5.3. Double-Angle and Half-Angle Formulas
Two very useful types of formulas follow easily from the trigonometric sum formulas, namely double-angle and half-angle formulas. If we apply the sine sum formula (5.2.2) with φ = θ, we get sin 2θ = sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ. Thus we have derived the following double-angle formula for sine:
- (5.3.1) sin 2θ = 2 sin θ cos θ
Similarly, we may derive the following double-angle formula for cosine:
- (5.3.2) cos 2θ = cos2θ - sin2θ
Actually, this formula has two additiona useful equivalent forms, which are easily derived using (5.1.1). Specifically we have
- (5.3.3) cos 2θ = 2 cos2θ - 1 = cos2θ - sin2θ = 1 - 2 sin2θ
Finally we have the double-angle formulas for the tangent and arctangent, namely
- (5.3.4) tan 2θ = 2 tan θ / (1 - tan2θ)
- (5.3.5) 2 arctan x = arctan(2x / (1 - x2))
All these double-angle formulas may be easily verified for known specific cases, as the following example shows:
Example 1: Verify formulas (5.3.1), (5.3.2) and (5.3.4) for θ = π/6.
Solution: From (5.3.1) we have sin(π/3) = 2 sin(π/6) cos(π/6). Now the left side is equal to √3/2 and the right side is equal to 2(1/2)(√3/2) = √3/2. Thus we have verified the double-angle formula for sine for θ = π/6. To verify (5.3.2) we check whether cos(π/3) = cos2(π/6) - sin2(π/6). The left side is 1/2 and the right side is (√3/2)2 - (1/2)2 = 3/4 - 1/4 = 1/2, verifying this formula. Finally, we check whether tan(π/3) = 2 tan(π/6) / (1 - tan2(π/6)). The left side is equal to √3 and the right side is equal to 2(√3/3) / (1 - 3/9) = (2√3/3) / (2/3) = (2√3) / 2 = √3, verifying (5.3.4) for this case.
Perhaps even more interesting than the double-angle formulas are the half-angle formulas, which allow us to compute trigonometric functioins of θ/2 from trigonometric functions of θ. One reason these formulas are of special interest is that they allow us to compute values of trigonometric functions for arbitrarily small arguments, simply by applying them as many times as desired. It was through the clever use of half-angle formulas, for instance, that the well-renowned Greek mathematician Archimedes was able to compute the value of π to three decimal places, an amazing feat for his day.
How do we derive half-angle formulas? The trick is to apply double-angle formulas with θ replaced by θ/2. Thus, for instance, Equation (5.3.1) implies sin θ = 2 sin(θ/2) cos(θ/2). Squaring both sides, we obtain
sin2θ = 4 sin2(θ/2) cos2(θ/2)
= 4 sin2(θ/2) [1 - sin2(θ/2)]
= 4 sin2(θ/2) - 4 sin4(θ/2)
Letting x = sin2(θ/2), this equation becomes sin2θ = 4x - 4x2, or equivalently, 4x2 - 4x + sin2θ = 0. But this is just a quadratic equation in x, which we can solve by means of the quadratic formula with a = 4, b = -4, and c = sin2θ. We have x =(-b±√D) / 2a, where D = b2-4ac = 16 - 16 sin2θ = 16 cos2θ. Thus we have x = (4 ± 4 cos θ) / 8 = (1 ± cos θ) / 2, whence
sin(θ/2) = ±√x = ±√(1 ± cos θ) / 2.
What signs should we take? To determine the appropriate signs, let us assume that θ is between 0 and π/2. This will imply that sin(θ/2) is positive and less than sin(π/4) = 1/√2. This means we must take the sign outside the radical to be pus and the sign inside to be minus. Thus we have derived the half-angle identity for sine, namely
- (5.3.6) sin(θ/2) = √(1 - cos θ) / 2
A very similar derivation leads to the following half-angle formula for cosine. We leave the proof as an exercise.
- (5.3.7) cos(θ/2) = √(1 + cos θ) / 2
Finally, Equations (5.3.6) and (5.3.7) together lead to the following half-angle formula for tangent:
- (5.3.8) tan(θ/2) = √(1 - cos θ) / (1 + cos θ)
Example 2: Use Equations (5.3.6) - (5.3.8) to compute the following:
- (a) sin π/8
- (b) cos π/8
- (c) tan π/8
Solution: From (5.3.6) we see that
Similarly, from (5.3.7) we have
Finally we have