prove the identity
sin2x-sin2y/sin2x+sin2y=tan(x-y)/tan(x+y)
This one is easier if you work on the right side.
(sin(x-y)/cos(x-y)) / (sin(x+y)/cos(x+y))
(sinxcosy-cosxsiny)/(cosxcosy+sinxsiny)
-------------------------------------
(sinxcosy+cosxsiny)/(cosxcosy-sinxsiny)
Expand those products, and you will be able to factor out some (sin^2+cos^2) and watch it become just like the left side.
Working on the left side first requires you to insert some canceling factors which are not obvious.
To prove the given identity:
Step 1: Rewrite the left side of the equation:
sin^2x - sin^2y
----------------- (1)
sin^2x + sin^2y
Step 2: Use the identity sin^2θ + cos^2θ = 1:
1) sin^2x = 1 - cos^2x
2) sin^2y = 1 - cos^2y
Rewriting equation (1) using the above identities and rearranging the terms:
(1 - cos^2x) - (1 - cos^2y)
-------------------------------
(1 - cos^2x) + (1 - cos^2y)
Step 3: Simplify the equation by removing the brackets:
- cos^2x + cos^2y
-------------------
- cos^2x + cos^2y
Step 4: Divide both the numerator and denominator by cos^2x + cos^2y:
(- cos^2x + cos^2y) / (cos^2x + cos^2y)
Step 5: The numerator and denominator are equal, so cancel out the common terms:
-1
Therefore, the left side of the equation simplifies to -1.
Step 6: Rewrite the right side of the equation:
tan(x-y)
---------
tan(x+y)
Step 7: Apply the identity:
tan(A-B) = (tanA - tanB) / (1 + tanA * tanB)
Using this identity, we can rewrite the right side of the equation as:
(tanx - tany) / (1 + tanx * tany)
Step 8: Divide the numerator and denominator by tany:
[(tanx / tany) - 1] / [(tanx / tany) + 1]
Step 9: Since tanx = sinx / cosx and tany = siny / cosy, we can substitute these values back into the equation:
[(sinx / cosx) / (siny / cosy) - 1] / [(sinx / cosx) / (siny / cosy) + 1]
Step 10: Multiply both the numerator and denominator by cosy * cosx:
[(sinx * cosy) / (cosx * siny) - cosx * cosy] / [(sinx * cosy) / (cosx * siny) + cosx * cosy]
Step 11: Multiply the numerator and denominator by siny:
[(sinx * cosy) - (cosx * siny)] / [(sinx * cosy) + (cosx * siny)]
Step 12: Apply the identity sin(A - B) = sinA * cosB - cosA * sinB:
sin(x-y) / sin(x+y)
Since sin(x-y) / sin(x+y) equals -1, the right side of the equation also equals -1.
Therefore, we have proven that:
(sin^2x - sin^2y) / (sin^2x + sin^2y) = tan(x-y) / tan(x+y)
To prove the given identity:
Start with the left-hand side (LHS):
LHS = (sin^2x - sin^2y) / (sin^2x + sin^2y)
Using the difference of squares identity:
LHS = [(sinx + siny)(sinx - siny)] / [(sin^2x + 2sinxcosx + sin^2y) + (sin^2x - 2sinycosy)]
Simplifying the denominator:
LHS = [(sinx + siny)(sinx - siny)] / [2sin^2x + 2sin^2y + 2sinxcosx - 2sinycosy] --(1)
Now let's move on to the right-hand side (RHS):
RHS = tan(x - y) / tan(x + y)
Using the trigonometric identity tan(A - B) = (tanA - tanB) / (1 + tanA * tanB), we have:
RHS = [(tanx - tany) / (1 + tanx * tany)] / [(tanx + tany) / (1 - tanx * tany)]
Simplifying the RHS:
RHS = [(tanx - tany) / (1 + tanx * tany)] * [(1 - tanx * tany) / (tanx + tany)]
Expanding and simplifying the numerator:
RHS = [(tanx - tany)(1 - tanx * tany)] / (1 - (tanx * tany)^2)
Using the identity 1 - (tanA)^2 = (secA)^2:
RHS = [(tanx - tany)(1 - tanx * tany)] / (sec(x+y))^2
Combining both denominators, we have:
RHS = [(tanx - tany)(1 - tanx * tany)] / (sec(x+y))^2 * (1 + tanx * tany) / (1 + tanx * tany)
Simplifying further:
RHS = [(tanx - tany)(1 - tanx * tany)] / [(1 + tanx * tany) * (sec(x+y))^2]
Applying the identity sec^2(A) = 1 + tan^2(A):
RHS = [(tanx - tany)(1 - tanx * tany)] / [(1 + tanx * tany) * (1 + tan(x+y))^2] --(2)
Comparing (1) and (2), we can see that the LHS is equal to the RHS:
LHS = [(sinx + siny)(sinx - siny)] / [2sin^2x + 2sin^2y + 2sinxcosx - 2sinycosy]
RHS = [(tanx - tany)(1 - tanx * tany)] / [(1 + tanx * tany) * (1 + tan(x+y))^2]
Therefore, we have successfully proved the given identity:
(sin^2x - sin^2y) / (sin^2x + sin^2y) = tan(x - y) / tan(x + y)