Guys please help.

[modest horizon]

something i find less confusing when it comes to trig values (cos & sin) in a complex algebraic expression is to:

re-write the cos( <theta> ) & sin( <theta> ) as an x & y

so instead of looking at m = a{ cos() }^3 + 3acos(){ sin() }^3

it becomes
m = ax^3 + 3axy^3

helps to hide some of the nasty parts haha

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hm

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yes fs

[modest horizon]

? fs?

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for sure

new to da platform huh

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new to THIS server

& only a passerby to a TON of internet jargon

so ... kind of

ive used Discord as a solo venture for approx 5[year] or so

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noiceez

ig

[modest horizon]

as for the (m +/- n) ^ (2/3)

REM: k^(2/3) = { k^2 }^(1/3)

thus finding what (m +/- n)^2 could be a decent work around (i havent written it for myself ... yet)

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hm

ive don m+n and m-n seperately

[modest horizon]

cool

nothing wrong with that

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so according to ur x nd y conversion

taking sin theeta as y
and cos theeta as x

(m+n)=(acosθ+asinθ)^3
(m-n)=(acosθ-asinθ)^3

now adding both of em

raised to 2/3

after that m stuck

[modest horizon]

the 1st ~issue~ that jumps out to me ... the a would not be raised to the third power

before u can condense into (x + y)^3 ... the a would need factored out as a common factor

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{(acosθ+asinθ)^3}^2/3+{(acosθ-asinθ)^3}^2/3

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hmm yes

[modest horizon]

ur so close!

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is it?

[modest horizon]

so consider ([k^3]^2)^(1/3)

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hm

[modest horizon]

since each of those exponents are applied to the base k, we can work them in any order that best suites our problem

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ok

so

will we factor out a?

as a^3

[modest horizon]

no

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outside da brackets

[modest horizon]

a never gets cubed

that was the mistake / issue in that step i was talking about

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why so ?

ohh

ok

but its asinθ nd acosθ ryt

[modest horizon]

where did the d & t come from??

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huh?

ohh

its and and right

nd and ryt

[modest horizon]

so ... ur answer will be in terms of the variables a & theta ... since we are not given values to "plug in" for those variables

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the answer is

2a^2/3

actually its a proof question

[modest horizon]

my bad!

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but i wasnt able to paste it

so this will do

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yeah ... the x^2 + y^2 cancel out

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no no its alr

[modest horizon]

well ...

resolve to 1

nicely done

!

that's the same answer i got

2a^(2/3)

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yeeees

bruh

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glad i could help (-

(=

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u gotta teach me sum math

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haha

what do u wanna know?

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ntng stuff lyk dis ig

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"math" is huge

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im not clear yet

dis ques

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"dis ques" ?= this question??

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haha yes

wait m still doin

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haha cool

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ill tell whn doubt will come

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ill keep an eye out on this server {=

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bruh i got smthn lyk
(acosθ+asinθ)^2 + (acosθ-asinθ)^2

i canceled out the 3

from cube nd 2/3

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hm

[modest horizon]

bummer ... i already threw out my scratch paper but im fairly certain the issue is here ... just before u did THIS step

the a should NOT be under the ^2

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hmm

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the a should be outside the parentheses in both

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yessssssssssssssssssssssssssssssssssssssssss

got ittttt

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haha cool

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a^2/3(cosθ+sinθ)+a^2/3(cosθ-sinθ)

smthn lyk dis

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not

no wait

[modest horizon]

what happened ur to ur ^2??

u should have a^(2/3)[ (x + y)^2 + (x - y)^2 ]

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a^2/3(cosθ+sinθ)^2+a^2/3(cosθ-sinθ)^2

yes

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yepper!

keep working from there

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i forgot about the square

i didnt typ

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happens haha

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donenez

ty