DWG NO. 5.3 — Lesson 3 of 5

Double-Angle and Half-Angle Formulas

Unit 5: Trigonometric Identities and Equations · ~25 min

Objective Derive the double-angle and half-angle formulas from the sum formulas, and use them to find exact values and simplify expressions.

Both formula families in this lesson are special cases of the sum and difference formulas from 5.2 — nothing new is being assumed, just A and B collapsing into a single angle. Setting B = A in the sum formulas gives the double-angle formulas; solving those same formulas backward for \(\frac{\theta}{2}\) gives the half-angle formulas. Keeping that connection in mind is more useful than memorizing four formulas as unrelated facts.

Double-angle formulas

Substitute B = A into sin(A + B) and cos(A + B):

sin2θ = 2 sinθ cosθ
cos2θ = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1
tan2θ = \(\frac{2tan\theta}{1 - tan^{2}\theta}\)

cos2θ has three equivalent forms because sin²θ + cos²θ = 1 lets you swap between them — pick whichever form matches what's given in a problem (an expression in sine only, cosine only, or mixed).

Half-angle formulas

Solving the cos2θ forms for sinθ and cosθ, then replacing θ with \(\frac{\theta}{2,}\) gives:

sin(\(\frac{\theta}{2}\)) = ±\(\sqrt{\frac{1 - cos\theta}{2}}\)
cos(\(\frac{\theta}{2}\)) = ±\(\sqrt{\frac{1 + cos\theta}{2}}\)
tan(\(\frac{\theta}{2}\)) = \(\frac{sin\theta}{1 + cos\theta}\) = \(\frac{1 - cos\theta}{sin\theta}\)

The ± sign isn't optional decoration — you have to determine it yourself, and it depends on which quadrant \(\frac{\theta}{2}\) lands in, not which quadrant θ is in.

Angle θ split into two equal halves — the half-angle formulas find sin and cos of just one of those halves

Worked Example 1 · Double-angle from a given ratio

ProblemGiven sinθ = \(\frac{3}{5,}\) with θ in Quadrant I, find sin2θ and cos2θ.

1Find cosθ: cos²θ = 1 − \(\frac{9}{25}\) = \(\frac{16}{25,}\) so cosθ = \(\frac{4}{5}\) (positive in QI).

2sin2θ = 2sinθcosθ = 2(\(\frac{3}{5}\))(\(\frac{4}{5}\)) = \(\frac{24}{25.}\)

3cos2θ = 1 − 2sin²θ = 1 − 2(\(\frac{9}{25}\)) = 1 − \(\frac{18}{25}\) = \(\frac{7}{25.}\)

sin2θ = \(\frac{24}{25}\) and cos2θ = \(\frac{7}{25}\)

Worked Example 2 · Exact value with a half-angle formula

ProblemFind the exact value of cos15° by treating 15° as \(\frac{30^\circ}{2.}\)

1cos(\(\frac{30^\circ}{2}\)) = \(\sqrt{\frac{1 + \cos 30^\circ}{2}}\) = \(\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}\).

2Simplify inside the radical: \(\frac{1 + \frac{\sqrt{3}}{2}}{2}\) = \(\frac{2 + \sqrt{3}}{4}\).

315° is in Quadrant I, so cosine is positive — keep the + sign.

cos15° = \(\frac{\sqrt{2 + \sqrt{3}}}{2}\)

Worked Example 3 · Half-angle with a sign check

ProblemGiven cosθ = \(\frac{-1}{3,}\) with θ in Quadrant II, find sin(\(\frac{\theta}{2}\)).

1θ between 90° and 180° means \(\frac{\theta}{2}\) is between 45° and 90° — Quadrant I, where sine is positive.

2sin(\(\frac{\theta}{2}\)) = \(\sqrt{\frac{1 - \cos\theta}{2}}\) = \(\sqrt{\frac{1 + \frac{1}{3}}{2}}\) = \(\sqrt{\frac{\frac{4}{3}}{2}}\) = \(\sqrt{\frac{2}{3}}\).

3Rationalize: \(\frac{\sqrt{\frac{2}{3}}\) = \(\sqrt{6}}{3.}\)

sin(\(\frac{\theta}{2}\)) = \(\frac{\sqrt{6}}{3}\)

Guided practice

1.Given cosθ = \(\frac{7}{25,}\) with θ in Quadrant I, find sin2θ.
Show answer

sinθ = \(\frac{24}{25.}\) sin2θ = 2(\(\frac{24}{25}\))(\(\frac{7}{25}\)) = \(\frac{336}{625}\).
2.Given cosθ = \(\frac{1}{4,}\) find cos2θ.
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cos2θ = 2cos²θ − 1 = 2(\(\frac{1}{16}\)) − 1 = \(\frac{1}{8}\) − 1 = \(\frac{-7}{8}\).
3.Given sinθ = \(\frac{5}{13}\) and cosθ = \(\frac{12}{13}\) (θ in QI), find tan(\(\frac{\theta}{2}\)).
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tan(\(\frac{\theta}{2}\)) = \(\frac{sin\theta}{1+cos\theta}\) = \(\frac{\frac{5}{13}}{1+\frac{12}{13}}\) = \(\frac{\frac{5}{13}}{\frac{25}{13}}\) = \(\frac{1}{5}\).
4.Simplify 2sin(x)cos(x).
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This is exactly the double-angle formula for sine: sin2x.
5.Explain why the ± sign in the half-angle formulas is decided by the quadrant of \(\frac{\theta}{2,}\) not the quadrant of θ.
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The formula is computing sin or cos of the angle \(\frac{\theta}{2}\) itself — its own sign depends on where that specific angle lands. Knowing θ's quadrant only tells you a range for \(\frac{\theta}{2,}\) which you then have to check against the standard quadrant sign rules.

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