sin(2x)cos(5x) + cos(2x)sin(5x) use identities to simplify

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sin(2x)cos(5x) + cos(2x)sin(5x) use identities to simplify

Mathematics
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sin(7x)
how did you get that?
sin(2x+5x)

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Other answers:

remember sin(z+y)=sinzcosy+sinycosz where in this case z=2x and y=5x
where did the cos go?
its an identity
this is so confusing.. :(
do you know the identity sin(z+y)=sinzcosy+sinycosz
yes
but i don't understand how it's used in this question because sin and cos are being multiplied together.
ok what if you let z=2x and y=5x what would you have?
i understand how to use the identity, i'm just confused as to where it's being used
ok we have [sin(2x)][cos(5x)]+[cos(2x)][sin(2x)]=[sin(z)][cos(y)]+[cos(y)][sin(z)]=sin(y+z) where y=2x and z=5x so this =sin(2x+5x)=sin(7x)
oops i wrote the samething twice and that post
ohhh my gosh. i get it now
i'm so sorry. haha
ok good i don't have to write it over correctly then?
no, that is good 1 / 1-cos(x) + 1 / 1 + cos(x)
i assume that (1-cosx) and (1+cosx) are in the denominator right?
yes
combine those fractions
okay
what do you get now?
numerator=(1+cosx)+(1-cosx)=2 and denominator=1-(cosx)^2=(sinx)^2
2/(sinx)^2=2(cscx)^2
i got a denominator of 1-cos(x)1+cos(x)
(1-cosx)(1+cosx) use foil here
ohh okay
or you could use the fact that (a-b)(a+b)=a^2-b^2
we haven't learned that one, but i'll note it
thats an algebra thing it should had been algebra you learned that not trig
hmm, i don't remember it thankyou though.
(x-3)(x+3)=x^2-9 and (x-5)(x+5)=x^2-25 and so on....
\[\tan(\pi/5) + \tan(\pi/10) = numerator. 1 - \tan(\pi/5)\tan(\pi/10)\]
sorry. let me re-write that
what you got the one before this one?
numerator = tan(pi/5) + tan(pi/10) denominator = 1 - tan(pi/5)tan(pi/10)
yes, i understood it.
Nice!
Do you remember the idenity tan(x+y)=[tanx+tany]/[1-tanxtany]
yes
can use that here?
x=pi/5 and y=pi/10
3pi / 10?
tan 3pi/10, i mean
yes!!! :))
ok if you want more help, i'm gonna be away from my computer so people will notice you you might want to start a new thread k? if you like me become a fan
okay

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