proving trig identities

[lethal current]

use (a + b)^2 = a^2 + 2ab + b^2

(cot(x)+1)^2

expand this

same with the other one

technically you could argue for taking square root of both sides

that would be simpler actually

sometimes first thing you think of is slower

(the new idea is changing a^2*b^2 to (ab)^2)

[lethal current]

do this instead

you have something squared times something squared

move both into one parentheses

when you bring it inside the squared

you unsquare it

otherwise when you multiply it out it will become to the 4th power

yep

then it wouldn't simplify

you can leave it as tanx sinx

remove the ^2 first then bring it in

also works

x^2 times (x+1)^2 is (xx + x)^2

so you can

[lethal current]

yeah if you divide both top and bottom by cos(y)cos(x) that is what you will get

you can multiply things

the other way to do this would be changing the top to sin(x + y) and the bottom to sin(x - y)

i think

but that wouldnt get you to the answer it seems

ah

where you get it from is you see that multiplying it would get you to the answer

you're trying to get from point A to point B

yeah

multiplying numerator and denominator by something is the same as multiplying by 1

well, if that thing isnt 0

which we ignore

because

ignore™️

[lyric vortex]

@glacial storm

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