
To work on one side only, we can simplify the equation using trigonometric identities and algebraic manipulations. Let's start with the left side of the equation:
sin^6 x + cos^6 x
We can rewrite sin^6 x as (sin^2 x)^3 and cos^6 x as (cos^2 x)^3:
(sin^2 x)^3 + (cos^2 x)^3
Now, we can apply the identity (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 to both terms:
(sin^2 x + cos^2 x)((sin^2 x)^2 - sin^2 x(cos^2 x) + (cos^2 x)^2)
Since sin^2 x + cos^2 x = 1 (based on the Pythagorean identity), we can replace it in the equation:
(1)((sin^2 x)^2 - sin^2 x(cos^2 x) + (cos^2 x)^2)
Now, let's simplify each term:
(sin^2 x)^2 = sin^4 x (product of the same base)
(cos^2 x)^2 = cos^4 x (product of the same base)
For the middle term, we have -sin^2 x(cos^2 x). We can rewrite this expression using the identity sin^2 x = 1 - cos^2 x:
-sin^2 x(cos^2 x) = -(1 - cos^2 x)(cos^2 x) = -cos^2 x + cos^4 x
Now, substituting the simplified terms back into the equation, we have:
(sin^4 x - cos^2 x + cos^4 x)
Now, we have an equation with terms of different powers of sin x and cos x. However, we can still simplify it further. We know that sin^2 x + cos^2 x = 1, so we can substitute 1 into the equation:
(sin^4 x - cos^2 x + cos^4 x) = (1 - cos^2 x - cos^2 x + cos^4 x) = (1 - 2cos^2 x + cos^4 x)
So, the simplified expression for the left side of the equation is:
sin^6 x + cos^6 x = 1 - 2cos^2 x + cos^4 x
Hence, we have now worked on one side of the equation only.