sin A/2 (Half-Angle Formula)

sin A/2 (Half-Angle Formula)

How to solve sin A/2 (Half-Angle Formula) problems: formula, proof, example, and its solution.

Formula

sin A/2 = +-sqrt[(1 - cos A) / 2]

sin A/2 = ±√(1 - cos A) / 2

Proof

sin A/2 (Half-Angle Formula): Proof of the Formula

cos (2⋅[A/2]) = 1 - 2 sin2 [A/2]

cos 2A (double-angle formula)

Move cos A to the right side.

And move -2 sin2 [A/2] to the left side.

Divide both sides by 2.

Square root both sides.

Then sin A/2 = ±√(1 - cos A) / 2.

Example

If sin theta = -3/5 and 3pi/2 <= theta <= 2pi, find the value of sin theta/2.

3π/2 ≦ θ ≦ 2π
So θ is in quadrant IV.

sin θ = -3/5
And sine: SOH.

So draw a right triangle in quadrant IV
whose opposite side is -3
and whose hypotenuse is 5.

This right triangle is a (3, 4, 5) right triangle.

So (purple side) = 4.

Pythagorean triples

The brown angle is the reference angle of ∠θ.

And cosine: CAH.

So cos θ = cos (brown)
= 4/5.

Next, find the sign of sin θ/2.

3π/2 ≦ θ ≦ 2π
So 3π/4 ≦ θ/2 ≦ π.

So θ/2 is in quadrant II.

Draw the axes of the coordinate plane
and write 'all, sin, tan, cos' like above.

This shows when the trigonometric function is (+):
for quadrant II, sine is (+).

So sin θ/2 is (+).

cos θ = 4/5
And sin θ/2 is (+).

So sin θ/2 = +√(1 - 4/5) / 2.

Complex fraction

Rationalizing a denominator