How to calculate B = root A^2 + A^2 + 2AA Cos theta?

[Indranil]

How to calculate B = root A^2 + A^2 + 2AA Cos theta?
The steps are below
B = root A^2 + A^2 + 2AA Cos theta = first step, root 2A^2 (1 + cos theta) = second step,root 4A^2 cos^2 theta/2 = third step, 2A cos theta/2
I don't understand the first, second and third steps. Could you please, get these steps easier so that I am able to understand?

 

[MarkFL]

Okay, let's begin with:

$\displaystyle B=\sqrt{A^2+A^2+2AA\cos(\theta)}$

Since $\displaystyle AA=A^2$ we have:

$\displaystyle B=\sqrt{A^2+A^2+2A^2\cos(\theta)}$

Now, $\displaystyle A^2+A^2=2A^2$, so we have:

$\displaystyle B=\sqrt{2A^2+2A^2\cos(\theta)}$

Factor out $\displaystyle 2A^2$:

$\displaystyle B=\sqrt{2A^2(1+\cos(\theta))}$

Next, use a form of the double angle identity for cosine $\displaystyle \cos(2\alpha)=2\cos^2(\alpha)-1\implies\cos(2\alpha)+1=2\cos^2(\alpha)$ to write:

$\displaystyle B=\sqrt{2A^2\left(2\cos^2\left(\frac{ \theta}{2}\right)\right)}$

Apply the commutative property of multiplication to write:

$\displaystyle B=\sqrt{4A^2\cos^2\left(\frac{\theta}{2}\right)}$

This can be written as:

$\displaystyle B=\sqrt{\left(2A\cos\left( \frac{\theta}{2}\right)\right)^2}= 2\left|A\cos\left(\frac{\theta}{2}\right)\right|$

Assuming $\displaystyle 0\le A\cos\left(\frac{\theta}{2}\right)$ we may write:

$\displaystyle B=2A\cos\left(\frac{\theta}{2}\right)$

Does that make sense?

 

[Dr.Peterson]

How to calculate B = root A^2 + A^2 + 2AA Cos theta?
The steps are below
B = root A^2 + A^2 + 2AA Cos theta = first step, root 2A^2 (1 + cos theta) = second step,root 4A^2 cos^2 theta/2 = third step, 2A cos theta/2
I don't understand the first, second and third steps. Could you please, get these steps easier so that I am able to understand?

This may be unnecessary, but I will add it in case it helps:

What they are doing is not "calculating", but "simplifying"; there is not one correct answer. They are doing what they do in order to put the result in a form that is meaningful, and has a geometric interpretation they presumably will be pointing out. So without being told the goal, you might not think to do the particular things they do. And if all you needed was a formula you could apply, none of it is really necessary.

But, in a word, what they do is (a) factor out A^2, (b) apply a half-angle formula, and (c) take the square root (using the fact that the root of a product is the product of the roots).

 

[Indranil]

Okay, let's begin with:

$\displaystyle B=\sqrt{A^2+A^2+2AA\cos(\theta)}$

Since $\displaystyle AA=A^2$ we have:

$\displaystyle B=\sqrt{A^2+A^2+2A^2\cos(\theta)}$

Now, $\displaystyle A^2+A^2=2A^2$, so we have:

$\displaystyle B=\sqrt{2A^2+2A^2\cos(\theta)}$

Factor out $\displaystyle 2A^2$:

$\displaystyle B=\sqrt{2A^2(1+\cos(\theta))}$

Next, use a form of the double angle identity for cosine $\displaystyle \cos(2\alpha)=2\cos^2(\alpha)-1\implies\cos(2\alpha)+1=2\cos^2(\alpha)$ to write:

$\displaystyle B=\sqrt{2A^2\left(2\cos^2\left(\frac{ \theta}{2}\right)\right)}$

Apply the commutative property of multiplication to write:

$\displaystyle B=\sqrt{4A^2\cos^2\left(\frac{\theta}{2}\right)}$

This can be written as:

$\displaystyle B=\sqrt{\left(2A\cos\left( \frac{\theta}{2}\right)\right)^2}= 2\left|A\cos\left(\frac{\theta}{2}\right)\right|$

Assuming $\displaystyle 0\le A\cos\left(\frac{\theta}{2}\right)$ we may write:

$\displaystyle B=2A\cos\left(\frac{\theta}{2}\right)$

Does that make sense?

Next, use a form of the double angle identity for cosine [FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]⟹[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT]cos⁡(2α)=2cos2⁡(α)−1⟹cos⁡(2α)+1=2cos2⁡(α)
I don't understand the part above. Could you simplify it please?

 

[MarkFL]

Next, use a form of the double angle identity for cosine [FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]⟹[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT]cos⁡(2α)=2cos2⁡(α)−1⟹cos⁡(2α)+1=2cos2⁡(α)
I don't understand the part above. Could you simplify it please?

What about it don't you understand? Are you not familiar with the double angle identities for cosine?

 

[Indranil]

What about it don't you understand? Are you not familiar with the double angle identities for cosine?

No sir, Could you discuss it or give links or video links so that I can understand?

 

[JeffM]

No sir, Could you discuss it or give links or video links so that I can understand?

http://mathhelpforum.com/trigonometry/185935-trigonometry-memorize-trigonometry-derive.html

Ignore the video. Read the pdf.

 

[Indranil]

Okay, let's begin with:

$\displaystyle B=\sqrt{A^2+A^2+2AA\cos(\theta)}$

Since $\displaystyle AA=A^2$ we have:

$\displaystyle B=\sqrt{A^2+A^2+2A^2\cos(\theta)}$

Now, $\displaystyle A^2+A^2=2A^2$, so we have:

$\displaystyle B=\sqrt{2A^2+2A^2\cos(\theta)}$

Factor out $\displaystyle 2A^2$:

$\displaystyle B=\sqrt{2A^2(1+\cos(\theta))}$

Next, use a form of the double angle identity for cosine $\displaystyle \cos(2\alpha)=2\cos^2(\alpha)-1\implies\cos(2\alpha)+1=2\cos^2(\alpha)$ to write:

$\displaystyle B=\sqrt{2A^2\left(2\cos^2\left(\frac{ \theta}{2}\right)\right)}$

Apply the commutative property of multiplication to write:

$\displaystyle B=\sqrt{4A^2\cos^2\left(\frac{\theta}{2}\right)}$

This can be written as:

$\displaystyle B=\sqrt{\left(2A\cos\left( \frac{\theta}{2}\right)\right)^2}= 2\left|A\cos\left(\frac{\theta}{2}\right)\right|$

Assuming $\displaystyle 0\le A\cos\left(\frac{\theta}{2}\right)$ we may write:

$\displaystyle B=2A\cos\left(\frac{\theta}{2}\right)$

Does that make sense?

[FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]⟹[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT]

1. why cos(2alfa) = 2cos^2(alfa)? as I have learned

[FONT=KaTeX_Main] cos 2α = cos[/FONT]²[FONT=KaTeX_Main] α − sin[/FONT]²[FONT=KaTeX_Main] α , [/FONT][FONT=KaTeX_Main]cos 2α = 1− 2 sin² α,[/FONT]
[FONT=KaTeX_Main]cos 2α = 2 cos² α − 1[/FONT][FONT=KaTeX_Main]
2. From what, you realized that there should be [/FONT][FONT=KaTeX_Main]2 cos[/FONT]² [FONT=KaTeX_Main]α in this equation [/FONT][FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Size3]√[/FONT]2A2(2cos2⁡(θ/2))?

3. [FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]⟹[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main]) in this equeation how you get '[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1'? Could you explain my above quearies? [/FONT]

 

[MarkFL]

Please use $\displaystyle \LaTeX$...when I quoted you post, this is what I see (ugh...thumbnails? really?):



Finding the needle I wish to address in that haystack is difficult.

1. why cos(2alfa) = 2cos^2(alfa)?

Please reread my post, I did not state the above...I stated:

$\displaystyle \cos(2\alpha)=2\cos^2(\alpha)-1$

 

[Indranil]

Please use $\displaystyle \LaTeX$...when I quoted you post, this is what I see (ugh...thumbnails? really?):

View attachment 9636

Finding the needle I wish to address in that haystack is difficult.



Please reread my post, I did not state the above...I stated:

$\displaystyle \cos(2\alpha)=2\cos^2(\alpha)-1$

Yes, I checked it. could you tell me please why did you put 2cos^2(theta/2) in the place of B= root 2A^2(1+cos(theta)?

 

[MarkFL]

Yes, I checked it. could you tell me please why did you put 2cos^2(theta/2) in the place of B= root 2A^2(1+cos(theta)?

I took the algebraic steps I needed to get the form indicated. In my first post in this thread, I explained every step taken.

 

[Dr.Peterson]

Yes, I checked it. could you tell me please why did you put 2cos^2(theta/2) in the place of B= root 2A^2(1+cos(theta)?

He didn't. He replaced 1+cos(theta) with 2cos^2(theta/2). This could be done because they are equal; in effect, this is the half-angle identity. It needed to be done because the goal was to use a half-angle.

Which of these two facts are you questioning? "Why" can mean different things, so you need to clarify.

 

[Indranil]

I took the algebraic steps I needed to get the form indicated. In my first post in this thread, I explained every step taken.

I checked your first thread. But I don't understand how you put the value '( 2cos² (θ⁄2) ) in the place of (1+cos (θ) ).
Still, I don't understand the steps
B = √2A² (1+cos (θ) ) to B = √2A² ( 2cos² (θ⁄2) ) Could you get these steps easier for me so that I can understand? I am confused, please help.

 

[MarkFL]

I checked your first thread. But I don't understand how you put the value '( 2cos² (θ⁄2) ) in the place of (1+cos (θ) ).
Still, I don't understand the steps
B = √2A² (1+cos (θ) ) to B = √2A² ( 2cos² (θ⁄2) ) Could you get these steps easier for me so that I can understand? I am confused, please help.

One double-angle identity for cosine is:

$\displaystyle \cos(2\alpha)=2\cos^2(\alpha)-1$

If we add 1 to both sides, we get:

$\displaystyle \cos(2\alpha)+1=2\cos^2(\alpha)$

Now, if we let:

$\displaystyle \theta=2\alpha\implies \alpha=\frac{\theta}{2}$

Then we have:

$\displaystyle \cos(\theta)+1=2\cos^2\left(\frac{\theta}{2}\right)$

Does that make sense?

 

[Indranil]

One double-angle identity for cosine is:

$\displaystyle \cos(2\alpha)=2\cos^2(\alpha)-1$

If we add 1 to both sides, we get:

$\displaystyle \cos(2\alpha)+1=2\cos^2(\alpha)$

Now, if we let:

$\displaystyle \theta=2\alpha\implies \alpha=\frac{\theta}{2}$

Then we have:

$\displaystyle \cos(\theta)+1=2\cos^2\left(\frac{\theta}{2}\right)$

Does that make sense?

Yes, I got it now. (1 + cosθ) and (cosθ + 1) is the same thing? Could you tell me, please?

 

[MarkFL]

Yes, I got it now. (1 + cosθ) and (cosθ + 1) is the same thing? Could you tell me, please?

Yes, the commutative property of addition tells us:

$\displaystyle a+b=b+a$

 

[Indranil]

Yes, the commutative property of addition tells us:

$\displaystyle a+b=b+a$

Could you simplify the second part of it, please?
tanα = A sinθ/A + A cosθ = 2A (sin θ/2) (cos θ/2) / (2A cos² θ/2) = tan θ/2or α = θ/2

 

[MarkFL]

Could you simplify the second part of it, please?
tanα = A sinθ/A + A cosθ = 2A (sin θ/2) (cos θ/2) / (2A cos² θ/2) = tan θ/2or α = θ/2

The best I can tell, you are essentially given to verify:

$\displaystyle \dfrac{A\sin(\theta)}{A+A\cos(\theta)}= \tan\left(\dfrac{\theta}{2}\right)$

I would first divide the numerator and denominator of the LHS by $\displaystyle A$ to get:

$\displaystyle \dfrac{\sin(\theta)}{1+\cos(\theta)}= \tan\left(\dfrac{\theta}{2}\right)$

The double-angle identity for sine is $\displaystyle \sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ which means:

$\displaystyle \sin(\theta)=2\sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right)$

And in this thread, we've already established that:

$\displaystyle 1+\cos(\theta)= 2\cos^2\left(\dfrac{\theta}{2}\right)$

And so applying these to the identity we are given to verify, we obtain:

$\displaystyle \dfrac{2\sin\left( \dfrac{\theta}{2}\right) \cos\left(\dfrac{ \theta}{2}\right)}{ 2\cos^2\left(\dfrac{ \theta}{2}\right)}= \tan\left(\dfrac{\theta}{2}\right)$

Divide the numerator and denominator of the LHS by $\displaystyle 2\cos\left(\dfrac{\theta}{2}\right)$ to get:

$\displaystyle \dfrac{ \sin\left(\dfrac{\theta}{2}\right)}{ \cos\left(\dfrac{\theta}{2}\right)}= \tan\left(\dfrac{\theta}{2}\right)$

Using the identity $\displaystyle \dfrac{\sin(\alpha)}{\cos(\alpha)}=\tan(\alpha)$ we have:

$\displaystyle \tan\left(\dfrac{\theta}{2}\right)= \tan\left(\dfrac{\theta}{2}\right) \quad\checkmark$

Shown as desired.

In the future, I suggest beginning a new thread for a new question. This way threads don't become potentially convoluted and hard to follow.

 

[Indranil]

The best I can tell, you are essentially given to verify:

$\displaystyle \dfrac{A\sin(\theta)}{A+A\cos(\theta)}= \tan\left(\dfrac{\theta}{2}\right)$

I would first divide the numerator and denominator of the LHS by $\displaystyle A$ to get:

$\displaystyle \dfrac{\sin(\theta)}{1+\cos(\theta)}= \tan\left(\dfrac{\theta}{2}\right)$

The double-angle identity for sine is $\displaystyle \sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ which means:

$\displaystyle \sin(\theta)=2\sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right)$

And in this thread, we've already established that:

$\displaystyle 1+\cos(\theta)= 2\cos^2\left(\dfrac{\theta}{2}\right)$

And so applying these to the identity we are given to verify, we obtain:

$\displaystyle \dfrac{2\sin\left( \dfrac{\theta}{2}\right) \cos\left(\dfrac{ \theta}{2}\right)}{ 2\cos^2\left(\dfrac{ \theta}{2}\right)}= \tan\left(\dfrac{\theta}{2}\right)$

Divide the numerator and denominator of the LHS by $\displaystyle 2\cos\left(\dfrac{\theta}{2}\right)$ to get:

$\displaystyle \dfrac{ \sin\left(\dfrac{\theta}{2}\right)}{ \cos\left(\dfrac{\theta}{2}\right)}= \tan\left(\dfrac{\theta}{2}\right)$

Using the identity $\displaystyle \dfrac{\sin(\alpha)}{\cos(\alpha)}=\tan(\alpha)$ we have:

$\displaystyle \tan\left(\dfrac{\theta}{2}\right)= \tan\left(\dfrac{\theta}{2}\right) \quad\checkmark$

Shown as desired.

In the future, I suggest beginning a new thread for a new question. This way threads don't become potentially convoluted and hard to follow.

As we know sin(2α) = 2 sin(α)cos(α) But could you please explain how you find sin(θ) = 2 sin(θ/2) cos(θ/2) ?

 

[Dr.Peterson]

As we know sin(2α) = 2 sin(α)cos(α) But could you please explain how you find sin(θ) = 2 sin(θ/2) cos(θ/2) ?

This is done by replacing α with θ/2: sin(θ) = sin(2(θ/2)) = 2 sin(θ/2) cos(θ/2).

Alternatively, you could say this: Let θ = 2α; then sin(θ) = sin(2α) = 2 sin(α)cos(α) = 2 sin(θ/2) cos(θ/2) because α = θ/2.

Such substitutions are common throughout trigonometry and especially in calculus, so it is important to become familiar with this process.