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Mathematics 15 Online
OpenStudy (anonymous):

verify an identity of sin5xcos3x = sin4xcos4x + sinxcosx

OpenStudy (anonymous):

I will post up what I know so far could you please check?

OpenStudy (anonymous):

kk ;) !

OpenStudy (anonymous):

OpenStudy (anonymous):

yes ! The first step is correct @lsugano

OpenStudy (anonymous):

But i dont know what to do next..><

OpenStudy (anonymous):

kk sin(2x) =??!

OpenStudy (valpey):

That works sin(8x) = sin(4x+4x) = 2sin(4x)cos(4x)

OpenStudy (anonymous):

sin(8x)=sin(2(4x))=???

OpenStudy (valpey):

Looks like you are almost there just use: sin(2x) = 2sin(x)cos(x)

OpenStudy (anonymous):

wait im really lost><

OpenStudy (valpey):

See the double angle formulas: http://www.sosmath.com/trig/Trig5/trig5/trig5.html

OpenStudy (anonymous):

just Use the double angle formula ! for sin(2x) and sin(8x)

OpenStudy (anonymous):

so like this: 8sinxcosx + 2sinxcosx ?

OpenStudy (anonymous):

you're really lost : ) ! you had this ; \[\sin(5x)\cos(3x)=\frac{1}{2}(\sin(\color{green}{8x})+\sin(\color{green}{2x}))\]

OpenStudy (anonymous):

okay!

OpenStudy (anonymous):

sin(8x)=...............(using the double angle formula....what we need to prove is something that contain 4x ....and we know that 8x=2(4x))

OpenStudy (anonymous):

sin(2a)=2sin(a)cos(a)

OpenStudy (anonymous):

why do we need to contain 4x?

OpenStudy (anonymous):

\[\sin5xcos3x = \sin(\color{red}{4x})\cos(\color{red}{4x}) + \sin(x)\cos(x)\]

OpenStudy (anonymous):

see now ?

OpenStudy (anonymous):

you need to change sin(8x)=............ using sin(2a)=2sin(a)cos(a)

OpenStudy (anonymous):

8/2 = 4 sin4a=4sin(a)cos(a)

OpenStudy (anonymous):

no ! just focus ! sin(2a)=2sin(a)cos(a) take a = 4x what will you get ?

OpenStudy (anonymous):

2sin4xcos4x

OpenStudy (anonymous):

yes ! now sin(2x)=......???

OpenStudy (anonymous):

2sin2xcos2x

OpenStudy (anonymous):

:'( !!! try again sin(2x)=......

OpenStudy (anonymous):

2sinxcosx

OpenStudy (anonymous):

yes ! Now can you finish this ? using sin(8x)=2sin(4x)cos(4x) and sin(2x)=2sin(x)cos(x) ???

OpenStudy (anonymous):

you add them right?

OpenStudy (anonymous):

Maybe :) ! Just do what you have to do ! and I'll check :)

OpenStudy (anonymous):

OHHH i got it okay i will write it on a tablet please check!!

OpenStudy (anonymous):

kk ! Sure :)

OpenStudy (anonymous):

OpenStudy (anonymous):

Yeeeees ! that's correct :) !

OpenStudy (anonymous):

ahh much more clear!!

OpenStudy (anonymous):

:) ! just practice more ! And you'll be fine :)

OpenStudy (anonymous):

so since 8x needed to be converted to the "sin2a" form, we divide 8xby2 and get 4x and we put a=4x correct?

OpenStudy (anonymous):

yes ! keep in mind the double angle formula sin(2a)=2sin(a)cos(a) or sin(something)=2sin(somthing/2)cos(something/2)

OpenStudy (anonymous):

wonderful advice!! thank you very much! you're a teacher right? cuz your way better than the math teachers id had in the past!!

OpenStudy (anonymous):

No ! I'm not a teacher !

OpenStudy (anonymous):

oh wow ! thank you so much!

OpenStudy (anonymous):

You're welcome ! Anytime :)

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