#Trig identity
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What is sin(2x) in terms of cos(x) and sin(x)?
2sin(x)cos(x) right?
[
\frac{\sin 2 \theta - \cos \theta}{2 \sin \theta - 1} \neq
\frac{\sin 2 \theta - \cos \theta}{\sin 2 \theta - 1} \neq
\frac{\sin 2 \theta}{\sin 2 \theta} \cdot \frac{- \cos \theta}{- 1}
]
Invariance
all of those are different
Invariance
okay thank you. i will try again after i finish eating lunch
Oh is that what’s written I was a bit confused 😅
Ight have a nice meal
hey i just retried it, just wanted to know if this looks right to you
hey i just retried it, just wanted to know if this looks right to you
sorry if it’s wrong yall i swear im trying😭
you need to take cos(x) common
cos(x) common? i’m sorry i don’t understand
like so
2cos(x)sin(x)-cos(x)=cos(x)(2sin(x)-1)
for example say, (a-1)b + b
this can be written as b{ (a-1) +1 }
could you tell me what would this equation look like when you take y common?
i know i’m supposed to factor but i have to remember how, give me a minute
never mind i have no idea. it’s okay if it’s too much too explain.
x(a+1) ?
ah good
now try this
you could also substitute (ax+4) with "t" if that makes you feel more comfortable
so would it y((ax+4)+k))
yeah good job :)
now can you see a " (x)y + y " here?
yea is would be the cosines. but if i tried it, it would come out different than the one you sent
give me a second let me see
id like to uh point that the operations you used here is wrong
what do you mean?
this is not correct
oh okay, i’m really bad with fractions. i’m guessing the a numerator cancels but the denominator a stays for x-x/a?
no see ax-x is really just (a-1)x
okay that makes sense
so you can not cut a from both the numerator and the denominator
even after i’ve simplified it this far? or is that just referring to my original attempt?
yeah i was referring to the original one
here you can see (2sinx - 1) in the numerator and in the denominator
so you can cancel them out to get your answer :)
np
+close