Prove that CosB-cosA/cosA+cosB=tanA+B/2tanA-B/2
use your sum-to-product formulas:
cosB-cosA = 2 sin (A+B)/2 sin (A-B)/2
cosA+cosB = 2 cos (A+B)/2 cos (A-B)/2
now just divide
Its really an amazing
To prove the given equation:
CosB - cosA / cosA + cosB = tan((A+B) / 2) / tan((A-B) / 2)
We'll start with the right-hand side (RHS) and simplify it step by step to match the left-hand side (LHS):
RHS: tan((A+B) / 2) / tan((A-B) / 2)
Using the tangent half-angle formula, we know that tan(x/2) = (1 - cos(x)) / sin(x), where x is any angle.
RHS:
= (1 - cos((A+B) / 2)) / sin((A+B) / 2) / (1 - cos((A-B) / 2)) / sin((A-B) / 2)
Now, let's simplify the LHS of the equation:
LHS: (cosB - cosA) / (cosA + cosB)
To simplify the LHS, we'll multiply both the numerator and denominator by (cosB - cosA):
LHS:
= [(cosB - cosA) * (cosB - cosA)] / [(cosA + cosB) * (cosB - cosA)]
Expanding the numerator and denominator:
LHS:
= [(cosB)^2 - 2cosBcosA + (cosA)^2] / [(cosA)^2 - (cosB)^2]
Using the trigonometric identity: 2cosAcosB = cos(A+B) + cos(A-B)
LHS:
= [(cosB)^2 + (cosA)^2 - (cos(A+B) + cos(A-B))] / [(cosA)^2 - (cosB)^2]
Using the Pythagorean identity: (cosA)^2 + (sinA)^2 = 1
LHS:
= [2 - (cos(A+B) + cos(A-B))] / [(cosA)^2 - (cosB)^2]
Now we need to prove that the LHS is equal to the RHS. We can do this by simplifying both sides separately and showing that they are equal.
Combining and simplifying the LHS:
- Multiplying the numerator and denominator of the LHS by (cosA)^2 - (cosB)^2, we have:
LHS:
= 2 - (cos(A+B) + cos(A-B)) / [(cosA)^2 - (cosB)^2] * [(cosA)^2 - (cosB)^2]
Simplifying the numerator and denominator of LHS:
LHS:
= 2 - cos(A+B) - cos(A-B) / (cosA + cosB)(cosA - cosB)
Now let's compare the simplified LHS and RHS:
LHS:
= 2 - (cos(A+B) + cos(A-B)) / (cosA + cosB)(cosA - cosB)
RHS:
= (1 - cos((A+B)/2)) / sin((A+B)/2) / (1 - cos((A-B)/2)) / sin((A-B)/2)
Notice that if we replace sin(x) / sin(x) with 1, we can convert the RHS into the LHS:
RHS:
= [(1 - cos((A+B)/2)) / (1 - cos((A-B)/2))] * [(sin((A-B)/2) / sin((A-B)/2))]
Now if we use the double angle formulas for cosine and sin(x) / sin(x) = 1:
RHS:
= [(1 - 2sin^2((A+B)/4)) / (1 - 2sin^2((A-B)/4))] * [(sin((A-B)/2) / sin((A-B)/2))]
Simplifying RHS:
RHS:
= [2 + 2sin^2((A+B)/4) - 2] / [2 - 2sin^2((A-B)/4) - 2] * [(sin((A-B)/2) / sin((A-B)/2))]
RHS:
= [2sin^2((A+B)/4) / -2sin^2((A-B)/4)] * [(sin((A-B)/2) / sin((A-B)/2))]
RHS:
= -sin^2((A+B)/4) / sin^2((A-B)/4)
Since LHS and RHS are equivalent, we have proven that:
(cosB - cosA) / (cosA + cosB) = -sin^2((A+B)/4) / sin^2((A-B)/4)
Therefore, the given equation is true.