Double-Angle & Half-Angle Identities

Q: Which version of $\cos 2\theta$ should I use?

Whichever one matches the value you were handed. All three are equal, so they cannot disagree. Given $\cos\theta$, use $2\cos^2\theta - 1$; given $\sin\theta$, use $1 - 2\sin^2\theta$; given both, $\cos^2\theta - \sin^2\theta$ is the quickest.

Q: How do I decide the sign on a half-angle answer?

Divide the given interval for $\theta$ by $2$, see which quadrant $\dfrac{\theta}{2}$ falls in, and use that quadrant's sign. The quadrant of $\theta$ is not the answer. An angle $\theta$ in Quadrant II has a negative cosine, but $\dfrac{\theta}{2}$ is in Quadrant I, where cosine is positive.

Q: Is $\sin 2\theta$ the same as $2\sin\theta$?

No. Try $\theta = 30^\circ$: $\sin 60^\circ = \dfrac{\sqrt{3}}{2} \approx 0.866$, while $2\sin 30^\circ = 1$. The $2$ is attached to the angle, not to the function value, and $\sin 2\theta = 2\sin\theta\cos\theta$ is what actually connects them.

The double-angle identities are the sum identities talking to themselves. Write $2\theta$ as $\theta + \theta$ , run it through $\sin(A + B)$ , and out falls $\sin 2\theta = 2\sin\theta\cos\theta$ . That is the whole story of where they come from, and it means you do not have to memorize anything genuinely new.

What they buy you is reach. Given one ratio at a single angle, you can find the value at twice that angle or at half of it, without ever knowing the angle itself. Half-angle identities also unlock exact values the special triangles never gave you, like $\sin 22.5^\circ$ .

The double-angle identities

Sine has one form: $\sin 2\theta = 2\sin\theta\cos\theta$ . It needs both the sine and the cosine of $\theta$ , so if a problem hands you only one of them, your first step is always the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to recover the other.

Cosine has three faces, all equal: $\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$ . They give the same number every time, so pick the one that matches what you already know. If you were given a cosine, use $2\cos^2\theta - 1$ and skip the extra work; if you were given a sine, use $1 - 2\sin^2\theta$ .

The half-angle identities and the sign decision

Going the other direction: $\sin\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1 - \cos\theta}{2}}$ and $\cos\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1 + \cos\theta}{2}}$ . Both need only $\cos\theta$ . Notice which sign goes with which function: sine takes the minus inside the fraction, cosine takes the plus.

The square root leaves a $\pm$ , and that is a decision you have to make, not a symbol you get to keep. The sign comes from the quadrant of $\dfrac{\theta}{2}$ , not the quadrant of $\theta$ . Halve the interval first. If $90^\circ < \theta < 180^\circ$ , then $45^\circ < \dfrac{\theta}{2} < 90^\circ$ , so $\dfrac{\theta}{2}$ sits in Quadrant I and both its sine and its cosine are positive, even though $\cos\theta$ itself was negative.

New exact values from old ones

The special triangles stop at $30^\circ$ , $45^\circ$ , and $60^\circ$ . Halving extends that list: $22.5^\circ$ is half of $45^\circ$ , $15^\circ$ is half of $30^\circ$ , and $75^\circ$ is half of $150^\circ$ . Feed the known cosine into the half-angle formula and simplify the nested radical.

Doubling extends it in the other direction and gives a fast check on your work. Since $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$ , the identity $\sin 2\theta = 2\sin\theta\cos\theta$ says $\sin 60^\circ = 2 \cdot \dfrac{1}{2} \cdot \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2}$ , which is the value you already knew. When an identity reproduces a fact you can verify, you are using it correctly.

Worked examples

Example 1: both double-angle values from one ratio

$\sin\theta = \dfrac{3}{5}$ and $0^\circ < \theta < 90^\circ$ . Find $\sin 2\theta$ and $\cos 2\theta$ .

Recover the cosine, positive because

\theta

is acute

\cos\theta = \sqrt{1 - \tfrac{9}{25}} = \dfrac{4}{5}

Double-angle identity for sine

\sin 2\theta = 2 \cdot \dfrac{3}{5} \cdot \dfrac{4}{5} = \dfrac{24}{25}

Double-angle identity for cosine

\cos 2\theta = \cos^2\theta - \sin^2\theta = \dfrac{16}{25} - \dfrac{9}{25}

Answer: $\sin 2\theta = \dfrac{24}{25}$ and $\cos 2\theta = \dfrac{7}{25}$

Example 2: choosing the right form of $\cos 2\theta$

$\cos\theta = \dfrac{5}{13}$ and $0^\circ < \theta < 90^\circ$ . Find $\cos 2\theta$ .

You were given a cosine, so use the cosine-only form

\cos 2\theta = 2\cos^2\theta - 1

Square and double

2\left(\dfrac{5}{13}\right)^2 = \dfrac{50}{169}

Write

1

\dfrac{169}{169}

and subtract

\dfrac{50}{169} - \dfrac{169}{169}

Answer: $\cos 2\theta = -\dfrac{119}{169}$ . The value is negative because $2\theta$ lands past $90^\circ$ .

Example 3: a new exact value

Find the exact value of $\sin 22.5^\circ$ .

22.5^\circ

is half of

45^\circ

, and

22.5^\circ

is in Quadrant I, so take the positive root

\sin 22.5^\circ = \sqrt{\dfrac{1 - \cos 45^\circ}{2}}

Substitute

\cos 45^\circ = \dfrac{\sqrt{2}}{2}

\sqrt{\dfrac{1 - \tfrac{\sqrt{2}}{2}}{2}}

Clear the inner fraction

\sqrt{\dfrac{2 - \sqrt{2}}{4}}

Answer: $\sin 22.5^\circ = \dfrac{\sqrt{2 - \sqrt{2}}}{2}$ , about $0.383$ .

Example 4: the sign comes from $\dfrac{\theta}{2}$

$\cos\theta = -\dfrac{7}{25}$ and $90^\circ < \theta < 180^\circ$ . Find $\cos\dfrac{\theta}{2}$ .

Halve the interval to place

\dfrac{\theta}{2}

45^\circ < \dfrac{\theta}{2} < 90^\circ

That is Quadrant I, so the cosine is positive

\cos\dfrac{\theta}{2} = +\sqrt{\dfrac{1 + \cos\theta}{2}}

Substitute the negative cosine

\sqrt{\dfrac{1 - \tfrac{7}{25}}{2}} = \sqrt{\dfrac{\tfrac{18}{25}}{2}}

Simplify

\sqrt{\dfrac{9}{25}} = \dfrac{3}{5}

Answer: $\cos\dfrac{\theta}{2} = \dfrac{3}{5}$

Try one yourself

Common questions

Which version of

\cos 2\theta

should I use?

Whichever one matches the value you were handed. All three are equal, so they cannot disagree. Given $\cos\theta$ , use $2\cos^2\theta - 1$ ; given $\sin\theta$ , use $1 - 2\sin^2\theta$ ; given both, $\cos^2\theta - \sin^2\theta$ is the quickest.

How do I decide the sign on a half-angle answer?

Divide the given interval for $\theta$ by $2$ , see which quadrant $\dfrac{\theta}{2}$ falls in, and use that quadrant's sign. The quadrant of $\theta$ is not the answer. An angle $\theta$ in Quadrant II has a negative cosine, but $\dfrac{\theta}{2}$ is in Quadrant I, where cosine is positive.

\sin 2\theta

the same as

2\sin\theta

No. Try $\theta = 30^\circ$ : $\sin 60^\circ = \dfrac{\sqrt{3}}{2} \approx 0.866$ , while $2\sin 30^\circ = 1$ . The $2$ is attached to the angle, not to the function value, and $\sin 2\theta = 2\sin\theta\cos\theta$ is what actually connects them.

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Double-Angle & Half-Angle Identities

The double-angle identities

The half-angle identities and the sign decision

New exact values from old ones

Worked examples

Example 1: both double-angle values from one ratio

Example 2: choosing the right form of $\cos 2\theta$

Example 3: a new exact value

Example 4: the sign comes from $\dfrac{\theta}{2}$

Try one yourself

Common questions

Keep learning

Sum & Difference Identities

Trigonometric Identities

Solving Trigonometric Equations

Double-Angle & Half-Angle Identities

The double-angle identities

The half-angle identities and the sign decision

New exact values from old ones

Worked examples

Example 1: both double-angle values from one ratio

Example 2: choosing the right form of cos⁡2θ\cos 2\thetacos2θ

Example 3: a new exact value

Example 4: the sign comes from θ2\dfrac{\theta}{2}2θ​

Try one yourself

Common questions

Keep learning

Sum & Difference Identities

Trigonometric Identities

Solving Trigonometric Equations

Example 2: choosing the right form of $\cos 2\theta$

Example 4: the sign comes from $\dfrac{\theta}{2}$