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December 18, 2014

December 18, 2014

Posted by **hayden** on Monday, February 23, 2009 at 7:58pm.

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 DONT AGREE WITH THIS

SHOULDNT IT BE

(sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + COS^4 X ??

- TRIG! -
**Reiny**, Monday, February 23, 2009 at 8:52pmof 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

- TRIG! -
**Anonymous**, Tuesday, December 13, 2011 at 9:49amcos^2x+3sinxcosx=2

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