Not sure how to approach problem

[The Bob]

$$\sin 2x=2\sin x\cos x$$ Sub in $$x = \frac{\pi}{4}$$

$$\sin (\frac{\pi}{4} \times \frac{4}{2}) = 2\sin \frac{\pi}{4} \cos \frac{\pi}{4}$$

$$\sin \frac{4\pi}{8} = 2\sin \frac{\pi}{4} \cos \frac{\pi}{4}$$

$$\sin \frac{\pi}{2} = 2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}$$

$$\sin \frac{\pi}{2} = 2 \times \frac{2}{4}$$

$$\sin \frac{\pi}{2} = 2 \times \frac{1}{2}$$

$$\sin \frac{\pi}{2} = 1$$

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

The Bob (2004 ©)

 

[quasar987]

In a sense, you are! :rofl:

http://mathworld.wolfram.com/Trigonometry.html

 

[dextercioby]

I'm sorry.Wrong Hint. :rofl: $$x\rightarrow \frac{\pi}{8}$$
U should be getting a biquadratic equation in $$\sin\frac{\pi}{8}$$

Daniel.

 

[The Bob]

In a sense, you are! :rofl:

I was hoping someone would get a small, unfunny joke.

http://mathworld.wolfram.com/Trigonometry.html

Been here. Doesn't help me much.

I will now try the equation again with $$x = \sin \frac{\pi}{8}$$

The Bob (2004 ©)

 

[The Bob]

$$\sin 2x=2\sin x\cos x$$ Sub in $$x = \frac{\pi}{8}$$

$$\sin (\frac{\pi}{8} \times \frac{4}{2}) = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}$$

$$\sin \frac{4\pi}{16} = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}$$

$$\sin \frac{\pi}{4} = 2 \times \sin\frac{\pi}{8} \times \cos \frac{\pi}{8}$$

But from here onwards I think it is wrong:

$$\frac{\sqrt{2}}{2} = (2 \times \sin \frac{\pi}{8}) \cos\frac{\pi}{8}$$

$$\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \times \cos\frac{\pi}{8}$$

$$\frac{\sqrt{2}}{2} \times \frac{2}{\sqrt{2}} = \cos\frac{\pi}{8}$$

$$\frac{4\sqrt{2}}{2} = \cos\frac{\pi}{8}$$

The Bob (2004 ©)

 

[BobG]

I'm not quite sure where you're all going with that.

The cosine gets you further:

$$cos(\frac{\pi}{4})=cos (\frac{\pi}{8} + \frac{\pi}{8})$$

$$cos(\frac{\pi}{4}) = cos (\frac{\pi}{8}) cos (\frac{\pi}{8}) - sin (\frac{\pi}{8}) sin (\frac{\pi}{8})$$

$$cos(\frac{\pi}{4}) = cos^2 (\frac{\pi}{8}) - sin^2 (\frac{\pi}{8})$$

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

$$cos(\frac{\pi}{4}) = 1 - 2 sin^2 (\frac{\pi}{8})$$

$$2 sin^2 (\frac{\pi}{8}) = 1 - cos (\frac{\pi}{4})$$

$$sin^2 (\frac{\pi}{8}) = \frac {1 - cos (\frac{\pi}{4})}{2}$$

$$sin (\frac{\pi}{8}) = \sqrt {\frac {1 - cos (\frac{\pi}{4})}{2}}$$

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.

 

[BobG]

$$\sin 2x=2\sin x\cos x$$ Sub in $$x = \frac{\pi}{8}$$

$$\sin (\frac{\pi}{8} \times \frac{4}{2}) = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}$$

$$\sin \frac{4\pi}{16} = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}$$

$$\sin \frac{\pi}{4} = 2 \times \sin\frac{\pi}{8} \times \cos \frac{\pi}{8}$$

But from here onwards I think it is wrong:

$$\frac{\sqrt{2}}{2} = (2 \times \sin \frac{\pi}{8}) \cos\frac{\pi}{8}$$

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

$$\frac{\sqrt{2}}{2} \times \frac{2}{\sqrt{2}} = \cos\frac{\pi}{8}$$

$$\frac{4\sqrt{2}}{2} = \cos\frac{\pi}{8}$$

The Bob (2004 ©)

Quote edited by me.

 

[dextercioby]

$$\sin 2x=2\sin x\cos x$$ Sub in $$x = \frac{\pi}{8}$$

$$\sin (\frac{\pi}{8} \times \frac{4}{2}) = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}$$

$$\sin \frac{4\pi}{16} = 2\sin \frac{\pi}{8} \cos \frac{\pi}{8}$$

$$\sin \frac{\pi}{4} = 2 \times \sin\frac{\pi}{8} \times \cos \frac{\pi}{8}$$

But from here onwards I think it is wrong:

$$\frac{\sqrt{2}}{2} = (2 \times \sin \frac{\pi}{8}) \cos\frac{\pi}{8}$$

$$\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \times \cos\frac{\pi}{8}$$

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

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.

 

[The Bob]

$$cos(\frac{\pi}{4}) = cos^2 (\frac{\pi}{8}) - sin^2 (\frac{\pi}{8})$$

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

My book says that: $$\cos^2A = \frac{1+\cos2A}{2}$$

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

The Bob (2004 ©)

 

[BobG]

My book says that: $$\cos^2A = \frac{1+\cos2A}{2}$$

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

The Bob (2004 ©)

Because I was solving for the sine of $$\frac{\pi}{8}$$

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

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

which gives you:

$$cos(\frac{\pi}{4}) = 2 cos^2(\frac{\pi}{8}) - 1$$
or
$$cos(\frac{\pi}{4})+1 = 2 cos^2(\frac{\pi}{8})$$

$$\frac{cos(\frac{\pi}{4})+1}{2} = cos^2(\frac{\pi}{8})$$

and $$\frac{\pi}{4}$$ is 2 times $$\frac{\pi}{8}$$

 

[The Bob]

Ok, you are trying to work out $$\sin(\frac{\pi}{8})$$ yet you need to replace the $$\cos^2(\frac{\pi}{8})$$. I think what is not going right in my head is that I am trying to work out $$\cos(\frac{\pi}{8})$$ or $$\sin(\frac{\pi}{8})$$ and I started with $$\cos(\frac{\pi}{4})$$ and I have ended up knowing $$\sin(\frac{\pi}{8})$$. 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.

 

[BobG]

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.

 

[The Bob]

Right so here I go (again :tongue2:):

I want to compute $$\sin\frac{3\pi}{8}$$

$$\sin\frac{3\pi}{8} = \sin(\frac{2\pi+\pi}{8}) = \sin(\frac{\pi}{4}+\frac{\pi}{8})$$

Therefore: $$\sin(\frac{\pi}{4}+\frac{\pi}{8}) = \sin\frac{\pi}{4}\cos\frac{\pi}{8}+\cos\frac{\pi}{4}\sin\frac{\pi}{8}$$

$$\sin\frac{\pi}{4}\cos\frac{\pi}{8}+\cos\frac{\pi}{4}\sin\frac{\pi}{8}$$ is equal to $$(\sin\frac{\pi}{8}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\cos\frac{\pi}{8})$$

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

Therefore: $$\sin \frac{\pi}{8} = \sqrt{\frac{1-\cos\frac{\pi}{4}}{2}}$$

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

$$= (\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\cos\frac{\pi}{8})$$

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

and so $$\cos\frac{\pi}{8} = \sqrt{\frac{1+\cos\frac{\pi}{4}}{2}} = \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}$$

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

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.

 

[BobG]

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.

 

[dextercioby]

Simple manipulations of radicals would yield
$$\sin\frac{3\pi}{8}=\frac{1}{4}(\sqrt{\sqrt{8}-2}+\sqrt{\sqrt{8}+2})$$
,which could become more 'simple',if one used the 'double-radicals' formulas.

Daniel.

 

[The Bob]

$$(\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}})$$

Completely stuck here now. I can't really see where to start. Do I square verything to remove the square root signs?

The Bob (2004 ©)

 

[dextercioby]

$$(\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}})$$

Completely stuck here now. I can't really see where to start. Do I square verything to remove the square root signs?

The Bob (2004 ©)

$$(\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}})=(\frac{\sqrt{2-\sqrt{2}}}{2}\cdot\frac{\sqrt{2}}{2})+(\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2}}{2})$$
$$=\frac{1}{4}(\sqrt{\sqrt{8}-2}+\sqrt{\sqrt{8}+2})$$

Daniel.

 

[The Bob]

$$(\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}})=(\frac{\sqrt{2-\sqrt{2}}}{2}\cdot\frac{\sqrt{2}}{2})+(\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2}}{2})$$
$$=\frac{1}{4}(\sqrt{\sqrt{8}-2}+\sqrt{\sqrt{8}+2})$$

The problem is, Dex, that I cannot see your stages of getting from one set of numbers to another. If I had the equation $$\frac{1}{x}=\frac{1}{2}$$ then I don't need to do any stages because it is so simple. The set of equations I have tried to simplify are simple to you but not to me. Therefore if I am to understand I need some stages in between e.g. I do not need to go:

$$\frac{1}{x}=\frac{1}{2}$$
=> $$1=\frac{1}{2} x$$
=> $$\frac{1}{\frac{1}{2}}= x$$

=> $$1 \times \frac{2}{1} = x$$

=> $$1 \times 2 = x = 2$$

Can you see what I mean? Both me and you and everyone on these forums can see that the answer to $$x$$ was 2 without doing the set of works I have just done.

This is the same for you with $$(\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\times\frac{\sqrt{2}}{2})+(\frac{\sqrt{2}}{2}\times\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}})$$ but not for me. It my be silly to you but it isn't to me. I need those stages to help me understand.

Thanks for all the help you have given me so for.

The Bob (2004 ©)

 

[dextercioby]

Okayn,let's calculate the first term VERY EXPLICITELY and i'll let u do the second,since it's very much similar.

$$I_{1}=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}} \times \frac{\sqrt{2}}{2}$$(1)

$$I_{1}=A\times B$$ (2)
,where
$$A=:\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}$$ (3)
$$B=:\frac{\sqrt{2}}{2}$$ (4)

Let's compute "A":
You know that:
$$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$$ (5)
So let's apply it in our case
$$A=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}} =\frac{\sqrt{1-\frac{\sqrt{2}}{2}}}{\sqrt{2}} =\frac{\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2}}}{\sqrt{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}$$ (6)

Then the product A*B becomes
$$I_{1}=\frac{\sqrt{2-\sqrt{2}}}{2}\times \frac{\sqrt{2}}{2} =\frac{1}{4} C$$ (7)
,where "C" is
$$C=:\sqrt{2-\sqrt{2}}\times \sqrt{2} =\sqrt{2\sqrt{2}-2}=\sqrt{\sqrt{8}-2}$$ (8)

Therefore
$$I_{1}=\frac{\sqrt{\sqrt{8}-2}}{4}$$

Daniel.

 

[The Bob]

Okayn,let's calculate the first term VERY EXPLICITELY and i'll let u do the second,since it's very much similar.

I see how it works now, therefore, I feel it is not necessary for me to do the second part as it simply means I change a subtract sign for an addition sign.

I thank you very much.

So $$\sin\frac{3\pi}{8}=\frac{1}{4}(\sqrt{\sqrt{8}-2}+\sqrt{\sqrt{8}+2})$$. This is the answer you wanted? Oh and I can take out a common factor so, again, I did not see the need to show it.

The Bob (2004 ©)

P.S. If that is the answer you wanted then can I have another one that will mean I do not need to learn more but can simply apply what is in this thread, just to make sure I know what I am doing.

 

[dextercioby]

This is a challanging problem.I haven't done it myself,so i don't know any solution and neither the answer.
Compute via trigonometry:
$$\sin\frac{\pi}{5}$$

Daniel.

PS.I would be grateful to the one that comes up with a solution.