verify an identity of sin5xcos3x = sin4xcos4x + sinxcosx
I will post up what I know so far could you please check?
kk ;) !
yes ! The first step is correct @lsugano
But i dont know what to do next..><
kk sin(2x) =??!
That works sin(8x) = sin(4x+4x) = 2sin(4x)cos(4x)
sin(8x)=sin(2(4x))=???
Looks like you are almost there just use: sin(2x) = 2sin(x)cos(x)
wait im really lost><
See the double angle formulas: http://www.sosmath.com/trig/Trig5/trig5/trig5.html
just Use the double angle formula ! for sin(2x) and sin(8x)
so like this: 8sinxcosx + 2sinxcosx ?
you're really lost : ) ! you had this ; \[\sin(5x)\cos(3x)=\frac{1}{2}(\sin(\color{green}{8x})+\sin(\color{green}{2x}))\]
okay!
sin(8x)=...............(using the double angle formula....what we need to prove is something that contain 4x ....and we know that 8x=2(4x))
sin(2a)=2sin(a)cos(a)
why do we need to contain 4x?
\[\sin5xcos3x = \sin(\color{red}{4x})\cos(\color{red}{4x}) + \sin(x)\cos(x)\]
see now ?
you need to change sin(8x)=............ using sin(2a)=2sin(a)cos(a)
8/2 = 4 sin4a=4sin(a)cos(a)
no ! just focus ! sin(2a)=2sin(a)cos(a) take a = 4x what will you get ?
2sin4xcos4x
yes ! now sin(2x)=......???
2sin2xcos2x
:'( !!! try again sin(2x)=......
2sinxcosx
yes ! Now can you finish this ? using sin(8x)=2sin(4x)cos(4x) and sin(2x)=2sin(x)cos(x) ???
you add them right?
Maybe :) ! Just do what you have to do ! and I'll check :)
OHHH i got it okay i will write it on a tablet please check!!
kk ! Sure :)
Yeeeees ! that's correct :) !
ahh much more clear!!
:) ! just practice more ! And you'll be fine :)
so since 8x needed to be converted to the "sin2a" form, we divide 8xby2 and get 4x and we put a=4x correct?
yes ! keep in mind the double angle formula sin(2a)=2sin(a)cos(a) or sin(something)=2sin(somthing/2)cos(something/2)
wonderful advice!! thank you very much! you're a teacher right? cuz your way better than the math teachers id had in the past!!
No ! I'm not a teacher !
oh wow ! thank you so much!
You're welcome ! Anytime :)
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