Trig Identity (1 Viewer)

[philly2000]

Prove

Thanks!

 

[lyounamu]

Prove

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]

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]

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]

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]

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)




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]

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)

lol woah, what happenned there

 

[iSplicer]

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)

elegant

 

[lyounamu]

lol woah, what happenned there

pretty smart way i thought.

he used cube rule then ye he "worked" it

 

[Drongoski]

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

 

[Timothy.Siu]

pretty smart way i thought.

he used cube rule then ye he "worked" it

lol cant see it haha

 

[lyounamu]

lol cant see it haha

wot?

 

[EvoRevolution]

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

 

[Timothy.Siu]

wot?

lol nvm i get it

 

[lyounamu]

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]

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]

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]

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

 

[Timothy.Siu]

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

 

[lyounamu]

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