if sinA+sinB=x and cosA-cosB=y then find tan[(A-B)/2]
I don't know how many trig ratios of half-angles you have developed or know but here is a page where many of them are done.
http://www.intmath.com/analytic-trigonometry/4-half-angle-formulas.php
about 2/3 of the way down you should find
tan [(A+B)/2) = (sinA +sinB)/(cosA + cosB)
= x/y
-y/x
I think you should use the Power Reducing or Half Angle Formula in Trigonometry.
Check out this link for the list of formulas - <a href="www.teachoo.com/9723/1412/Trigonometry-Formulas/category/2-sin-x-sin-y-formula/"> Trigonometry Formulas </a>
To find the value of tan[(A-B)/2], we first need to express sin(A-B) and cos(A-B) in terms of sinA, sinB, cosA, and cosB.
Using the formula for the difference of two sines:
sin(A-B) = sinA*cosB - cosA*sinB
Using the formula for the difference of two cosines:
cos(A-B) = cosA*cosB + sinA*sinB
Now, we can express tan[(A-B)/2] in terms of sinA, sinB, cosA, and cosB.
Using the half-angle formula for tangent:
tan[(A-B)/2] = sin((A-B)/2) / cos((A-B)/2)
Let's calculate sin((A-B)/2):
sin((A-B)/2) = ± sqrt((1 - cos(A-B)) / 2)
Since we have cos(A-B), we can substitute it in the formula:
sin((A-B)/2) = ± sqrt((1 - (cosA*cosB + sinA*sinB)) / 2)
Now, let's calculate cos((A-B)/2):
cos((A-B)/2) = ± sqrt((1 + cos(A-B)) / 2)
Substituting cos(A-B) in the formula:
cos((A-B)/2) = ± sqrt((1 + (cosA*cosB + sinA*sinB)) / 2)
Finally, we can now calculate tan[(A-B)/2]:
tan[(A-B)/2] = sin((A-B)/2) / cos((A-B)/2) = ± sqrt((1 - (cosA*cosB + sinA*sinB)) / (1 + (cosA*cosB + sinA*sinB)))
Thus, the solution for tan[(A-B)/2] in terms of x and y is ± sqrt((1 - x*y) / (1 + x*y)).