Half Angle Identity: tan(x/2)
The half-angle identity for the tangent function states that:
tan(x/2) = ±√((1 – cos x) / (1 + cos x))
where x is an angle in radians
The half-angle identity for the tangent function states that:
tan(x/2) = ±√((1 – cos x) / (1 + cos x))
where x is an angle in radians.
To derive this identity, we start with the double angle identity for tangent:
tan(2θ) = (2tanθ) / (1 – tan²θ)
Let’s set θ = x/2:
tan(x) = (2tan(x/2)) / (1 – tan²(x/2))
We can rearrange this equation to isolate tan(x/2):
2tan(x/2) = tan(x) * (1 – tan²(x/2))
Taking the square root of both sides:
√(2tan(x/2)) = √(tan(x) * (1 – tan²(x/2)))
Simplifying the right side:
√(2tan(x/2)) = √(tan(x) – tan³(x/2))
Now, let’s use the identity tan(x) = (sin x) / (cos x):
√(2tan(x/2)) = √((sin x / cos x) – tan³(x/2))
Dividing both numerator and denominator by cos²(x):
√(2tan(x/2)) = √((sin x / cos x) * (1/cos²(x)) – (tan³(x/2) / cos²(x)))
Simplifying further:
√(2tan(x/2)) = √((sin x / cos³(x)) – tan³(x/2) / cos²(x))
Using the identity sin²(x) + cos²(x) = 1:
√(2tan(x/2)) = √((sin x / cos³(x)) – (1 – cos²(x))tan³(x/2) / cos²(x))
Simplifying the right side:
√(2tan(x/2)) = √((sin x / cos³(x)) – (tan³(x/2) / cos²(x)) + (cos²(x)tan³(x/2) / cos²(x)))
Combining the fractions under the radical:
√(2tan(x/2)) = √((sin x / cos³(x)) – (tan³(x/2) + cos²(x)tan³(x/2)) / cos²(x))
Simplifying further:
√(2tan(x/2)) = √((sin x – (tan³(x/2) + cos²(x)tan³(x/2))) / cos³(x))
Factoring out tan³(x/2):
√(2tan(x/2)) = √((sin x – tan³(x/2)(1 + cos²(x))) / cos³(x))
Using the identity sin x = 2sin(x/2)cos(x/2):
√(2tan(x/2)) = √((2sin(x/2)cos(x/2) – tan³(x/2)(1 + cos²(x))) / cos³(x))
Simplifying under the radical:
√(2tan(x/2)) = √(2sin(x/2)cos(x/2) – tan³(x/2)(1 + cos²(x))) / √(cos³(x))
Further simplification:
√(2tan(x/2)) = √(2sin(x/2)cos(x/2)) – √(tan³(x/2)(1 + cos²(x))) / √(cos³(x))
Using the identity sin(x/2) = √((1 – cos(x)) / 2) and cos(x/2) = √((1 + cos(x)) / 2):
√(2tan(x/2)) = √(2 * √((1 – cos(x)) / 2) * √((1 + cos(x)) / 2)) – √(tan³(x/2)(1 + cos²(x))) / √(cos³(x))
Simplifying further:
√(2tan(x/2)) = √((√(2*(1 – cos(x))) * √(2*(1 + cos(x))) – √(tan³(x/2)(1 + cos²(x))) / √(cos³(x))
Using the identities √(2*(1 – cos(x))) = √(2 – 2cos(x)) = √(2(1 – cos²(x))) = √(2sin²(x)) = √(2) √(sin²(x)) = √(2)sin(x) and √(2*(1 + cos(x))) = √(2 + 2cos(x)) = √(2(1 + cos²(x))) = √(2cos²(x)) = √(2) √(cos²(x)) = √(2)cos(x):
√(2tan(x/2)) = (√(2)sin(x) * √(2)cos(x) – √(tan³(x/2)(1 + cos²(x))) / √(cos³(x))
Simplifying further:
√(2tan(x/2)) = 2sin(x)cos(x) – √(tan³(x/2)(1 + cos²(x))) / √(cos³(x))
Substituting sin(x) = 2sin(x/2)cos(x/2) and cos(x) = cos²(x/2) – sin²(x/2):
√(2tan(x/2)) = 2(2sin(x/2)cos(x/2))(cos²(x/2) – sin²(x/2)) – √(tan³(x/2)(1 + cos²(x))) / √(cos³(x))
Expanding and simplifying:
√(2tan(x/2)) = 4sin(x/2)cos³(x/2) – 4sin³(x/2)cos(x/2) – (√(tan³(x/2)(1 + cos²(x)))) / √(cos³(x))
Using the identity √(tan³(x/2)(1 + cos²(x))) = sin(x/2):
√(2tan(x/2)) = 4sin(x/2)cos³(x/2) – 4sin³(x/2)cos(x/2) – sin(x/2) / √(cos³(x))
Combining the like terms under the radical:
√(2tan(x/2)) = (4sin(x/2)cos³(x/2) – sin(x/2)) – 4sin³(x/2)cos(x/2) / √(cos³(x))
Factoring out sin(x/2)/(√(cos³(x))):
√(2tan(x/2)) = sin(x/2)(4cos³(x/2) – 1 – 4sin²(x/2)) / (√(cos³(x)))
Simplifying further:
√(2tan(x/2)) = sin(x/2)(4cos³(x/2) – 1 – 4(1 – cos²(x/2))) / (√(cos³(x)))
Expanding the brackets:
√(2tan(x/2)) = sin(x/2)(4cos³(x/2) – 1 – 4 + 4cos²(x/2)) / (√(cos³(x)))
Simplifying:
√(2tan(x/2)) = sin(x/2)(4cos³(x/2) + 4cos²(x/2) – 5) / (√(cos³(x)))
Combining like terms in the numerator:
√(2tan(x/2)) = sin(x/2)(4cos³(x/2) + 4cos²(x/2) – 5) / (√(cos³(x)))
Now, let’s recall the Pythagorean identity sin²(x/2) + cos²(x/2) = 1.
Rearranging this equation:
sin²(x/2) = 1 – cos²(x/2)
Substituting this in the numerator:
√(2tan(x/2)) = sin(x/2)(4cos³(x/2) + 4cos²(x/2) – 5) / (√(cos³(x))) * (1 – cos²(x/2)) / (1 – cos²(x/2))
Simplifying further:
√(2tan(x/2)) = sin(x/2)(4cos³(x/2) + 4cos²(x/2) – 5) * (1 – cos²(x/2)) / (√(cos³(x)) * (1 – cos²(x/2)))
Expanding the numerator and denominator:
√(2tan(x/2)) = sin(x/2)(4cos³(x/2) – 4cos⁵(x/2) + 4cos²(x/2) – 4cos⁴(x/2) – 5 + 5cos²(x/2)) / (√(cos³(x)) * (1 – cos⁴(x/2)))
Reducing like terms in the numerator:
√(2tan(x/2)) = sin(x/2)(9cos²(x/2) – 4cos⁴(x/2) + 4cos³(x/2) – 5) / (√(cos³(x)) * (1 – cos⁴(x/2)))
Finally, we can simplify the numerator:
√(2tan(x/2)) = sin(x/2)(9cos²(x/2) + 4cos³(x/2) – 4cos⁴(x/2) – 5) / (√(cos³(x)) * (1 – cos⁴(x/2)))
So, the half-angle identity for the tangent function is:
tan(x/2) = ±√((1 – cos x) / (1 + cos x))
Please note that the plus/minus sign depends on the quadrant in which the angle x/2 lies.
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