chels
  • chels
sin(2x)cos(5x) + cos(2x)sin(5x) use identities to simplify
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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myininaya
  • myininaya
sin(7x)
chels
  • chels
how did you get that?
myininaya
  • myininaya
sin(2x+5x)

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More answers

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

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