Half angle identities in trigonometry

If you find the trigonometric expressions having angles in multiples of half, then half angle formulas can be used for reducing them to basic form. This half angle identities are very easy to mug up and keeping them in mind is highly beneficial to you. So lets list out a few important trig half angle formulas and understand the logic behind them along with examples.

Very important trig half angle formulas

The half angle trig identities given below are frequently used in problems and should be mugged up.

  • sin(θ/2) = ± √((1-cosθ)/2).

  • cos(θ/2) = ± √((1+cosθ)/2).

  • tan(θ/2) = cosecθ – cotθ.

  • cot(θ/2) = cosecθ + cotθ.

The trig half angle identities given above provides effective way of solving the equations. Lets derive the first one to understand the logic behind the origin of formula. From the double angle identity, we have cos2θ = 1 – 2sin2θ or 2sin2θ = 1 – cos2θ i.e. sin2θ = (1 – cos2θ)/2. Lets take square-root on both sides. Doing so, we get sinθ = ± √((1 – cos2θ)/2). Putting θ/2 instead of θ, we get the required half angle identity. Similarly other trig identities half angle can be derived. A couple of half angle identities examples will sound perfect at this time.

(1) Derive the half angle formula cos(θ/2) = ± √((1+cosθ)/2):

We know that cos2θ = 2cos2θ – 1 => cos2θ = (1 + cos2θ)/2. Taking the roots on both sides, we get cosθ = ± √((1+cos2θ)/2). Replacing θ with θ/2 yields cos(θ/2) = ± √((1+cosθ)/2). Thus the half angle equation is derived.

(2) Given cosθ = 1/3, find sin(θ/2) in first quadrant:
From the trig formula for half angle, sin(θ/2) = ± √((1-cosθ)/2) = ± √((1 – 1/3)/2) = ± √(1/3) = ± 1/√3. Since θ is an angle in first quadrant, value of sine function has to be positive. Hence the answer is +1/√3.

This is how the half angle identities can be applied to solve various complex trigonometric functions with minimal efforts.

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