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Helpp c3

Okay so by using trig double angles, how do i turn this:
sin^2 y cos^2 y
into one trigonometric function?

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Reply 1
Consider the square root of your function.

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Reply 2
Original post by Krollo
Consider the square root of your function.

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what do you mean?
thank you.
Reply 3
Original post by hajs
Okay so by using trig double angles, how do i turn this:
sin^2 y cos^2 y
into one trigonometric function?


i am not sure if that is what you mean but

sin(4x) = sin(2 x 2x) = 2sin(2x)cos(2x)

thus

.... your is half of that...

does this help?
Reply 4
Original post by TeeEm
i am not sure if that is what you mean but

sin(4x) = sin(2 x 2x) = 2sin(2x)cos(2x)

thus

.... your is half of that...

does this help?

i dont know,
according to the answer is is:

1/4( 4sin^2 ycos^2 y ) = ( 2sinycosy ) 2 = sin^2 2y
not sure how they got that?
Reply 5
Original post by hajs
Okay so by using trig double angles, how do i turn this:
sin^2 y cos^2 y
into one trigonometric function?


sin2ycos2y=(sinycosy)2\sin^2y \cos^2y = (\sin y \cos y)^2

Do you recognize what's in the brackets on the RHS?
Reply 6
Original post by davros
sin2ycos2y=(sinycosy)2\sin^2y \cos^2y = (\sin y \cos y)^2

Do you recognize what's in the brackets on the RHS?


yes i get that bit, but then how do i get the answer from thereon?
Reply 7
Original post by hajs
what do you mean?
thank you.


What can you do with sinycosy

(or indeed double that? )

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Reply 8
Original post by hajs
i dont know,
according to the answer is is:

1/4( 4sin^2 ycos^2 y ) = ( 2sinycosy ) 2 = sin^2 2y
not sure how they got that?


do you mean simplify sin2ycos2y to something else?
or sin(2y)cos(2y)?
Reply 9
Original post by hajs
yes i get that bit, but then how do i get the answer from thereon?


Well what is sinycosy in terms of a well-known identity?
Reply 10
Original post by TeeEm
do you mean simplify sin2ycos2y to something else?
or sin(2y)cos(2y)?

this
Reply 11
Original post by hajs
this


cos2x = 1/2 +1/2cos(2x) by rearranging the double angle for cos2x


similarly

sin2x = 1/2 -1/2cos(2x) by rearranging the double angle for cos2x

....
Reply 12
Original post by davros
Well what is sinycosy in terms of a well-known identity?


that's were i am confused, there's no identity in the book?:confused: maybe im just not picking it up
can you please explain? :redface:
Reply 13
Original post by hajs
that's were i am confused, there's no identity in the book?:confused: maybe im just not picking it up
can you please explain? :redface:


look at my last post

see if this helps

tells us what are you meant to get

then reply just to one of us so only one can help you ( better this way)
Reply 14
Original post by hajs
that's were i am confused, there's no identity in the book?:confused: maybe im just not picking it up
can you please explain? :redface:


So you've never seen the following identity before:

sin2A = 2sinAcosA

?
Reply 15
Original post by TeeEm
cos2x = 1/2 +1/2cos(2x) by rearranging the double angle for cos2x


similarly

sin2x = 1/2 -1/2cos(2x) by rearranging the double angle for cos2x

....


Original post by TeeEm
look at my last post

see if this helps

tells us what are you meant to get

then reply just to one of us so only one can help you ( better this way)


okay.
so im meant to be getting: 1/4( 4sin2θcos2θ ) = ( 2sinθcosθ )^2 = 1/4sin2

i really dont understand so please really dumb it down haha

i get it except for the bold bit?
why a quarter?
(edited 9 years ago)
Reply 16
Original post by hajs
okay.
so im meant to be getting: 1/4( 4sin2θcos2θ ) = ( 2sinθcosθ )^2 = 1/4sin2

i really dont understand so please really dumb it down haha



no problem

It looks that you have an identity to prove ...

write down just the Left hand side

then the just the right hand side

then I will explain step by step
Reply 17
Original post by TeeEm
no problem

It looks that you have an identity to prove ...

write down just the Left hand side

then the just the right hand side

then I will explain step by step


i dont have a right hand side.
all im given is sin^2 ycos^2 y
Reply 18
Original post by hajs
i dont have a right hand side.
all im given is sin^2 ycos^2 y


ok I think you have done most of the work already

sin[SUP]2ycos2y =1/4(4sin2ycos2y)

=1/4(2sinycosy)2.


do you follow this so far?
(edited 9 years ago)
Reply 19
[QUOTE="TeeEm;51161301"]ok I think you have done most of the work already

sin2ycos2y =1/4(4sin2ycos2y)

=1/4(2sinycosy)2.


do you follow this so far?


I will rewrite it bad typing

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