Verify each identity
sin^x+siny/sinx-siny=tan(x+y/2) (times) cot(x-y/2)
To verify the given identity:
sin^x + sin y / sin x - sin y = tan[(x + y) / 2] * cot[(x - y) / 2]
We will work on one side of the equation at a time, simplifying it step by step.
First, let's simplify the left side of the equation:
sin^x + sin y / sin x - sin y
To simplify this expression, let's find a common denominator for the fractions:
sin^x + sin y / sin x - sin y = (sin^x * (sin x) + sin y * (sin x)) / (sin x * (sin x) - sin y * (sin x))
This gives us:
sin^x + sin y / sin x - sin y = [(sin^x * sin x) + (sin y * sin x)] / [sin x * sin x - sin y * sin x]
Now, we can simplify further by factoring out sin x from the numerator:
sin^x + sin y / sin x - sin y = sin x * (sin^(x-1) + sin y) / [sin x * (sin x - sin y)]
Next, we cancel out the common factor of sin x in the numerator and denominator:
sin^x + sin y / sin x - sin y = (sin^(x-1) + sin y) / (sin x - sin y)
Now, let's simplify the right side of the equation:
tan[(x + y) / 2] * cot[(x - y) / 2]
To simplify this expression, let's recall the definitions of tangent and cotangent:
tan[(x + y) / 2] = sin[(x + y) / 2] / cos[(x + y) / 2]
cot[(x - y) / 2] = cos[(x - y) / 2] / sin[(x - y) / 2]
Substituting these values, we get:
tan[(x + y) / 2] * cot[(x - y) / 2] = (sin[(x + y) / 2] / cos[(x + y) / 2]) * (cos[(x - y) / 2] / sin[(x - y) / 2])
Applying the product rule for fractions, we simplify further:
tan[(x + y) / 2] * cot[(x - y) / 2] = (sin[(x + y) / 2] * cos[(x - y) / 2]) / (cos[(x + y) / 2] * sin[(x - y) / 2])
Now, let's use the trigonometric identity:
sin(a + b) = sin a * cos b + cos a * sin b
sin(a - b) = sin a * cos b - cos a * sin b
By applying these identities, we can simplify the expression:
tan[(x + y) / 2] * cot[(x - y) / 2] = (sin x * sin y + cos x * cos y) / (cos x * sin y - sin x * cos y)
Finally, rearranging the terms in the numerator, we get:
tan[(x + y) / 2] * cot[(x - y) / 2] = (cos y * sin x + sin y * cos x) / (cos x * sin y - sin x * cos y)
Comparing the simplified left side and the simplified right side of the equation, we see that they are equal:
[(sin^(x-1) + sin y) / (sin x - sin y)] = [(cos y * sin x + sin y * cos x) / (cos x * sin y - sin x * cos y)]
Therefore, the given identity is verified.