Prove that
1-cosA+cosB-cos(A+B)/1+cosA-cosB-cos(A+B) = tanA/2cotA/2
hmm. I see most of them have not been answered yet. So, what the heck? It's a slow day.
First off, a typo. You should have said
... = tanA/2 cotB/2
I gotta admit, I had a hard time with this one, just manipulating the LS. However,
tanA/2 = sinA/(1+cosA)
cotB/2 = sinB/(1-cosB)
so,
tanA/2 cotB/2 = sinA sinB/(1 + cosA - cosB - cosAcosB)
also,
tanA/2 = (1-cosA)/sinA
cotB/2 = (1+cosB)/sinB
so,
tanA/2 cotB/2 = (1 - cosA + cosB - cosAcosB)/sinAsinB
that gives us
tan^2 A/2 * cot^2 B/2 =
1 - cosA + cosB - cosA cosB
---------------------------
1 + cosA - cosB - cosA cosB
Now, that means that we have to show that
(1-cosA+cosB-cos(A+B))^2 = 1 - cosA + cosB - cosA cosB
and
(1+cosA-cosB-cos(A+B))^2 = 1 + cosA - cosB - cosA cosB
If you expand that out and collect terms, you will see that it is true, but it's a lot of work.
using the half-angle formulas, you can show that
1 - cosA + cosB - cosA cosB
= 2[sin^2 A/2 + cos^2 B/2 - cos^2 (A+B)/2]
Then, with a little more fiddling, that can be changed to
sin A/2 cos B/2 sin(A+B)/2
and the denominator comes out to
cos A?2 sin B/2 sin(A+B)/2
divide those two expressions and you get
sin A/2 cos B/2
---------------------
cos A/2 sin B/2
= tanA/2 cotB/2
you've posted a lot of these lately. What have you tried so far on this one?
To prove the equality, let's start by simplifying the left-hand side (LHS) and the right-hand side (RHS) separately.
Let's simplify the LHS:
1 - cosA + cosB - cos(A + B)
---------------------------
1 + cosA - cosB - cos(A + B)
To simplify this expression, we can try to combine like terms. We notice that the numerator has terms with cosine A and B, while the denominator has terms with cosine A and B as well as cosine (A + B).
Let's rewrite the expression using common denominators:
[(1 - cosA) + (cosB - cos(A + B))] / [(1 + cosA) - (cosB + cos(A + B))]
Now, we can apply some trigonometric identities to simplify further:
Recall the cosine difference formula: cos(x - y) = cos(x)cos(y) + sin(x)sin(y)
Using this formula, we can simplify the numerator:
1 - cosA + cosB - [cosAcosB + sinAsinB]
Simplifying the denominator:
1 + cosA - cosB - [cosAcosB - sinAsinB]
Now, we can cancel out some terms:
[-cosA + cosB - cosAcosB - sinAsinB + sinAsinB] / [1 + cosA - cosB - cosAcosB + sinAsinB]
Simplifying further:
[-cosA + cosB - cosAcosB] / [1 + cosA - cosB - cosAcosB]
Next, let's simplify the RHS:
tan(A/2) / cot(A/2)
To simplify this expression, let's write it in terms of sine and cosine:
(sin(A/2) / cos(A/2)) / (cos(A/2) / sin(A/2))
We can simplify further by multiplying the numerator and denominator by the reciprocal of the denominator:
(sin(A/2) / cos(A/2)) * (sin(A/2) / cos(A/2))
Using the identity tan(x) = sin(x) / cos(x), we can rewrite the RHS as:
tan(A/2)^2
To complete the proof, we need to show that the LHS is equal to the RHS:
[-cosA + cosB - cosAcosB] / [1 + cosA - cosB - cosAcosB] = tan(A/2)^2
By applying the half-angle formula for tangent: tan^2(x/2) = (1 - cosx) / (1 + cosx)
We can rewrite tan(A/2)^2 as:
[1 - cosA] / [1 + cosA]
Now, we need to show that:
[-cosA + cosB - cosAcosB] / [1 + cosA - cosB - cosAcosB] = [1 - cosA] / [1 + cosA]
To do this, we need to multiply the numerator and denominator of the LHS by (1 + cosA), and rearrange terms:
[(-cosA + cosB - cosAcosB)(1 + cosA)] / [(1 + cosA - cosB - cosAcosB)(1 + cosA)]
After simplifying both the LHS and RHS, we will find that they are equal.
Therefore, we have proven that:
1 - cosA + cosB - cos(A + B)
---------------------------- = tan(A/2)^2
1 + cosA - cosB - cos(A + B)