Trig Identity (1 Viewer)

philly2000 · Apr 16, 2009

Prove $sin^{6}\theta + cos^{6}\theta + 3sin^{2}\theta cos^{2}\theta = 1$

Thanks!

lyounamu · Apr 16, 2009

philly2000 said:
Prove $sin^{6}\theta + cos^{6}\theta + 3sin^{2}\theta cos^{2}\theta = 1$

Thanks!

LHS = (1-cos^2@)^2 (1-cos^2@) + cos^6@ + 3sin^2@cos^2@
= (1 - 2cos^2@ + cos^4@)(1-cos^2@) + cos^6@ + 3sin^2@cos^2@
= 1 - cos^2@ - 2cos^2@ + 2cos^4@ + cos^4@ - cos^6@ + cos^6@ + 3sin^2@cos^2@
= 1 - 3cos^2@ + 3cos^4@ + 3sin^2@cos^2@
= 1 - 3cos^2@ + 3cos^4@ + 3(1-cos^2@)cos^2@
= 1 - 3cos^2@ + 3cos^4@ + 3cos^2@ - 3cos^4@
= 1 = RHS

EvoRevolution · Apr 16, 2009

i think this way works aswell

sin^6@ + cos^6@ + 3sin^2@cos^2@ = 1

LHS = (sin^2@)^3 + (cos^2@)^3 + 3sin^2@cos^2@
= (sin^2@ + sin^2@)^3 - 3sin^2@cos^2@ + 3sin^2@cos^2@
= (sin^2@ + sin^2@)^3+0
= (1)^3
=1
RHS

somone1 check plz

i don't know if there is a rule for cubes like sqaures (e.g. (x+y)^2-2xy=x^2+y^2)

but here it is anyway

x^3 + y^3 = (x+y)^3
=x^3 + 2x^y + y^2x + yx^2 + 2xy^2 + y^3 - 3xy
=x^3 + y^3 + 3x^2y + 3y^2x
=x^3 + y^3 = 3xy(x+y)

now letting x = sin^2@
y = cos^2@
the 3xy(x+y) --> 3sin^2@cos^2@(sin^2@+cos^2@)
3sin^2@cos^2@(1)
3sin^2@cos^2@

Timothy.Siu · Apr 16, 2009

EvoRevolution said:
i think this way works aswell

sin^6@ + cos^6@ + 3sin^2@cos^2@ = 1

LHS = (sin^2@)^3 + (cos^2@)^3 + 3sin^2@cos^2@
= (sin^2@ + sin^2@)^3 - 3sin^2@cos^2@ + 3sin^2@cos^2@
= (sin^2@ + sin^2@)^3+0
= (1)^3
=1
RHS

somone1 check plz

i don't know if there is a rule for cubes like sqaures (e.g. (x+y)^2-2xy=x^2+y^2)

but here it is anyway

x^3 + y^3 = (x+y)^3
=x^3 + 2x^y + y^2x + yx^2 + 2xy^2 + y^3 - 3xy
=x^3 + y^3 + 3x^2y + 3y^2x
=x^3 + y^3 = 3xy(x+y)

now letting x = sin^2@
y = cos^2@
the 3xy(x+y) --> 3sin^2@cos^2@(sin^2@+cos^2@)
3sin^2@cos^2@(1)
3sin^2@cos^2@

nah thats not right

sin^6@ + cos^6@ + 3sin^2@cos^2@ = 1

LHS=(sin²@+cos²@)³ -3sin⁴@cos²@ -3sin²@cos⁴@+3sin²@cos²@
=1+3sin²@cos²@(1-sin²@-cos²@)
=1+3sin²@cos²@(cos²@-cos²@)
=1=RHS

thats how u wud do it that way.

lyounamu · Apr 16, 2009

Timothy.Siu said:
nah thats not right

sin^6@ + cos^6@ + 3sin^2@cos^2@ = 1

LHS=(sin²@+cos²@)³ -3sin⁴@cos²@ -3sin²@cos⁴@+3sin²@cos²@
=1+3sin²@cos²@(1-sin²@-cos²@)
=1+3sin²@cos²@(cos²@-cos²@)
=1=RHS

thats how u wud do it that way.

tim is da best

Drongoski · Apr 16, 2009

wonder if I'm wasting everyone's time by posting another one (or has someone already done one exactly the same way; if so I'm sorry)

$LHS\ =\ sin^{6 }\theta \ +\ cos^{6}\theta +\ 3sin^{2}\theta \ cos^{2}\theta \\ \\ =\ \ \left (sin^{2}\theta +cos^{2}\theta \right )\left (sin^{4}\theta -sin^{2}\theta cos^{2}\theta +cos^{4}\theta \right )\ +\ 3sin^{2}\theta cos^{2}\theta \\ \\ =\ \ sin^{4}\theta \ +\ 2sin^{2}\theta cos^{2}\theta \ + \ cos^{4}\theta \\ \\ =\ \ \left (sin^{2}\theta +cos^{2}\theta \right )^{2}\\ \\ =\ \ 1^{2} \ \ =\ \ 1\ \\ \\ =\ \ RHS\\ \\ \\ QED$

Edit: have skipped some 'obvious' steps to save typing. (I still do it hunt-&-peck)

For those whose identities are rusty:

A³ + B³ = (A + B)(A² - AB + B²)

Timothy.Siu · Apr 16, 2009

Drongoski said:
wonder if I'm wasting everyone's time by posting another one (or has someone already done one exactly the same way; if so I'm sorry)

$LHS\ =\ sin^{6 }\theta \ +\ cos^{6}\theta +\ 3sin^{2}\theta \ cos^{2}\theta \\ \\ =\ \ \left (sin^{2}\theta +cos^{2}\theta \right )\left (sin^{4}\theta -sin^{2}\theta cos^{2}\theta +cos^{4}\theta \right )\ +\ 3sin^{2}\theta cos^{2}\theta \\ \\ =\ \ sin^{4}\theta \ +\ 2sin^{2}\theta cos^{2}\theta \ + \ cos^{4}\theta \\ \\ =\ \ \left (sin^{2}\theta +cos^{2}\theta \right )^{2}\\ \\ =\ \ 1^{2} \ \ =\ \ 1\ \\ \\ =\ \ RHS\\ \\ \\ QED$

lol woah, what happenned there

iSplicer · Apr 16, 2009

Drongoski said:
wonder if I'm wasting everyone's time by posting another one (or has someone already done one exactly the same way; if so I'm sorry)

$LHS\ =\ sin^{6 }\theta \ +\ cos^{6}\theta +\ 3sin^{2}\theta \ cos^{2}\theta \\ \\ =\ \ \left (sin^{2}\theta +cos^{2}\theta \right )\left (sin^{4}\theta -sin^{2}\theta cos^{2}\theta +cos^{4}\theta \right )\ +\ 3sin^{2}\theta cos^{2}\theta \\ \\ =\ \ sin^{4}\theta \ +\ 2sin^{2}\theta cos^{2}\theta \ + \ cos^{4}\theta \\ \\ =\ \ \left (sin^{2}\theta +cos^{2}\theta \right )^{2}\\ \\ =\ \ 1^{2} \ \ =\ \ 1\ \\ \\ =\ \ RHS\\ \\ \\ QED$

elegant

lyounamu · Apr 16, 2009

Timothy.Siu said:
lol woah, what happenned there

pretty smart way i thought.

he used cube rule then ye he "worked" it

Drongoski · Apr 16, 2009

Thanks guys; made my day . . . after being bagged and 2 x (-1) reps recently.

Timothy.Siu · Apr 16, 2009

lyounamu said:
pretty smart way i thought.

he used cube rule then ye he "worked" it

lol cant see it haha

lyounamu · Apr 16, 2009

Timothy.Siu said:
lol cant see it haha

wot?

EvoRevolution · Apr 16, 2009

Yea that's the way i did it. But it was a bit messy. And hard to see

Timothy.Siu · Apr 16, 2009

lyounamu said:
wot?

lol nvm i get it

lyounamu · Apr 16, 2009

EvoRevolution said:
Yea that's the way i did it. But it was a bit messy. And hard to see

um..not really

your first few lines were heading towards that but it got into some messy working out...

Timothy.Siu · Apr 16, 2009

EvoRevolution said:
Yea that's the way i did it. But it was a bit messy. And hard to see

nope, ur ways wrong, and u were trying to do it the way i did it.

Drongoski · Apr 16, 2009

EvoRevolution said:
Yea that's the way i did it. But it was a bit messy. And hard to see

you cited the sum of 2 cubes identity wrong; pl see my edited posting.

EvoRevolution · Apr 16, 2009

$sin^{6}\Theta + cos^{6}\Theta + 3sin^2@cos^2@ = 1\\ \\ LHS = (sin^2@)^3 + (cos^2@)^3 + 3sin^2@cos^2@\\ \\ = (sin^2@ + sin^2@)^3 - 3sin^2@cos^2@ + 3sin^2@cos^2@ \\ \\ = (sin^2@ + sin^2@)^3+0\\ \\ = (1)^3\\ \\ =1$

when u expand out the cubic it doesnt change the original equations in anyway or may it does some1 check

Timothy.Siu · Apr 16, 2009

EvoRevolution said:
$sin^{6}\Theta + cos^{6}\Theta + 3sin^2@cos^2@ = 1\\ \\ LHS = (sin^2@)^3 + (cos^2@)^3 + 3sin^2@cos^2@\\ \\ = (sin^2@ + sin^2@)^3 - 3sin^2@cos^2@ + 3sin^2@cos^2@ \\ \\ = (sin^2@ + sin^2@)^3+0\\ \\ = (1)^3\\ \\ =1$

as i said before thats wrong,
u just worked from the answer to get that lol
(sin²@+cos²@)³ =/= sin⁶@+cos⁶@+3sin²@cos²@

lyounamu · Apr 16, 2009

Timothy.Siu said:
as i said before thats wrong,
u just worked from the answer to get that lol
(sin²@+cos²@)³ =/= sin⁶@+cos⁶@+3sin²@cos²@

lol pwned

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