TriG QueStiOn.. (1 Viewer)

[+GriM ReApeR+]

a) Write an expression for sin^4(2x) in terms of cos4x
b) Hence show that (integration sign)(sin^4(2x))dx=3x/8-1/8sin4x+1/64sin8x+c

ok... I dont understand part a) so I cant do part b)...
does part a) mean that I have to find sin^4(2x) as the suject of the equation with all terms in cos4x...or somthing like that...pls help...

 

[CM_Tutor]

It means you need to find real numbers A, B and C such that

sin⁴2x = A + Bcos4x + Ccos8x

Note - it's not very clear, but if you look at the integration you need to get in (b), it is clear that this is what is needed.

 

[+GriM ReApeR+]

Originally posted by CM_Tutor
It means you need to find real numbers A, B and C such that

sin⁴2x = A + Bcos4x + Ccos8x

Note - it's not very clear, but if you look at the integration you need to get in (b), it is clear that this is what is needed.

so that means i have to expand sin^4(2x)?

 

[CM_Tutor]

You will need to get an expression for sin²2x by expanding cos4x. Then square - you can eliminate the cos²4x term that results by looking at cos8x.

 

[AGB]

Originally posted by CM_Tutor
You will need to get an expression for sin²2x by expanding cos4x. Then square - you can eliminate the cos²4x term that results by looking at cos8x.

cant you just use that formula for sin²x = (.5 - .5cos2x) i.e. re write the equation as (sin²x)² then do it normally??

 

[CM_Tutor]

Originally posted by AGB
cant you just use that formula for sin²x = (.5 - .5cos2x) i.e. re write the equation as (sin²x)² then do it normally??

Remember that the question asked for sin⁴2x in terms of cos4x

 

[AGB]

Originally posted by CM_Tutor
Remember that the question asked for sin⁴2x in terms of cos4x

sorry... typo...

but i got the right answer using that identity (you have to use it twice):

sin⁴x = (sin²x)²

= (.5 - .5cos4x)²

= .25 - .5cos4x + .25(cos²4x)

integrate that and you get....

= .25x - .125sin4x + .25[.5x + .125sin8x]

= 3x/8 - sin4x/8 + sin8x/64 + C

as required...

 

[CM_Tutor]

You mean sin⁴2x = (sin²2x)² = (.5 - .5cos4x)² = .25 - .5cos4x + .25(cos²4x)
which is the same as expanding cos4x and then squaring.

In other words, our methods are the same.