Not sure how to approach problem

[Markd]

Hi,

During a review this question has popped up (not sure what unit)

"Determine the exact value*

````5pi``````11pi
sin ___ - cos ___
````4`````` 6
I am not sure what to do with the sin and cos?

Thanks

 

[dextercioby]

Hi,

During a review this question has popped up (not sure what unit)

"Determine the exact value*

````5pi``````11pi
sin ___ - cos ___
````4`````` 6
I am not sure what to do with the sin and cos?

Thanks

So you have to find this value...
\sin\frac{5\pi}{4}-\cos\frac{11\pi}{6}...??

Okay.
\sin\frac{5\pi}{4}=\sin(\pi+\frac{\pi}{4}})=-\sin\frac{\pi}{4}
\cos\frac{11\pi}{6}=\cos(2\pi-\frac{\pi}{6})=\cos\frac{\pi}{6}

Can u take from here?Can u see the properties of sine and cosine i used to get to my results??

Daniel.

 

[The Bob]

Hi,

During a review this question has popped up (not sure what unit)

"Determine the exact value*

````5pi``````11pi
sin ___ - cos ___
````4`````` 6
I am not sure what to do with the sin and cos?

Thanks

Can you not factorise the \pi out for a start and then see if there is a relationship between sin and cos. I know that one is a translation of the other and that this can be shown by sin(x+90) = cos(x).

This might help you but I think some maths brainbox will come along and say it is wrong.

The Bob (2004 ©)

 

[The Bob]

\sin\frac{5\pi}{4}=\sin(\pi+\frac{\pi}{4}})=-\sin\frac{\pi}{4}
\cos\frac{11\pi}{6}=\cos(2\pi-\frac{\pi}{6})=\cos\frac{\pi}{6}

How did you get to these conclusions? (I know they must be right but I cannot see how).

The Bob (2004 )

 

[quasar987]

Simpler (or at least more fundamental) is to just look at the trigonometric circle (this one's not very good because it doesn't show the angles) and notice that for every integer multiple of pi, the sine of that value is 0, and for every odd integer multiple of pi, the cosine of that value is -1. In mathematical language,

sin(n\pi) = 0 \ \forall n \in \mathbb{Z}

and

cos((2n+1)\pi)=-1 \ \forall n \in \mathbb{Z}

Now, 5 is an integer ==> sin(5pi) = 0. 11 is an odd integer ==> cos (11pi) = -1.


(TheBob: look at the trigonometric circle, you will see that they are true.)

 

[The Bob]

(TheBob: look at the trigonometric circle, you will see that they are true.)

Unfortunately time is not on my side for tonight but I will look properly tomorrow. However on my scanning of the web page I saw these functions:

sec and csc. Could you explain what they are please?

The Bob (2004 ©)

 

[dextercioby]

Unfortunately time is not on my side for tonight but I will look properly tomorrow. However on my scanning of the web page I saw these functions:

sec and csc. Could you explain what they are please?

The Bob (2004 ©)

They are the 2 less used circular trigonometric functions:

\sec x=:\frac{1}{\cos x}
\csc x=:\frac{1}{\sin x}

I would rather not those two functions.Sine and cosine are much easier to work with.

Daniel.

PS.If u can't manage with the trigonometric circle (which would be really bad),u can prove my formulas using the addition formulas for sine and cosine.

 

[The Bob]

\sec x=:\frac{1}{\cos x}
\csc x=:\frac{1}{\sin x}

Cheers for this.

(which would be really bad)

Going for the vote of confidence I see. Thanks for this as well.

My problem is that I need some numbers in place of the x's and the i's.

If anyone could do that so I can see how you all got from:
\sin\frac{5\pi}{4}=\sin(\pi+\frac{\pi}{4}})=-\sin\frac{\pi}{4}

And:\cos\frac{11\pi}{6}=\cos(2\pi-\frac{\pi}{6})=\cos\frac{\pi}{6}

then that woul be a big help. I have tried to see for myself but nothing I do makes any sense.

Cheers.

The Bob (2004 ©)

 

[dextercioby]

\sin\frac{5\pi}{4}=\sin(\frac{4\pi+\pi}{4})=\sin(\pi+\frac{\pi}{4})=\sin\pi\cos\frac{\pi}{4}+\sin\frac{\pi}{4}\cos{\pi}

Now you know that \sin\pi=0;\cos\pi=-1
Then
\sin\frac{5\pi}{4}=-\sin\frac{\pi}{4}


And following the same pattern,u prove the other formula as well.

Daniel.

 

[The Bob]

I wish either I had your understanding or this countries educations system was better.

\sin\frac{5\pi}{4}=\sin(\frac{4\pi+\pi}{4})=\sin(\pi+\frac{\pi}{4})

This now makes sense. but the next stage I do not understand. I cannot see how you get from \sin(\pi+\frac{\pi}{4}) to \sin\pi\cos\frac{\pi}{4}+\sin\frac{\pi}{4}\cos{\pi}.

\cos\frac{11\pi}{6}=\cos(\frac{12\pi-\pi}{6})=\cos(2\pi-\frac{\pi}{6}) but the next bit \cos\frac{\pi}{6} does not follow like the \sin bit does. So cosine is different to sine in some respect and so I cannot even try that final stage like the sine section.

As you can see, without this understanding here it is impossible for me to understand how to answer the question. I will be honest and say that none of this has been covered yet at my college but it interests me and so if you have the time and the patience I would like to learn this please.

Cheers.

The Bob (2004 ©)

 

[dextercioby]

The fundamental formula is
\sin(x+y)=\sin x\cos y+\sin y\cos x(1)
I know a beautiful geometrical proof to it,but unfortunately i can't share it with you.
It's the formula i applied for those 2 angles (\pi \text{and} \frac{\pi}{4}).

Use formula (1) to prove that:
\sin(x+\frac{\pi}{2})=\cos x (2)
Use formula (1) to prove that:
\sin(x+\pi)=-\sin x (3)

Use formulas (1),(2) and (3)to prove that:
\cos(x+y)=\cos x\cos y-\sin x\sin y(4)

'Sine'is an odd function:
\sin(-x)=-\sin x(5)
'Cosine'is an even function:
\cos(-x)=\cos(x) (6)

Use formulas (1),(4),(5) and (6) to prove:
\sin(x-y)=\sin x\cos y-\sin y\cos x (7)
\cos(x-y)=\cos x\cos y+\sin x\cos y (8)

Daniel.

 

[The Bob]

Right so:
\sin(x+\frac{\pi}{2}) = \sin x \cos\frac{\pi}{2}+\sin\frac{\pi}{2} \cos x but the stage that allows you to get to \cos x is missing in my mind. Is it:

\cos \pi = -1 = \cos 180 therefore \cos \frac{\pi}{2} = \cos 90 = 0

\sin \pi = 0 = \sin 180 therefore \sin \frac{\pi}{2} = \sin 90 = 1

Therefore \sin x \cos\frac{\pi}{2}+\sin\frac{\pi}{2} \cos x = (\sin x \times 0)+(1 \times \cos x) = 0+\cos x = \cos x

And: \sin(x+\pi) = \sin x \cos\pi+\sin\pi \cos x = (\sin x \times -1)+(0 \times \cos x) = -\sin x + 0 = -\sin x

The next bit is going to take a little more thinking.

The Bob (2004 ©)

 

[The Bob]

\cos(x+y)=\cos x\cos y-\sin x\sin y

Therefore: \cos(x+y)=\sin (x + \frac{\pi}{2})\cos y+\sin (x+\pi) \sin y but then what?

The Bob (2004 ©)

EDIT: This is the wrong way round.

 

[dextercioby]

Well,did u prove the 'cos' addition formula??
HINT:
\cos(x+y)=\sin(x+y+\frac{\pi}{2})=\sin[x+(y+\frac{\pi}{2})]=...
,and make use of the 'sin' addition formula for
x=x;y=y+\frac{\pi}{2}

Daniel.

 

[The Bob]

Well,did u prove the 'cos' addition formula??
HINT:
\cos(x+y)=\sin(x+y+\frac{\pi}{2})=\sin[x+(y+\frac{\pi}{2})]=...

But if \cos x=\sin(x+\frac{\pi}{2}) then \cos y=\sin(y+\frac{\pi}{2})

which means that \cos x \cos y = \sin(x+\frac{\pi}{2}) \sin(y+\frac{\pi}{2})


\cos (x+y) = \cos x \cos y - \sin x \sin y therefore:

\cos (x+y) = \sin(x+\frac{\pi}{2}) \sin(y+\frac{\pi}{2}) - \sin x \sin y

But this must be wrong because it does not leave what I need.

It isn't making much sense to me.

The Bob (2004 ©)

 

[The Bob]

I'm tired and need sleep so I am off. I will think about it but I am really sorry: I do not get it. I can get parts but not the rest.

The Bob (2004 ©)

 

[BobG]

The unit circle (or trigonometric circle) is the easiest method. And you really have very few numbers to memorize.

sin 0 = \frac {\sqrt{0}}{2}

sin \frac{\pi}{6} = \frac {\sqrt{1}}{2}

sin \frac{\pi}{4} = \frac {\sqrt{2}}{2}

sin \frac{\pi}{3} = \frac {\sqrt{3}}{2}

sin \frac{\pi}{2} = \frac {\sqrt{4}}{2}

Obviously, it's a little over the top to use the square root of zero or the square root of one, but it shows the progression very clearly. In other words, in practice:

sin 0 = 0

sin \frac{\pi}{6} = \frac {1}{2}

sin \frac{\pi}{4} = \frac {\sqrt{2}}{2}

sin \frac{\pi}{3} = \frac {\sqrt{3}}{2}

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

Cosine regresses in the same way, starting from 1 for cosine 0 going to 0 for cosine pi/2.

If you've got one quarter of the circle down, the only thing to visualize is the sign of each as you move to different quadrants of the trig circle and how far the radius is from the x-axis for the sine or from the y-axis for the cosine.

If you have a good grasp of the basics, you don't actually have to get into the sum/difference laws until you start getting the 'tougher' angles, like \frac{\pi}{12} and so on.

A good look at the unit circle and you'll see that the end point of \frac{\pi}{6} is just as far away from the y-axis as the end point of -\frac{\pi}{6}.

The end point of \frac{\pi}{6} is just as far away from the x-axis as \frac{5 \pi}{6}.

Being able to visualize that makes the odd/even identities for sine and cosine intuitively obvious even before you prove them (and, yes, in trig class, you get to step along proving one identity after another, but it's a lot easier if you can already visualize these things in your head before you start).

Edit: And in light of dextercioby's brilliant observation about the difference in time zones, I've deleted that insult about being such a slacker you have to take a nap in the middle of the afternoon.

 

[dextercioby]

Trigonometry,whether circular,elliptic or hyperbolic,is wonderful and not really difficult.Since it's almost midnight over the Channel,i'll let u sleep on them...

Daniel.

 

[Markd]

OK I have received the answer but I am still don't understand the problem
``````````__ ___
Answer: -(\/2 + \/3 )
````````` __________
`````````````2

Please Help!

by the way the teacher said you use the trignometric circle

 

[BobG]

Hi,

During a review this question has popped up (not sure what unit)

"Determine the exact value*

````5pi``````11pi
sin ___ - cos ___
````4`````` 6
I am not sure what to do with the sin and cos?

Thanks

\pi radians is halfway around the circle, or along the negative x-axis.

\frac {5 \pi}{4} radians is \frac {\pi}{4} radians past halfway. If you know your sines for one quarter of the circle, and can visualize where \frac {5 \pi}{4} radians is on the unit circle, you know that the sine of \frac {5 \pi}{4} has to be - \frac{\sqrt {2}}{2} .

You do the same for the cosine of the second angle - you learn the unit circle well enough that you know what the cosine of \frac{11 \pi}{6} has to be.

Substitute the values for the sine and the cosine into the original problem. Since they both have a denominator of 2, you can combine them into the answer you received:

\frac{-\sqrt{2} - \sqrt{3}}{2} = \frac{-(\sqrt{2} + \sqrt{3})}{2}

 

[Stargate]

That's Physics for you and welcome!

 

[The Bob]

Edit: And in light of dextercioby's brilliant observation about the difference in time zones, I've deleted that insult about being such a slacker you have to take a nap in the middle of the afternoon.

You're too kind.

My maths teacher caught me trying this problem in Maths today and he actually tried to help me. It turns out that this sort of question is in the hardest module of Pure and Applied Maths for the last year of college in England. As I have just started college I hope you can all understand why I did/do not get it straight away.

Anyway back to the question :

\sin\frac{5\pi}{4} = \sin(\frac{4\pi+\pi}{4}) = \sin(\pi+\frac{\pi}{4})

This end result can be then applied in the same way as \sin(x+y)=\sin x\cos y+\sin y\cos x is:

\sin(\pi+\frac{\pi}{4}) = \sin\pi\cos\frac{\pi}{4}+\sin\frac{\pi}{4}\cos\pi = (0\times\cos\frac{\pi}{4})+(\sin\frac{\pi}{4}\times -1) = 0 + -\sin\frac{\pi}{4} = -\sin\frac{\pi}{4}

\cos\frac{11\pi}{6} = \cos(\frac{12\pi-\pi}{6}) = \cos(2\pi-\frac{\pi}{6})

This end result can be then applied in the same way as \cos(x-y)=\cos x\cos y+\sin x\cos y is:

\cos(2\pi-\frac{\pi}{6}) = \cos2\pi\cos\frac{\pi}{6}+\sin2\pi\sin\frac{\pi}{6} = (1\times\cos\frac{\pi}{6})+(0\times\sin\frac{\pi}{6}) = \cos\frac{\pi}{6}+0 = \cos\frac{\pi}{6}

\sin\frac{5\pi}{4}-\cos\frac{11\pi}{6} = -\sin\frac{\pi}{4}-\cos\frac{\pi}{6} = -\frac{\sqrt{2}}{2}-\frac {\sqrt{3}}{2} = \frac{-\sqrt{2}-\sqrt{3}}{2} = \frac{-(\sqrt{2}+\sqrt{3})}{2}

So this is all there is to it?

Can anyone give me another one to do please?

The Bob (2004 ©)

 

[dextercioby]

Okay,if you're really "fired up and ready to go",there's a simple one
Compute \sin\frac{3\pi}{8}
,without a calculator,tables of functions and stuff like that.Just plain simple circular trigonometry.

Daniel.

 

[The Bob]

Compute \sin\frac{3\pi}{8}

I have started trying but does compute mean simplify for this type of question (the same as evaluate)?

The Bob (2004 ©)

 

[dextercioby]

That "compute" means "bring a mathematically rigurous proof that
\sin\frac{3\pi}{8}\sim 0.92365
",using trigonometry and algebra calculations only.

Daniel.

 

[The Bob]

\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})

Then I thought about the \sin(x+y) and so:

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

There must be a simple say to do it.

The Bob (2004 ©)

 

[dextercioby]

\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})

Then I thought about the \sin(x+y) and so:

\sin(\frac{\pi}{4}+\frac{\pi}{8}) = \sin \frac{\pi}{8}\cos \frac{\pi}{4}+ \cos \frac{\pi}{8} \sin \frac{\pi}{4} 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... U know that
\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}
U need to compute
\sin\frac{\pi}{8}=...?? \cos\frac{\pi}{8}=...??

Daniel.

 

[BobG]

\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})

Then I thought about the \sin(x+y) and so:

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

 

[dextercioby]

Who said it was pretty??Is math always pretty??Is that the reason we invented calculators and aproximation methods?

Daniel.

 

[The Bob]

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

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

The Bob (2004 ©)

 

[dextercioby]

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

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.

 

[Markd]

Thankyou so much for your help everyone!

 

[The Bob]

(\frac{\sqrt{2-\sqrt{2}}}{2} \times \frac{\sqrt{2}}{2})+ (\frac{\sqrt{2}}{2} \times \frac{\sqrt{2+\sqrt{2}}}{2}) but I found the answers to \sin\frac{\pi}{8} and \cos\frac{\pi}{8} here.

How should I have got to them?

The Bob (2004 ©)

 

[dextercioby]

You cheated... Okay:
HINT:
\sin 2x=2\sin x\cos x

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

Daniel.

 

[BobG]

Or, you could use:

sin \frac{x}{2} = \sqrt{\frac{1-cos x}{2}}

cos \frac{x}{2} = \sqrt{\frac{1+cos x}{2}}

with

x=\frac{\pi}{4}

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:

sin x = \sqrt{\frac{1-cos 2x}{2}}

you substitute the half-angle for x.

 

[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!

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

 

[dextercioby]

I'm sorry.Wrong Hint. 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!

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.

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 ):

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.