Posted by hayden on Monday, February 23, 2009 at 4:05pm.

sin^6 x + cos^6 x=1 - (3/4)sin^2 2x

work on one side only!

Responses

Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm
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.



Trig please help! - hayden, Monday, February 23, 2009 at 7:57pm
(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 ??

of course, you are right,

also check up on this part:

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)

should say:
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) + cos^4x)
= sin^4x + 2(sin^2x)(cos^2x) + cos^4x - 3(sin^2x)(cos^2x)
= (sin^2x + cos^2x)^2 - 3(sin^2x)(cos^2x)

sorry about the typos, one has to be so careful with this crazy trig stuff

cos^2x+3sinxcosx=2

The result given by Reiny is correct. Let's go step by step to clarify it:

Starting with:
sin^6 x + cos^6 x

We can rewrite this expression as the sum of cubes:
(sin^2x)^3 + (cos^2x)^3

Using the formula for the sum of cubes:
a^3 + b^3 = (a + b)(a^2 - ab + b^2), we have:
sin^6 x + cos^6 x = (sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + cos^4 x)

Now, let's simplify the expression inside the parentheses:
sin^4x - (sin^2x)(cos^2x) + cos^4 x

We can see that this expression looks similar to (sin^2x + cos^2x)^2, except for the middle term. So, let's try to rewrite it in that form:
sin^4x + 2(sin^2x)(cos^2x) + cos^4x - 4(sin^2x)(cos^2x)

Simplifying further:
(sin^2x + cos^2x) - 3(sin^2x)(cos^2x)

We know that sin^2x + cos^2x = 1, so we can substitute that in:
1 - 3(sin^2x)(cos^2x)

Finally, we notice that 3(sin^2x)(cos^2x) is equal to (3/4)sin^2 2x, so we can substitute that as well:
1 - (3/4)sin^2 2x

Therefore, we have shown that:
sin^6 x + cos^6 x = 1 - (3/4)sin^2 2x

Regarding your question about the expression inside the parentheses, it should indeed be:
sin^4x - (sin^2x)(cos^2x) + cos^4 x

The expression you suggested, cos^4 x, is not correct because we are dealing with sin^6 x + cos^6 x, not sin^2 x + cos^2 x.

I hope this clarifies any confusion.

In order to understand the proof that was given by Reiny, let's break it down step by step:

The left side (LS) of the equation is sin^6 x + cos^6 x. Reiny noticed that this expression resembles the sum of cubes: (a^3 + b^3) = (a + b)(a^2 - ab + b^2).

So, applying this concept to LS, we have:
LS = (sin^2x)^3 + (cos^2x)^3
= (sin^2x + cos^2x)(sin^4x - (sin^2x)(cos^2x) + cos^4x)

Now, let's take an "aside" and expand (sin^2x + cos^2x)^2, which would be sin^4x + 2(sin^2x)(cos^2x) + cos^4x. We notice that the expression we obtained above is almost the same, but with a difference in the coefficient of the middle term.

To fix that difference, we can rewrite the expression as:
(sin^4x - (sin^2x)(cos^2x) + sin^4x) = sin^4x + 2(sin^2x)(cos^2x) + cos^4x - 4(sin^2x)(cos^2x)

Now, let's simplify further:
= (sin^2x + cos^2x) - 3(sin^2x)(cos^2x)
= 1 - 3(sin^2x)(cos^2x)

We're almost there! Now, we recall the identity sin 2A = 2(sinA)(cosA). We can use this to rewrite (sin^2x)(cos^2x) as (1/4)sin^2 2x.

So, 3(sin^2x)(cos^2x) = 3((1/4)sin^2 2x) = (3/4)sin^2 2x.

Substituting this back into the previous expression, we have:
1 - 3(sin^2x)(cos^2x) = 1 - (3/4)sin^2 2x.

Therefore, the left side (LS) is equal to the right side (RS).

As for your question, you asked if it should be cos^4x instead of sin^4x in the expression (sin^2x + cos^2x)(sin^4x - (sin^2x)(cos^2x) + sin^4x). According to the proof provided by Reiny, it should indeed be cos^4x.