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Cosxsinx P4

Hey guys sorry but i'm stuck in a bit of a :eek:

with this q on my P4 paper...

Ok i've got a curve y = 3sin2x + cos2x and prior to doing a volume of revolution bit i need to square y and write it in the form Asin4x + Bcos4x + c

With me so far?

Now this is how far i got

(3sin2x + cos2x)^2 = 9sin^2(4x) + cos^2(4x) + 6sin2xcos2x

By using 2sinxcosx = sin 2A i can get that to:

9sin^2(4x) + cos^2(4x) + 3sin4x

But where do i go from here? Please help!!
Love Boo xx

Reply 1

kikzen

why do you need to get it in that form?

Reply 2

kikzen

also when you square, the first line should read

(3sin2x + cos2x)^2 = 9sin^2(2x) + cos^2(2x) + 6sin2xcos2x

which will lead to

3sin4x - 4cos4x + 5

or something like that :smile:

Reply 3

Cirsium

Because the exam question tells me to...

Reply 4

Fermat

(3sin2x + cos2x)² = 9sin²(2x) + cos²(2x) + 6sin(2x)cos(2x)

now use,

sin²(2x) = ½(1 - cos4x)
cos²(2x) = ½(1 + cos4x)
2sin2xcos2x = sin4x

Reply 5

m:)ckel

(3sin2x + cos2x)^2 = 9(sin^2 2x) + (cos^2 x) + 6sin2xcos2x

= 9[(0.5)(1-cos4x)] + (0.5)(1+cos4x) + 3[2sin2xcos2x] <---using half angle identities

= 9/2 - (9/2)cos4x + (1/2) + (1/2)cos4x + 3sin4x

= 3sin4x - (9/2)cos4x + 5

Reply 6

evariste_3086

Bekaboo

Hey guys sorry but i'm stuck in a bit of a :eek:

with this q on my P4 paper...

Ok i've got a curve y = 3sin2x + cos2x and prior to doing a volume of revolution bit i need to square y and write it in the form Asin4x + Bcos4x + c

y=9sin^2(2x)+cos^2(2x)+6cos2xsin2x
sin 2x=2cosxsinx so 3sin 4x=6cos2xsin2x
sin^2 (2x)+cos^2 (2x)=1
so 9sin^2(2x)+cos^2(2x)=8 sin^2 2x+1
but cos 4x=cos^2 2x-sin ^2 2x=(1-2sin^2 (2x))
so
4(1-cos 4x)=8sin^2 (2x)