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Trigonometric Function Toolkit · Tutorial 142 of 1495

Deriving the Half-Angle Identities

Learn how to derive the half-angle formulas and decide whether each square root represents a positive or negative trigonometric value.

Beginner 8 min read

What You'll Learn

  • Derive the squared half-angle identities from two forms of the cosine double-angle identity
  • Rewrite the identities using an angle divided by two
  • Distinguish the nonnegative principal square root from a signed trigonometric value
  • Determine the signs of sine and cosine from the quadrant of the half-angle
  • Apply the formulas and verify results using the double-angle identity
  • Recognize why a square root without quadrant information may give the wrong answer

Why Half-Angle Formulas Need a Sign

The cosine double-angle identity relates the cosine of an angle to the sine and cosine of half that angle. Rearranging it gives formulas for the squares of those half-angle values. To obtain the values themselves, we take square roots—but a square root alone does not tell us whether the original value was positive or negative.

That distinction is essential. The principal square root is always nonnegative, while sine and cosine can be positive, negative, or zero depending on the angle. The half-angle identities therefore have two parts: an algebraic formula for the square, and a sign choice determined by the quadrant of the half-angle.

Deriving the Squared Identities

The cosine double-angle identity has two useful forms:

$$ \cos(2x)=1-2\sin^2 x \qquad\text{and}\qquad \cos(2x)=2\cos^2 x-1. $$

To isolate \(\sin^2 x\), begin with the first form. Add \(2\sin^2 x\) to both sides and subtract \(\cos(2x)\):

$$ \begin{aligned} \cos(2x)&=1-2\sin^2 x,\\ 2\sin^2 x&=1-\cos(2x),\\ \sin^2 x&=\frac{1-\cos(2x)}{2}. \end{aligned} $$

To isolate \(\cos^2 x\), use the second form. Add \(1\) to both sides and divide by \(2\):

$$ \begin{aligned} \cos(2x)&=2\cos^2 x-1,\\ \cos(2x)+1&=2\cos^2 x,\\ \cos^2 x&=\frac{1+\cos(2x)}{2}. \end{aligned} $$

Now set \(x=\theta/2\), so \(2x=\theta\). This gives the squared half-angle identities.

Theorem: For every real angle \(\theta\), $$ \sin^2\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{2}, \qquad \cos^2\left(\frac{\theta}{2}\right)=\frac{1+\cos\theta}{2}. $$

The identities hold for every real \(\theta\), because they follow by algebraic rearrangement of the double-angle identity, which itself holds for every real angle. The right-hand sides are nonnegative: since \(-1\leq \cos\theta\leq 1\), both \(1-\cos\theta\) and \(1+\cos\theta\) are at least zero.

From Squares to Signed Values

Taking square roots of the squared identities gives the possible magnitudes of the half-angle sine and cosine. The square-root symbol \(\sqrt{\phantom{x}}\) means the nonnegative principal square root. Consequently, the signed formulas need a separate sign choice:

$$ \sin\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1-\cos\theta}{2}}, \qquad \cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1+\cos\theta}{2}}. $$

The sign is not chosen by the value of \(\cos\theta\) alone. It is chosen by the signs of sine and cosine at \(\theta/2\). In particular, locate the half-angle in a quadrant, using any information given about \(\theta\). If the angle is outside a standard \(0\) to \(2\pi\) cycle, reduce \(\theta/2\) by a whole-number multiple of \(2\pi\) to identify its position on the unit circle. Such a reduction does not change its sine or cosine.

Quadrant containing \(\theta/2\)Sign of \(\sin(\theta/2)\)Sign of \(\cos(\theta/2)\)
IPositivePositive
IIPositiveNegative
IIINegativeNegative
IVNegativePositive

If \(\theta/2\) lies on an axis rather than in a quadrant, one or both values may be zero. The squared identities still apply; the sign choice is then unambiguous because the corresponding square root is zero.

A Complete Justification for the Sign Rule

The squared identities determine the absolute values of the half-angle sine and cosine. For any real number \(a\), the nonnegative square root of \(a^2\) is \(|a|\), not necessarily \(a\). Thus:

$$ \sqrt{\sin^2(\theta/2)}=\left|\sin(\theta/2)\right|, \qquad \sqrt{\cos^2(\theta/2)}=\left|\cos(\theta/2)\right|. $$

The quadrant tells us whether each original value equals its absolute value or the negative of its absolute value. This gives the following procedure.

1
Find the squared value.
Use the appropriate half-angle identity and simplify the right-hand side.
2
Take the nonnegative square root.
This gives the magnitude of the sine or cosine, not yet its signed value.
3
Locate the half-angle.
Use the given interval or angle information to determine the quadrant of \(\theta/2\).
4
Apply the quadrant sign and check.
Choose the sign for each function separately, then verify by squaring or using the double-angle identity.

Worked Examples

Worked Example: Both Half-Angle Values in Quadrant I

Suppose \(\cos\theta=5/13\) and \(0<\theta<\pi\). Find \(\sin(\theta/2)\) and \(\cos(\theta/2)\).

First divide the interval for \(\theta\) by \(2\). It follows that \(0<\theta/2<\pi/2\), so \(\theta/2\) is in Quadrant I. Both its sine and cosine are positive.

Use the sine identity:

$$ \begin{aligned} \sin^2(\theta/2) &=\frac{1-\cos\theta}{2}\\ &=\frac{1-5/13}{2}\\ &=\frac{8/13}{2}\\ &=\frac{4}{13}. \end{aligned} $$

The nonnegative square root is \(2/\sqrt{13}\), and the Quadrant I sign is positive. Therefore \(\sin(\theta/2)=2/\sqrt{13}\). For cosine:

$$ \begin{aligned} \cos^2(\theta/2) &=\frac{1+\cos\theta}{2}\\ &=\frac{1+5/13}{2}\\ &=\frac{18/13}{2}\\ &=\frac{9}{13}. \end{aligned} $$

The nonnegative square root is \(3/\sqrt{13}\), also positive in Quadrant I. Thus \(\cos(\theta/2)=3/\sqrt{13}\).

Check with the cosine double-angle identity:

$$ \cos\theta =2\cos^2(\theta/2)-1 =2\left(\frac{9}{13}\right)-1 =\frac{18}{13}-\frac{13}{13} =\frac{5}{13}. $$

This reproduces the given cosine, and both signs agree with the stated interval.

Worked Example: A Negative Cosine in Quadrant II

Suppose \(\cos\theta=-3/5\) and \(\pi<\theta<3\pi/2\). Find \(\sin(\theta/2)\) and \(\cos(\theta/2)\).

Dividing the interval by \(2\) gives \(\pi/2<\theta/2<3\pi/4\). This places \(\theta/2\) in Quadrant II, where sine is positive and cosine is negative.

For the sine:

$$ \begin{aligned} \sin^2(\theta/2) &=\frac{1-\cos\theta}{2}\\ &=\frac{1-(-3/5)}{2}\\ &=\frac{8/5}{2}\\ &=\frac{4}{5}. \end{aligned} $$

The magnitude is \(\sqrt{4/5}=2/\sqrt{5}\). Sine is positive in Quadrant II, so \(\sin(\theta/2)=2/\sqrt{5}\). For the cosine:

$$ \begin{aligned} \cos^2(\theta/2) &=\frac{1+\cos\theta}{2}\\ &=\frac{1-3/5}{2}\\ &=\frac{2/5}{2}\\ &=\frac{1}{5}. \end{aligned} $$

The magnitude is \(1/\sqrt{5}\). Cosine is negative in Quadrant II, so \(\cos(\theta/2)=-1/\sqrt{5}\).

Check both signs and the given cosine by computing:

$$ \cos\theta =\cos^2(\theta/2)-\sin^2(\theta/2) =\frac{1}{5}-\frac{4}{5} =-\frac{3}{5}. $$

The result matches the given value, and the signs place the half-angle in Quadrant II.

Worked Example: A Half-Angle in Quadrant III

Suppose \(\cos\theta=3/5\) and \(2\pi<\theta<3\pi\). Find \(\sin(\theta/2)\) and \(\cos(\theta/2)\).

Here \(\pi<\theta/2<3\pi/2\), so the half-angle is in Quadrant III. Both its sine and cosine are negative. In particular, the positive value of \(\cos\theta\) does not imply that \(\cos(\theta/2)\) is positive.

For the sine:

$$ \begin{aligned} \sin^2(\theta/2) &=\frac{1-\cos\theta}{2}\\ &=\frac{1-3/5}{2}\\ &=\frac{2/5}{2}\\ &=\frac{1}{5}. \end{aligned} $$

Its magnitude is \(1/\sqrt{5}\), and its Quadrant III sign is negative. Hence \(\sin(\theta/2)=-1/\sqrt{5}\). For cosine:

$$ \begin{aligned} \cos^2(\theta/2) &=\frac{1+\cos\theta}{2}\\ &=\frac{1+3/5}{2}\\ &=\frac{8/5}{2}\\ &=\frac{4}{5}. \end{aligned} $$

Its magnitude is \(2/\sqrt{5}\), and its Quadrant III sign is negative. Hence \(\cos(\theta/2)=-2/\sqrt{5}\).

As a check:

$$ \cos\theta =\cos^2(\theta/2)-\sin^2(\theta/2) =\frac{4}{5}-\frac{1}{5} =\frac{3}{5}. $$

The computed cosine agrees with the given value, and both negative signs are consistent with Quadrant III.

Common Pitfall: Treating the Square Root as the Answer

A frequent error is to take the positive square root and assume it is automatically the value of the trigonometric function. For example, in the second worked example, the cosine identity gives:

$$ \cos^2(\theta/2)=\frac{1}{5}. $$

The calculation \(\cos(\theta/2)=1/\sqrt{5}\) is wrong for that example: the half-angle is in Quadrant II, where cosine is negative. The correction is to use the principal square root only for the magnitude, then apply the quadrant sign:

$$ \left|\cos(\theta/2)\right|=\frac{1}{\sqrt{5}}, \qquad \cos(\theta/2)=-\frac{1}{\sqrt{5}}. $$

A second related mistake is to try to decide the half-angle sign from \(\cos\theta\) alone. The given cosine helps evaluate the squared identity, but the interval or other angle information is what locates \(\theta/2\) and determines its signs.

Why These Identities Are Useful

The identities convert a trigonometric value at \(\theta\) into the squares of sine and cosine at \(\theta/2\). That makes them useful whenever a problem supplies information about \(\cos\theta\) but asks about half the angle. The formulas also illustrate a general point in algebra: an equation for a square determines a magnitude, while additional information is needed to recover the sign.

Keep the two tasks separate. First calculate the squared value accurately. Then use the location of the half-angle to choose the sign. If either task is skipped, a numerically plausible square root may still be the wrong trigonometric value.

Check Your Understanding

For each question, distinguish the squared identity from the sign information needed to obtain a signed value.

  1. Derive \(\sin^2(\theta/2)=(1-\cos\theta)/2\) from the cosine double-angle identity.
  2. If \(\cos\theta=-1/3\) and \(0<\theta<\pi\), which signs should \(\sin(\theta/2)\) and \(\cos(\theta/2)\) have? Explain using the half-angle interval.
  3. Why is \(\sqrt{\cos^2(\theta/2)}\) equal to \(\left|\cos(\theta/2)\right|\), rather than always equal to \(\cos(\theta/2)\)?
  4. If \(\theta/2\) is in Quadrant IV, which sign is used for each square root in the signed half-angle formulas?
  5. What information, beyond the value of \(\cos\theta\), may be needed to select the signs of the half-angle values?

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