Question

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 ??

Answer 1

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) + <b>cos</b>^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

Answer 2

cos^2x+3sinxcosx=2