Not sure how to approach problem

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  • #26
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[tex] \sin\frac{3\pi}{8} = \sin(\frac{2\pi+\pi}{8}) = \sin(\frac{2\pi}{8}+\frac{\pi}{8}) = \sin(\frac{\pi}{4}+\frac{\pi}{8})[/tex]

Then I thought about the [tex]\sin(x+y)[/tex] and so:

[tex]\sin(\frac{\pi}{4}+\frac{\pi}{8}) = \sin \frac{\pi}{8}\cos \frac{\pi}{4}+ \cos \frac{\pi}{8} \sin \frac{\pi}{4}[/tex] but I do not know what this fractions make.

There must be a simple say to do it.

The Bob (2004 ©)
 
  • #27
dextercioby
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The Bob said:
[tex] \sin\frac{3\pi}{8} = \sin(\frac{2\pi+\pi}{8}) = \sin(\frac{2\pi}{8}+\frac{\pi}{8}) = \sin(\frac{\pi}{4}+\frac{\pi}{8})[/tex]

Then I thought about the [tex]\sin(x+y)[/tex] and so:

[tex]\sin(\frac{\pi}{4}+\frac{\pi}{8}) = \sin \frac{\pi}{8}\cos \frac{\pi}{4}+ \cos \frac{\pi}{8} \sin \frac{\pi}{4}[/tex] but I do not know what this fractions make.

There must be a simple say to do it.

The Bob (2004 ©)
Okay,u're on the right track... :approve: U know that
[tex] \sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2} [/tex]
U need to compute
[tex] \sin\frac{\pi}{8}=...??[/tex] [tex] \cos\frac{\pi}{8}=...?? [/tex]

Daniel.
 
  • #28
BobG
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The Bob said:
[tex] \sin\frac{3\pi}{8} = \sin(\frac{2\pi+\pi}{8}) = \sin(\frac{2\pi}{8}+\frac{\pi}{8}) = \sin(\frac{\pi}{4}+\frac{\pi}{8})[/tex]

Then I thought about the [tex]\sin(x+y)[/tex] and so:

[tex]\sin(\frac{\pi}{4}+\frac{\pi}{8}) = \sin \frac{\pi}{8}\cos \frac{\pi}{4}+ \cos \frac{\pi}{8} \sin \frac{\pi}{4}[/tex] but I do not know what this fractions make.

There must be a simple say to do it.

The Bob (2004 ©)
Not extremely simple, but there is a half angle formula for sines and cosines.

Actually not difficult, either, but the answer just doesn't look all that pretty.
 
  • #29
dextercioby
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Who said it was pretty??Is math always pretty??Is that the reason we invented calculators and aproximation methods???

Daniel.
 
  • #30
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[tex](\sin \frac{\pi}{8}\times \frac{\sqrt{2}}{2})+ (\frac{\sqrt{2}}{2} \times \cos \frac{\pi}{8})[/tex]

[tex]\sin \frac{\pi}{8} = \tan \frac{\pi}{8} \cos \frac{\pi}{8}[/tex] but I do not think this will get me anywhere at all.

The Bob (2004 ©)
 
  • #31
dextercioby
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The Bob said:
[tex](\sin \frac{\pi}{8}\times \frac{\sqrt{2}}{2})+ (\frac{\sqrt{2}}{2} \times \cos \frac{\pi}{8})[/tex]
As the other Bob on this forum suggested,u must use some formula giving u the 'sine' of the 'angle/2' in terms of the sine of 'angle'.

Daniel.
 
  • #32
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Thankyou so much for your help everyone!
 
  • #33
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[tex](\frac{\sqrt{2-\sqrt{2}}}{2} \times \frac{\sqrt{2}}{2})+ (\frac{\sqrt{2}}{2} \times \frac{\sqrt{2+\sqrt{2}}}{2})[/tex] but I found the answers to [tex]\sin\frac{\pi}{8}[/tex] and [tex]\cos\frac{\pi}{8}[/tex] here.

How should I have got to them???

The Bob (2004 ©)
 
  • #34
dextercioby
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You cheated... :mad: :tongue2: Okay:
HINT:
[tex]\sin 2x=2\sin x\cos x [/tex]

Make [itex] x\rightarrow \frac{\pi}{4} [/itex]
in the formula i've given u and a little bit of algebra will lead you to the result.

Daniel.
 
  • #35
BobG
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Or, you could use:

[tex]sin \frac{x}{2} = \sqrt{\frac{1-cos x}{2}}[/tex]

[tex]cos \frac{x}{2} = \sqrt{\frac{1+cos x}{2}}[/tex]

with

[tex]x=\frac{\pi}{4}[/tex]

If you know the sum identity for cosines, you can figure out the double angle identity [it's just cos (A+A)]. If you modify the double angle identity (cos^2 x = 1 - sin^2 x), you can get to the half angle identities used above (in other words, instead of:

[tex]sin x = \sqrt{\frac{1-cos 2x}{2}}[/tex]

you substitute the half-angle for x.
 
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  • #36
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[tex]\sin 2x=2\sin x\cos x [/tex] Sub in [tex]x = \frac{\pi}{4}[/tex]

[tex]\sin (\frac{\pi}{4} \times \frac{4}{2}) = 2\sin \frac{\pi}{4} \cos \frac{\pi}{4}[/tex]

[tex]\sin \frac{4\pi}{8} = 2\sin \frac{\pi}{4} \cos \frac{\pi}{4}[/tex]

[tex]\sin \frac{\pi}{2} = 2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}[/tex]

[tex]\sin \frac{\pi}{2} = 2 \times \frac{2}{4}[/tex]

[tex]\sin \frac{\pi}{2} = 2 \times \frac{1}{2}[/tex]

[tex]\sin \frac{\pi}{2} = 1[/tex]

So now what??? AAAAAARRRRRRRRRGGGGGGGHHHHHHHHHHH!!!!!!!!!!!!!!! I feel like I am going around in circles.

The Bob (2004 ©)
 
  • #38
dextercioby
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I'm sorry.Wrong Hint. :rofl: [tex] x\rightarrow \frac{\pi}{8} [/tex]
U should be getting a biquadratic equation in [tex] \sin\frac{\pi}{8} [/tex]

Daniel.
 
  • #39
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quasar987 said:
In a sense, you are! :rofl:
I was hoping someone would get a small, unfunny joke. :smile:

quasar987 said:
Been here. Doesn't help me much.

I will now try the equation again with [tex]x = \sin \frac{\pi}{8}[/tex]

The Bob (2004 ©)
 
  • #40
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[tex]\sin 2x=2\sin x\cos x [/tex] Sub in [tex]x = \frac{\pi}{8}[/tex]

[tex]\sin (\frac{\pi}{8} \times \frac{4}{2}) = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}[/tex]

[tex]\sin \frac{4\pi}{16} = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}[/tex]

[tex]\sin \frac{\pi}{4} = 2 \times \sin\frac{\pi}{8} \times \cos \frac{\pi}{8}[/tex]

But from here onwards I think it is wrong:

[tex]\frac{\sqrt{2}}{2} = (2 \times \sin \frac{\pi}{8}) \cos\frac{\pi}{8}[/tex]

[tex]\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \times \cos\frac{\pi}{8}[/tex]

[tex]\frac{\sqrt{2}}{2} \times \frac{2}{\sqrt{2}} = \cos\frac{\pi}{8}[/tex]

[tex]\frac{4\sqrt{2}}{2} = \cos\frac{\pi}{8}[/tex]

The Bob (2004 ©)
 
  • #41
BobG
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I'm not quite sure where you're all going with that.

The cosine gets you further:

[tex]cos(\frac{\pi}{4})=cos (\frac{\pi}{8} + \frac{\pi}{8})[/tex]

[tex]cos(\frac{\pi}{4}) = cos (\frac{\pi}{8}) cos (\frac{\pi}{8}) - sin (\frac{\pi}{8}) sin (\frac{\pi}{8})[/tex]

[tex]cos(\frac{\pi}{4}) = cos^2 (\frac{\pi}{8}) - sin^2 (\frac{\pi}{8})[/tex]

[tex]cos(\frac{\pi}{4}) = [1 - sin^2 (\frac{\pi}{8})] - sin^2 (\frac{\pi}{8})[/tex]

[tex] cos(\frac{\pi}{4}) = 1 - 2 sin^2 (\frac{\pi}{8})[/tex]

[tex]2 sin^2 (\frac{\pi}{8}) = 1 - cos (\frac{\pi}{4})[/tex]

[tex] sin^2 (\frac{\pi}{8}) = \frac {1 - cos (\frac{\pi}{4})}{2}[/tex]

[tex] sin (\frac{\pi}{8}) = \sqrt {\frac {1 - cos (\frac{\pi}{4})}{2}}[/tex]

You can use a similar method to get the half angle formula for the cosine. Instead of substituting (1- sin^2 x) for cos^2 x, you substitute (1 - cos^2 x) for sin^2 x.
 
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  • #42
BobG
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The Bob said:
[tex]\sin 2x=2\sin x\cos x [/tex] Sub in [tex]x = \frac{\pi}{8}[/tex]

[tex]\sin (\frac{\pi}{8} \times \frac{4}{2}) = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}[/tex]

[tex]\sin \frac{4\pi}{16} = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}[/tex]

[tex]\sin \frac{\pi}{4} = 2 \times \sin\frac{\pi}{8} \times \cos \frac{\pi}{8}[/tex]

But from here onwards I think it is wrong:

[tex]\frac{\sqrt{2}}{2} = (2 \times \sin \frac{\pi}{8}) \cos\frac{\pi}{8}[/tex]

[tex]\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \times \cos\frac{\pi}{8}[/tex]
Right here is where it's wrong. (2 sin x) is not the same as (sin 2x).

[tex]\frac{\sqrt{2}}{2} \times \frac{2}{\sqrt{2}} = \cos\frac{\pi}{8}[/tex]

[tex]\frac{4\sqrt{2}}{2} = \cos\frac{\pi}{8}[/tex]

The Bob (2004 ©)
Quote edited by me.
 
  • #43
dextercioby
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The Bob said:
[tex]\sin 2x=2\sin x\cos x [/tex] Sub in [tex]x = \frac{\pi}{8}[/tex]

[tex]\sin (\frac{\pi}{8} \times \frac{4}{2}) = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}[/tex]

[tex]\sin \frac{4\pi}{16} = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}[/tex]

[tex]\sin \frac{\pi}{4} = 2 \times \sin\frac{\pi}{8} \times \cos \frac{\pi}{8}[/tex]

But from here onwards I think it is wrong:

[tex]\frac{\sqrt{2}}{2} = (2 \times \sin \frac{\pi}{8}) \cos\frac{\pi}{8}[/tex]

[tex]\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \times \cos\frac{\pi}{8}[/tex]

[tex]\frac{\sqrt{2}}{2} \times \frac{2}{\sqrt{2}} = \cos\frac{\pi}{8}[/tex]
It's wrong here as well.You simplified wrongly.
[tex]\frac{4\sqrt{2}}{2} = \cos\frac{\pi}{8}[/tex]

The Bob (2004 ©)
Quote edited by moi.
My contribution is marked with 'red' and appears as a follow-up to the message by Bob.The 'right' Bob. :tongue2:

Daniel.
 
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  • #44
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BobG said:
[tex]cos(\frac{\pi}{4}) = cos^2 (\frac{\pi}{8}) - sin^2 (\frac{\pi}{8})[/tex]

[tex]cos(\frac{\pi}{4}) = [1 - sin^2 (\frac{\pi}{8})] - sin^2 (\frac{\pi}{8})[/tex]
My book says that: [tex]\cos^2A = \frac{1+\cos2A}{2}[/tex]

Therefore why is it not: [tex]\cos^2(\frac{\pi}{8}) = \frac{1+\cos2(\frac{\pi}{8})}{2}[/tex] ?

The Bob (2004 ©)
 
  • #45
BobG
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The Bob said:
My book says that: [tex]\cos^2A = \frac{1+\cos2A}{2}[/tex]

Therefore why is it not: [tex]\cos^2(\frac{\pi}{8}) = \frac{1+\cos2(\frac{\pi}{8})}{2}[/tex] ?

The Bob (2004 ©)
Because I was solving for the sine of [tex]\frac{\pi}{8}[/tex]

If solving for the cosine, you'd make the following substitution instead:

[tex]cos(\frac{\pi}{4}) = cos^2(\frac{\pi}{8}) - [1 - cos^2 (\frac{\pi}{8})][/tex]

which gives you:

[tex]cos(\frac{\pi}{4}) = 2 cos^2(\frac{\pi}{8}) - 1[/tex]
or
[tex]cos(\frac{\pi}{4})+1 = 2 cos^2(\frac{\pi}{8})[/tex]

[tex]\frac{cos(\frac{\pi}{4})+1}{2} = cos^2(\frac{\pi}{8})[/tex]

and [tex]\frac{\pi}{4}[/tex] is 2 times [tex]\frac{\pi}{8}[/tex]
 
  • #46
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Ok, you are trying to work out [tex]\sin(\frac{\pi}{8})[/tex] yet you need to replace the [tex]\cos^2(\frac{\pi}{8})[/tex]. I think what is not going right in my head is that I am trying to work out [tex]\cos(\frac{\pi}{8})[/tex] or [tex]\sin(\frac{\pi}{8})[/tex] and I started with [tex]\cos(\frac{\pi}{4})[/tex] and I have ended up knowing [tex]\sin(\frac{\pi}{8})[/tex]. I can sort of see how the maths works but I do not see how I was meant to know there was link.

The Bob (2004 ©)

P.S. Dex say it was easy. :frown:
 
  • #47
BobG
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Unless you take a trig class, it is hard to see the link. For some of the identities, you almost have to know the end formula you want in order to figure out how to get there.

They're supposed to be shortcuts and they are if you use them a lot and memorize them.

The idea is to memorize as few identities as possible and knowing there is a link allows you to do that. For example, if you know the cosine and sine sum identities, you know the double angle formulas for both with almost no thought, even if you haven't used them in a long time.

There's a few you're just better off memorizing. Considering the number of steps you need to get from the correct double angle formula or from the correct sum identity to the half angle formula and considering the similarity between the half angle formulas for cosine and sine, the half angle formula is one worth remembering.
 
  • #48
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Right so here I go (again :tongue2:):

I want to compute [tex]\sin\frac{3\pi}{8}[/tex]

[tex]\sin\frac{3\pi}{8} = \sin(\frac{2\pi+\pi}{8}) = \sin(\frac{\pi}{4}+\frac{\pi}{8})[/tex]

Therefore: [tex]\sin(\frac{\pi}{4}+\frac{\pi}{8}) = \sin\frac{\pi}{4}\cos\frac{\pi}{8}+\cos\frac{\pi}{4}\sin\frac{\pi}{8}[/tex]

[tex]\sin\frac{\pi}{4}\cos\frac{\pi}{8}+\cos\frac{\pi}{4}\sin\frac{\pi}{8}[/tex] is equal to [tex](\sin\frac{\pi}{8}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\cos\frac{\pi}{8})[/tex]

BobG said that: [tex]\sin x = \sqrt{\frac{1-\cos 2x}{2}}[/tex] so substitute [tex]x = \frac{\pi}{8}[/tex]

Therefore: [tex]\sin \frac{\pi}{8} = \sqrt{\frac{1-\cos\frac{\pi}{4}}{2}}[/tex]

So now: [tex](\sin\frac{\pi}{8}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\cos\frac{\pi}{8}) = (\sqrt{\frac{1-\cos\frac{\pi}{4}}{2}}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\cos\frac{\pi}{8})[/tex]

[tex]= (\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\cos\frac{\pi}{8})[/tex]

For [tex]\cos\frac{\pi}{8}[/tex] am I to assume that [tex]\cos\frac{x}{2} = \sqrt{\frac{1+\cos x}{2}}[/tex] is equal to [tex]\cos x = \sqrt{\frac{1+\cos 2x}{2}}[/tex]

and so [tex]\cos\frac{\pi}{8} = \sqrt{\frac{1+\cos\frac{\pi}{4}}{2}} = \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}[/tex]

So now: [tex](\sin\frac{\pi}{8}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\cos\frac{\pi}{8}) = (\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}})[/tex]

This is the bit were I get stuck. I can sort of see how to simplify it but not very well.

The Bob (2004 ©)

P.S. Just making sure everything is fine to date. :smile:
 
  • #49
BobG
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Yes. Simplifying it isn't horrible (you have a square root of two in both the numerator and denominator for each term), but the end answer still doesn't look very pretty.
 
  • #50
dextercioby
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Simple manipulations of radicals would yield
[tex] \sin\frac{3\pi}{8}=\frac{1}{4}(\sqrt{\sqrt{8}-2}+\sqrt{\sqrt{8}+2}) [/tex]
,which could become more 'simple',if one used the 'double-radicals' formulas.

Daniel.
 

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