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Find Sin3A when SinA=1/4

Hi guys, got a question i just can figure out :frown:

"Find Sin3A when SinA=1/4"

I know i need to use Sin3A=Sin(2A+A)

And from there Sin(2A+A)=Sin2ACosA+Cos2ASinA

But from here i really dont know where to progress from?

Cheers in advance
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by Ramjam
Hi guys, got a question i just can figure out :frown:

"Find Sin3A when SinA=1/4"

I know i need to use Sin3A=Sin(2A+A)

And from there Sin(2A+A)=Sin2ACosA+Cos2ASinA

But from here i really dont know where to progress from?

Cheers in advance


use the double angle formula
Reply 2
Avatar for Lulu24
Lulu24
13
Original post by Ramjam
Hi guys, got a question i just can figure out :frown:

"Find Sin3A when SinA=1/4"

I know i need to use Sin3A=Sin(2A+A)

And from there Sin(2A+A)=Sin2ACosA+Cos2ASinA

But from here i really dont know where to progress from?

Cheers in advance


I might be wrong but i imagine you would use the fact that Cos + Sin add up to 90 degree's or pie/2

and maybe substitute in Sin2A as 2SinACosA and the Cos2A as a cos^2

then i imagine you may be able to substitute in SinA as 1/4 and CosA....?

lol long shot but who knows :smile: not my strongest part of maths but i was just doin this C4 stuff too.
Reply 3
Avatar for davros
davros
16
Original post by Ramjam
Hi guys, got a question i just can figure out :frown:

"Find Sin3A when SinA=1/4"

I know i need to use Sin3A=Sin(2A+A)

And from there Sin(2A+A)=Sin2ACosA+Cos2ASinA

But from here i really dont know where to progress from?

Cheers in advance


Write sin2A as 2sinAcosA in the 1st term which will give you a factor of Cos^2A which you can rewrite in terms of sinA

Write cos2A = cos^2A - sin^2A in the 2nd term, which you can then rewrite solely in terms of sinA.

Then plug in the value of sinA you are given :smile:
Reply 4
Avatar for Ramjam
Ramjam
OP
1
Original post by Lulu24
I might be wrong but i imagine you would use the fact that Cos + Sin add up to 90 degree's or pie/2

and maybe substitute in Sin2A as 2SinACosA and the Cos2A as a cos^2

then i imagine you may be able to substitute in SinA as 1/4 and CosA....?

lol long shot but who knows :smile: not my strongest part of maths but i was just doin this C4 stuff too.


Ill take a look into that bud thankyou
Reply 5
Avatar for Ramjam
Ramjam
OP
1
Original post by davros
Write sin2A as 2sinAcosA in the 1st term which will give you a factor of Cos^2A which you can rewrite in terms of sinA

Write cos2A = cos^2A - sin^2A in the 2nd term, which you can then rewrite solely in terms of sinA.

Then plug in the value of sinA you are given :smile:


I think i get that but still not 100% sure i understand :frown:
Reply 6
Avatar for Hasufel
Hasufel
16
you can translate this into the well known identity for sin(3A) which just involves sine`s:

sin⁡(A+2A) \sin(A+2A) expand this.

substitute the first cos⁡(2A)\cos(2A) with the identity cos⁡2(A)−sin⁡2(A)\cos^{2}(A)- \sin^{2}(A)

Then, the sin⁡(2A)\sin(2A) with 2sin⁡(A)cos⁡(A)2 \sin(A) \cos(A)

Multiply out - but -

Notice where you`d have cos^{2}(A)`s and substitute them with (1-sin^{2}(A))`s

Then simplify..

Questions like this are all about using idientities to simplify what you have into something you can use without having to "figure out" something else.
(edited 9 years ago)
Reply 7
Avatar for davros
davros
16
Original post by Ramjam
I think i get that but still not 100% sure i understand :frown:


Post some working based on the suggestions you've had so far and show us how far you've got.
Reply 8
Avatar for Ramjam
Ramjam
OP
1
Original post by Hasufel
you can translate this into the well known identity for sin(3A) which just involves sine`s:

sin⁡(A+2A) \sin(A+2A) expand this.

substitute the first cos⁡(2A)\cos(2A) with the identity cos⁡2(A)−sin⁡2(A)\cos^{2}(A)- \sin^{2}(A)

Then, the sin⁡(2A)\sin(2A) with 2sin⁡(A)cos⁡(A)2 \sin(A) \cos(A)

Multiply out - but -

Notice where you`d have cos^{2}(A)`s and substitute them with (1-sin^{2}(A))`s

Then simplify..

Questions like this are all about using idientities to simplify what you have into something you can use without having to "figure out" something else.


Original post by davros
Post some working based on the suggestions you've had so far and show us how far you've got.


From what you guys have said ive managed to get this far, not sure how to progress from here or even if what ive done is correct?

IMAG1037.jpg398.4KB
Reply 9
Avatar for davros
davros
16
Original post by Ramjam
From what you guys have said ive managed to get this far, not sure how to progress from here or even if what ive done is correct?



Your first term is OK but you've gone wrong with your second bracket - you should have ended up with sinA(1 - 2sin^2A) which you can then expand to combine with the first term.
Reply 10
Avatar for Ramjam
Ramjam
OP
1
Original post by davros
Your first term is OK but you've gone wrong with your second bracket - you should have ended up with sinA(1 - 2sin^2A) which you can then expand to combine with the first term.


so it should be Sin(2A+A)=(2SinA)(1-Sin^2A)+sinA(1 - 2sin^2A)
Original post by Ramjam
so it should be Sin(2A+A)=(2SinA)(1-Sin^2A)+sinA(1 - 2sin^2A)


YES.

Now expand, and collect, and you have the identity:

sin⁡(3A)≡3sin⁡(A)−4sin⁡3(A)\sin(3A) \equiv 3 \sin(A)- 4 \sin^{3}(A)

and just plug in sin(A)..
(edited 9 years ago)
Reply 12
Avatar for Ramjam
Ramjam
OP
1
Original post by Hasufel
YES.

Now expand, and collect, and you have the identity:

sin⁡(3A)≡3sin⁡(A)−4sin⁡3(A)\sin(3A) \equiv 3 \sin(A)- 4 \sin^{3}(A)

and just plug in sin(A)..


Sin3A= (3x(1/4)) - (4x(1/4^3))

Sin3A = (0.75) - (0.0625)

Sin3A = 0.6875
Original post by Ramjam
Sin3A= (3x(1/4)) - (4x(1/4^3))

Sin3A = (0.75) - (0.0625)

Sin3A = 0.6875


YUP.. or just as a quotient... 1116 \displaystyle \frac{11}{16}

(3/4 - 1/16, = 12/16 - 1/16)

i think the fractional form might be better, as sin(a) is 1/4 in the question...
(edited 9 years ago)
Reply 14
Avatar for Ramjam
Ramjam
OP
1
Original post by Hasufel
YUP.. or just as a quotient... 1116 \displaystyle \frac{11}{16}


Thank you all for this help you have given me, think ive actually understood this.
Original post by Ramjam
Thank you all for this help you have given me, think ive actually understood this.


:smile:

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