I need to know the sines, no?

[allegansveritatem]

Here is the identity I am asked to verify:

I tried to work this out in a bunch of ways the most promising of which is the following:
I this the way to proceed here? If so...I wonder what to do when I only know the cosines...I mean is there a way to find the sines when you know the cosines...seems to me there is but I can't seem to find it anywhere in my text. I don't think the author would stick a problem like this in without introducing the means to solve it. But...he likes to throw people into the deep end sometimes. Actually,a lot of times!

 

[Deleted member 4993]

Here is the identity I am asked to verify:View attachment 19394

I tried to work this out in a bunch of ways the most promising of which is the following:View attachment 19395
I this the way to proceed here? If so...I wonder what to do when I only know the cosines...I mean is there a way to find the sines when you know the cosines...seems to me there is but I can't seem to find it anywhere in my text. I don't think the author would stick a problem like this in without introducing the means to solve it. But...he likes to throw people into the deep end sometimes. Actually,a lot of times!

While proving IDENTITY, you should start from one side and ARRIVE at the other side to have a "clean" proof.

Hint: Let:

2 * Cos^-1(x)= Θ

Cos(Θ/2) = x

you know:

Cos(2α) = 2*Cos²(α) -1

No need to deal with Sine functions......

 

[LCKurtz]

You almost have it. Note that both [MATH]x[/MATH] and [MATH]2x^2-1[/MATH] are in [MATH][-1,1][/MATH]. So their inverse cosine angles [MATH]\alpha[/MATH] and [MATH]\beta[/MATH] are in [MATH][0,\pi][/MATH]. Consequently if the cosine of the left side = the cosine of the right side, the angles themselves are equal and you are done.

 

[Dr.Peterson]

I hope you can see at least some echo of the double-angle formula hidden in there; that's what this is all about.

You've made a good start; I like your approach of naming the two angles whose cosines are known. So you have [MATH]\cos(\alpha) = x[/MATH] and [MATH]\cos(\beta) = 2x^2 - 1[/MATH]. Now think about what the goal looks like in terms of these variables: [MATH]2\alpha = \beta[/MATH]. So this is what you have to show.

So what is [MATH]\cos(2\alpha)[/MATH], in terms of x? And what is [MATH]\cos(\beta)[/MATH], in terms of x?

 

[allegansveritatem]

While proving IDENTITY, you should start from one side and ARRIVE at the other side to have a "clean" proof.

Hint: Let:

2 * Cos^-1(x)= Θ

Cos(Θ/2) = x

you know:

Cos(2α) = 2*Cos²(α) -1

No need to deal with Sine functions......

2 arccos(x)=theta, from which we can say cos(theta/2)=x : that is something new to me. But having checked it out on my calculator i see that it is true. I will have to think the rest of this through. I will come back tomorrow. Thanks for the tips.

 

[allegansveritatem]

You almost have it. Note that both [MATH]x[/MATH] and [MATH]2x^2-1[/MATH] are in [MATH][-1,1][/MATH]. So their inverse cosine angles [MATH]\alpha[/MATH] and [MATH]\beta[/MATH] are in [MATH][0,\pi][/MATH]. Consequently if the cosine of the left side = the cosine of the right side, the angles themselves are equal and you are done.

yes, I have graphed both sides of this equation and know that for that range they are identical...but I want to show it algebraically, or should I say trigonometrically? Thanks

 

[allegansveritatem]

I hope you can see at least some echo of the double-angle formula hidden in there; that's what this is all about.

You've made a good start; I like your approach of naming the two angles whose cosines are known. So you have [MATH]\cos(\alpha) = x[/MATH] and [MATH]\cos(\beta) = 2x^2 - 1[/MATH]. Now think about what the goal looks like in terms of these variables: [MATH]2\alpha = \beta[/MATH]. So this is what you have to show.

So what is [MATH]\cos(2\alpha)[/MATH], in terms of x? And what is [MATH]\cos(\beta)[/MATH], in terms of x?

actually I was thinking in terms of the subtraction formula, i.e., cos (alpha- beta). Probably would have come a cropper though because no value is give to check by. I did see that 2 alpha equals beta but didn't know how to get from that to a double angle formula. I will ponder this tomorrow and see what I can come up with. I will be back. Thanks

 

[Deleted member 4993]

2 arccos(x)=theta, from which we can say cos(theta/2)=x : that is something new to me. But having checked it out on my calculator i see that it is true. I will have to think the rest of this through. I will come back tomorrow. Thanks for the tips.

let:

cos^-1(x) = t

Then by definition of "inverse function":

cos(t) = x

Then, if

2 * cos^-1(x) = Θ

cos^-1(x) = Θ/2

cos(Θ/2) = x

 

[Deleted member 4993]

While proving IDENTITY, you should start from one side and ARRIVE at the other side to have a "clean" proof.

Hint: Let:

2 * Cos^-1(x)= Θ...........................................(1)

Cos(Θ/2) = x

you know:

Cos(2α) = 2*Cos²(α) -1

No need to deal with Sine functions......

2 * Cos^-1(x)= Θ.....................................................(1)

Cos(Θ/2) = x

2x² - 1 = 2*Cos²(Θ/2) - 1 = Cos[2 * (Θ/2)] = Cos(Θ) .................. So

Cos-1[2x² - 1] = Θ

using eqn (1) into above

2 * Cos^-1(x) = Cos^-1[2x² - 1]..................................QED

 

[allegansveritatem]

While proving IDENTITY, you should start from one side and ARRIVE at the other side to have a "clean" proof.

Hint: Let:

2 * Cos^-1(x)= Θ

Cos(Θ/2) = x

you know:

Cos(2α) = 2*Cos²(α) -1

No need to deal with Sine functions......

my main problem seems gto be that I

2 * Cos^-1(x)= Θ.....................................................(1)

Cos(Θ/2) = x

2x² - 1 = 2*Cos²(Θ/2) - 1 = Cos[2 * (Θ/2)] = Cos(Θ) .................. So

Cos-1[2x² - 1] = Θ

using eqn (1) into above

2 * Cos^-1(x) = Cos^-1[2x² - 1]..................................QED

I read over this post this morning and copied it out. Later when math mode kicked in I took it and tried to figure it out. After a few false starts I saw what was being done:We find the value of x on one side and plug that value into the appropriate place on the other side. Believe it or not,. it took me a long time to get this straight. Here is what I did:

I'm still not sure that this is correct but it is my best shot so far.