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Double Angle and Half Angle Identities

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Double Angle and Half Angle Identities - Complete Interactive Lesson

Part 1: Where the Double-Angle Formulas Come From

🔁 Double Angle & Half Angle Identities

Part 1 of 7 — Where the Double-Angle Formulas Come From

Topics in This Part

Section
The Sum Formulas We Build On
Deriving $\sin(2\theta)$ and $\cos(2\theta)$
The Three Faces of $\cos(2\theta)$

🔑 Key Concept: A "double angle" formula rewrites $\sin(2\theta)$ , $\cos(2\theta)$ , or $\tan(2\theta)$ in terms of $\sin\theta$ , , and alone. Every one of them is just an .

We Start From the Sum Formulas

You already know the angle-sum identities:

$\sin(A + B) = \sin A\cos B + \cos A\sin B$ $\cos (A +$

Concept Check 🎯

The Three Faces of $\cos(2\theta)$

The cosine double angle is special: using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ , you can write it .

Pick the Right Face 🔽

Choose the most efficient form of $\cos(2\theta)$ for each situation.

Compute a Double Angle 🧮

Suppose $\sin\theta = \tfrac{3}{5}$ and $\cos\theta = \tfrac{4}{5}$ .

What You've Got So Far

$\boxed{\sin(2\theta) = 2\sin\theta\cos\theta}$ $\cos ($

Part 2: Tangent & Evaluating From a Given Ratio

🔁 Double Angle & Half Angle Identities

Part 2 of 7 — Tangent & Evaluating From a Given Ratio

🔑 The Idea: If you know $\sin\theta$ and $\cos\theta$ (or can find them from one), you can compute $\sin(2\theta)$ , $\cos(2\theta)$ , and exactly — no calculator, no decimals.

Part 3: Reading Double Angles Backwards (Recognition)

🔁 Double Angle & Half Angle Identities

Part 3 of 7 — Reading Double Angles Backwards (Recognition)

🔑 The Skill: The exam rarely says "use the double-angle formula." Instead it shows you an expression like $2\sin(3x)\cos(3x)$ and expects you to recognize it as $\sin(6x)$ . This is the reverse direction, and it is worth easy points.

Pattern Recognition

Read each formula right-to-left. The angle inside doubles.

Part 4: The Half-Angle Formulas

🔁 Double Angle & Half Angle Identities

Part 4 of 7 — The Half-Angle Formulas

Topics in This Part

Section
Solving $\cos(2\theta)$ for $\sin^2$ and $\cos^2$

Part 5: Power-Reduction Formulas

🔁 Double Angle & Half Angle Identities

Part 5 of 7 — Power-Reduction Formulas

🔑 The Idea: The same algebra that built the half-angle formulas gives the power-reduction formulas, which trade a squared trig function for a first-power one. This is the single most important identity for integrating in Calculus BC.

The Power-Reduction Formulas

Stopping one step before the square root in Part 4's derivation leaves:

$\boxed{\sin^2\theta = \frac{1 - \cos(2\theta)}{2}}$

Part 6: Solving Equations & Real Applications

🔁 Double Angle & Half Angle Identities

Part 6 of 7 — Solving Equations & Real Applications

🔑 The Idea: When an equation mixes $\sin(2x)$ with $\sin x$ (or $\cos(2x)$ with $\cos x$ ), you can't combine them — until you use a double-angle identity to make . Then it factors.

Part 7: Mixed Mastery & Exit Quiz

🔁 Double Angle & Half Angle Identities

Part 7 of 7 — Mixed Mastery & Exit Quiz

You can now (1) derive and apply the double-angle formulas, (2) recognize them backwards, (3) use the half-angle formulas with the right sign, (4) reduce powers, and (5) solve equations. Let's pull it all together.

Master Reference Card

Identity	Formula
$\sin(2\theta)$	$2\sin\theta\cos\theta$

angle-sum formula with both angles equal

cos (A + B) = cos A cos B - sin A sin B

A double angle is just the special case $A = B = \theta$ , because $\theta + \theta = 2\theta$ .

Derivation of $\sin(2\theta)$

$\sin(2\theta) = \sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta$

Derivation of $\cos(2\theta)$

$\cos(2\theta) = \cos(\theta + \theta) = \cos\theta\cos\theta - \sin\theta\sin\theta = \cos^2\theta - \sin^2\theta$

🔑 Take-away: You never have to memorize these in isolation. If you know the sum formulas, you can rebuild every double-angle identity in two lines.

three equivalent ways

Form	How to get it
$\cos(2\theta) = \cos^2\theta - \sin^2\theta$	the direct derivation
$\cos(2\theta) = 2\cos^2\theta - 1$	replace $\sin^2\theta$ with
$\cos(2\theta) = 1 - 2\sin^2\theta$	replace $\cos^2\theta$ with

💡 Why three forms? You pick whichever one matches what you're given. Know only $\cos\theta$ ? Use $2\cos^2\theta - 1$ . Know only $\sin\theta$ ? Use $1 - 2\sin^2\theta$ . This choice saves enormous time on the exam.

1) $\sin(2\theta) = 2\sin\theta\cos\theta = \,?$ (enter as a fraction) 2) Using $\cos(2\theta) = 1 - 2\sin^2\theta$ , what is $\cos(2\theta)$ ? (enter as a fraction)

cos (2 θ) = cos^{2} θ - sin^{2} θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ

In Part 2 we add the tangent double angle and start using these to compute exact values. The whole rest of the lesson rests on the boxed formulas above — make sure they feel automatic.

The Tangent Double Angle

Divide $\sin(2\theta)$ by $\cos(2\theta)$ (or set $A=B=\theta$ in the tangent-sum formula) to get:

$\boxed{\tan(2\theta) = \dfrac{2\tan\theta}{1 - \tan^2\theta}}$

Quick Example: $\tan\theta = \tfrac{1}{2}$

$\tan(2\theta) = \frac{2\cdot\frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

⚠️ Watch the denominator. The $1 - \tan^2\theta$ is undefined when $\tan\theta = \pm 1$ (i.e. $\theta = 45^\circ$ ), which makes sense: is undefined.

Tangent Double Angle 🧮

Use $\tan(2\theta) = \dfrac{2\tan\theta}{1 - \tan^2\theta}$ . Enter each answer as a fraction like 8/15.

1) $\tan\theta = \tfrac{1}{3} \;\Rightarrow\; \tan(2\theta) = \,?$

Worked Example: Build Everything From One Ratio

Given: $\sin\theta = \dfrac{3}{5}$ with $\theta$ in Quadrant I. Find $\sin(2\theta)$ , $\cos(2\theta)$ , and $\tan(2\theta)$ .

Step 1 — Find $\cos\theta$ . Since $\theta$ is in QI, cosine is positive: $\cos\theta = \sqrt{1 - \left(\tfrac{3}{5}\right)^2} = \sqrt{1 - \tfrac{9}{25}} = \sqrt{\tfrac{16}{25}} = \frac{4}{5}$

Step 2 — Double angles. $\sin(2\theta) = 2\sin\theta\cos\theta = 2\cdot\tfrac{3}{5}\cdot\tfrac{4}{5} = \frac{24}{25}$

✅ Sanity check: $\left(\tfrac{24}{25}\right)^2 + \left(\tfrac{7}{25}\right)^2 = \tfrac{576 + 49}{625} = \tfrac{625}{625} = 1$ ✓

Concept Check 🎯

Build From a Ratio 🧮

Given $\cos\theta = -\dfrac{3}{5}$ with $\theta$ in Quadrant II (so $\sin\theta = \tfrac{4}{5}$ ):

1) $\sin(2\theta) = \,?$ (enter as a fraction) 2) $\cos(2\theta) = \,?$ (enter as a fraction)

$2\sin(3x)\cos(3x) = \sin(2\cdot 3x) = \sin(6x)$

Here $\alpha = 3x$ , so $2\alpha = 6x$ .

💡 Whatever is inside becomes the " $\theta$ ". Don't be thrown by $3x$ , $5\beta$ , or $\tfrac{x}{2}$ — the structure is all that matters.

Concept Check 🎯

Sometimes You Must Factor First

The pattern is often hidden until you pull out a common factor.

Example: $6\sin\theta\cos\theta$

Factor out the $2$ that the formula needs: $6\sin\theta\cos\theta = 3\cdot(2\sin\theta\cos\theta) = 3\sin(2\theta)$

Example: $4\cos^2\theta - 2$

Factor out a $2$ to expose $2\cos^2\theta - 1$ : $4\cos^2\theta - 2 = 2(2\cos^2\theta - 1) = 2\cos(2\theta)$

⚠️ Don't force it. $2\cos^2\theta - 1$ is a double angle, but $2\cos^2\theta + 1$ is — the is essential. Check the exact constant before you simplify.

Expose the Pattern 🔽

Factor, then rewrite each expression as a single double angle.

Double the Inside Angle 🧮

Each expression is a double-angle pattern in disguise. Rewrite it as a single trig function of the form $\sin(k x)$ or $\cos(k x)$ , then enter just the integer $k$ .

1) $2\sin(6x)\cos(6x) = \sin(k x)$ . Find $k$ . 2) $1 - 2\sin^2(9x) = \cos(k x)$ . Find $k$ . 3) $\cos^2(x) - \sin^2(x) = \cos(k x)$ . Find $k$ .

Recap

Recognition is a two-step habit:

Factor out whatever constant the formula requires (a $2$ for $\cos$ / $\sin$ patterns).
Match the structure and double the inside angle.

🔑 Reading identities backwards is exactly the move you'll need when integrating $\sin^2 x$ or $\cos^2 x$ in calculus — which is where Part 5's power-reduction formulas come in.

🔑 Key Concept: Half-angle formulas come from running the cosine double angle in reverse. They let you find $\sin\left(\tfrac{\theta}{2}\right)$ from $\cos\theta$ — perfect for exact values like $\sin 15^\circ$ .

Deriving the Half-Angle Formulas

Start with two faces of the cosine double angle and solve for the squared term:

$\cos(2\theta) = 1 - 2\sin^2\theta \;\Rightarrow\; \sin^2\theta = \frac{1 - \cos(2\theta)}{2}$ $\cos(2\theta) = 2\cos^2\theta - 1 \;\Rightarrow\; \cos^2\theta = \frac{1 + \cos(2\theta)}{2}$

Now replace $\theta$ with $\tfrac{\theta}{2}$ (so $2\theta$ becomes $\theta$ ) and take a square root:

$\boxed{\sin\!\left(\tfrac{\theta}{2}\right) = \pm\sqrt{\dfrac{1 - \cos\theta}{2}}}$

💡 Memory hook: sine goes with subtraction ( $1 - \cos\theta$ ); cosine goes with… the + ( $1 + \cos\theta$ ). The two $\tan$ forms are always positive-or-negative without a $\pm$ to choose — the sign comes out automatically.

Choosing the $\pm$ Sign

The square root forces a decision: is the half-angle's sine (or cosine) positive or negative?

The answer depends on the quadrant of $\tfrac{\theta}{2}$ — not the quadrant of $\theta$ .

Example: $\theta = 300^\circ$

Then $\tfrac{\theta}{2} = 150^\circ$ , which lands in Quadrant II. In QII:

$\sin\left(\tfrac{\theta}{2}\right) > 0$ → choose +
$\cos\left(\tfrac{\theta}{2}\right) < 0$ → choose

⚠️ The #1 mistake is copying the sign from $\theta$ . Always halve the angle first, locate that quadrant, then pick the sign.

Pick the Sign 🔽

For each $\theta$ , decide the quadrant of $\tfrac{\theta}{2}$ and the sign of $\sin\left(\tfrac{\theta}{2}\right)$ .

Worked Example: Exact Value of $\sin 15^\circ$

Notice $15^\circ = \tfrac{30^\circ}{2}$ , and we know $\cos 30^\circ = \tfrac{\sqrt{3}}{2}$ .

Step 1 — Pick the sign. $15^\circ$ is in Quadrant I, so $\sin 15^\circ > 0$ → use +.

Step 2 — Apply the formula. $\sin 15^\circ = \sqrt{\frac{1 - \cos 30^\circ}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$

✅ Check: $\dfrac{\sqrt{2 - \sqrt{3}}}{2} \approx 0.2588$ , and a calculator gives ✓. (This also equals the "nested-radical-free" form .)

Concept Check 🎯

Half-Angle Setup 🧮

Given $\cos\theta = \dfrac{7}{25}$ with $\theta$ in Quadrant I (so $\tfrac{\theta}{2}$ is also in QI — all positive):

1) Inside the sine formula, $\dfrac{1 - \cos\theta}{2} = \,?$ (enter as a fraction) 2) Therefore $\sin\left(\tfrac{\theta}{2}\right) = \,?$ And

\boxed{\cos^2\theta = \frac{1 + \cos(2\theta)}{2}}

The exponent $2$ on the left becomes a $2$ inside the cosine on the right — the power "moves into the angle."

A squared function oscillates in a way that's hard to work with, but $\cos(2\theta)$ is a single, clean wave. For example, the average value of $\sin^2\theta$ over a full period is $\tfrac{1}{2}$ , which you can read straight off the formula (the $\cos(2\theta)$ term averages to $0$ ).

💡 In calculus you can't integrate $\sin^2 x$ directly, but you can integrate $\tfrac{1}{2} - \tfrac{1}{2}\cos(2x)$ . That swap is this formula.

Concept Check 🎯

Reducing Higher Powers

For a fourth power, apply power reduction twice. Reduce $\sin^4\theta = (\sin^2\theta)^2$ :

$\sin^4\theta = \left(\frac{1 - \cos(2\theta)}{2}\right)^2 = \frac{1 - 2\cos(2\theta) + \cos^2(2\theta)}{4}$

Now reduce the leftover $\cos^2(2\theta) = \tfrac{1 + \cos(4\theta)}{2}$ and combine:

$\sin^4\theta = \frac{3 - 4\cos(2\theta) + \cos(4\theta)}{8}$

⚠️ Each pass doubles the angle ( $2\theta$ , then $4\theta$ ). Track those carefully — a missed doubling is the usual error here.

Power Reduction 🧮

Use $\sin^2\theta = \dfrac{1-\cos(2\theta)}{2}$ and $\cos^2\theta = \dfrac{1+\cos(2\theta)}{2}$ .

1) The average value of $\cos^2\theta$ over one full period equals $\,?$ (enter as a fraction or decimal) 2) In $\sin^2(5\theta) = \dfrac{1 - \cos(k\theta)}{2}$ , the value of is In the reduced form of , the angle inside the cosine term is . Find .

Reduce the Power 🔽

Choose the correct power-reduction rewrite for each expression.

Recap

Power	Reduced form
$\sin^2\theta$	$\dfrac{1 - \cos(2\theta)}{2}$
$\cos^2\theta$	$\dfrac{1 + \cos(2\theta)}{2}$
$\sin^2\theta\cos^2\theta$	$\dfrac{1 - \cos(4\theta)}{8}$

🔑 The third row is a bonus: $\sin^2\theta\cos^2\theta = \left(\tfrac{1}{2}\sin 2\theta\right)^2 = \tfrac{1}{4}\sin^2(2\theta) = \tfrac{1}{4}\cdot\tfrac{1 - \cos(4\theta)}{2}$ . Notice it's built from the double-angle identity you started with in Part 1.

every term the same angle

Solving $\sin(2x) = \sin x$ on $[0, 2\pi)$

Step 1 — Expand the double angle so all terms use angle $x$ : $2\sin x\cos x = \sin x$

Step 2 — Move everything to one side and factor. (Never divide by $\sin x$ — you'd lose roots.) $2\sin x\cos x - \sin x = 0 \;\Rightarrow\; \sin x\,(2\cos x - 1) = 0$

Step 3 — Solve each factor. $\sin x = 0 \;\Rightarrow\; x = 0,\ \pi$ $2\cos x - 1 = 0 \;\Rightarrow\; \cos x = \tfrac{1}{2} \;\Rightarrow\; x = \tfrac{\pi}{3},\ \tfrac{5\pi}{3}$

Solutions: $x = 0,\ \tfrac{\pi}{3},\ \pi,\ \tfrac{5\pi}{3}$ .

⚠️ Do not divide both sides by $\sin x$ ! That throws away $x = 0$ and $x = \pi$ . Always factor instead.

Concept Check 🎯

A Real Application: Projectile Range

The horizontal range of a projectile launched at speed $v$ and angle $\theta$ is

$R(\theta) = \frac{v^2 \sin(2\theta)}{g}$

The double angle $\sin(2\theta)$ is the whole story:

$\sin(2\theta)$ is largest when $2\theta = 90^\circ$ , i.e. $\theta = 45^\circ$ . That's why gives maximum range.

💡 The double-angle identity isn't just algebra — it explains a physical fact you may have seen in physics class.

Apply It 🔽

Reason about the range model $R(\theta) = \dfrac{v^2\sin(2\theta)}{g}$ and the equation work above.

Solve It 🧮

Solve on $[0, 2\pi)$ . Give answers as multiples of $\pi$ where asked.

1) $\sin(2x) = 0$ has how many solutions on $[0, 2\pi)$ ? (count them) 2) Solving $\cos(2x) + \cos x = 0$ gives $\cos x = \tfrac{1}{2}$ or $\cos x = -1$ . The solution from $\cos x = -1$ is $x = k\pi$ ; find $k$ .

⚠️ Three habits that save points: match the $\cos(2\theta)$ form to what you're given; halve the angle before picking a $\pm$ sign; and factor (never divide) when solving equations.

Mixed Warm-Up 🔽

One quick item from each skill before the quiz.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.

\tan\theta = 2 \;\Rightarrow\; \tan(2\theta) = \,?

\tan\theta = \tfrac{3}{4} \;\Rightarrow\; \tan(2\theta) = \,?

\cos(2\theta) = 1 - 2\sin^2\theta = 1 - 2\cdot\tfrac{9}{25} = 1 - \tfrac{18}{25} = \frac{7}{25}

\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{24/25}{7/25} = \frac{24}{7}

\boxed{\cos\!\left(\tfrac{\theta}{2}\right) = \pm\sqrt{\dfrac{1 + \cos\theta}{2}}}

\boxed{\tan\!\left(\tfrac{\theta}{2}\right) = \dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1 + \cos\theta}}

\sin 15^\circ \approx 0.2588

\tfrac{\sqrt{6} - \sqrt{2}}{4}

(enter as a fraction)

\cos\left(\tfrac{\theta}{2}\right) = \,?

(enter as a fraction)

41−2cos(2θ)+cos2(2θ)

2 cos x - 1 = 0 \Rightarrow cos x = \frac{1}{2} \Rightarrow x = \frac{π}{3}, \frac{5 π}{3}

Two angles that add to

90^\circ

(like

30^\circ

and

60^\circ

) give the same range, because

\sin(2\cdot 30^\circ) = \sin 60^\circ = \sin 120^\circ = \sin(2\cdot 60^\circ)

$2\sin\alpha\cos\alpha$	$\sin(2\alpha)$	the $\sin(2\theta)$ pattern with $\theta = \alpha$
$\cos^2\alpha - \sin^2\alpha$	$\cos(2\alpha)$	first form of
$1 - 2\sin^2\alpha$	$\cos(2\alpha)$	third form of $\cos(2\theta)$
$2\cos^2\alpha - 1$	$\cos(2\alpha)$	second form of $\cos(2\theta)$
$\dfrac{2\tan\alpha}{1-\tan^2\alpha}$	$\tan(2\alpha)$

Double Angle and Half Angle Identities

Double Angle and Half Angle Identities - Complete Interactive Lesson

Part 1: Where the Double-Angle Formulas Come From

🔁 Double Angle & Half Angle Identities

Topics in This Part

We Start From the Sum Formulas

The Three Faces of cos⁡(2θ)\cos(2\theta)cos(2θ)

What You've Got So Far

Part 2: Tangent & Evaluating From a Given Ratio

🔁 Double Angle & Half Angle Identities

Part 3: Reading Double Angles Backwards (Recognition)

🔁 Double Angle & Half Angle Identities

Pattern Recognition

Part 4: The Half-Angle Formulas

🔁 Double Angle & Half Angle Identities

Topics in This Part

Part 5: Power-Reduction Formulas

🔁 Double Angle & Half Angle Identities

The Power-Reduction Formulas

Part 6: Solving Equations & Real Applications

🔁 Double Angle & Half Angle Identities

Part 7: Mixed Mastery & Exit Quiz

🔁 Double Angle & Half Angle Identities

Master Reference Card

Derivation of sin⁡(2θ)\sin(2\theta)sin(2θ)

Derivation of cos⁡(2θ)\cos(2\theta)cos(2θ)

The Tangent Double Angle

Quick Example: tan⁡θ=12\tan\theta = \tfrac{1}{2}tanθ=21​

Worked Example: Build Everything From One Ratio

Example

Sometimes You Must Factor First

Example: 6sin⁡θcos⁡θ6\sin\theta\cos\theta6sinθcosθ

Example: 4cos⁡2θ−24\cos^2\theta - 24cos2θ−2

Recap

Deriving the Half-Angle Formulas

Choosing the ±\pm± Sign

Example: θ=300∘\theta = 300^\circθ=300∘

Worked Example: Exact Value of sin⁡15∘\sin 15^\circsin15∘

Why this is useful

Reducing Higher Powers

Recap

Solving sin⁡(2x)=sin⁡x\sin(2x) = \sin xsin(2x)=sinx on [0,2π)[0, 2\pi)[0,2π)

A Real Application: Projectile Range

The Three Faces of $\cos(2\theta)$

Derivation of $\sin(2\theta)$

Derivation of $\cos(2\theta)$

Quick Example: $\tan\theta = \tfrac{1}{2}$

Example: $6\sin\theta\cos\theta$

Example: $4\cos^2\theta - 2$

Choosing the $\pm$ Sign

Example: $\theta = 300^\circ$

Worked Example: Exact Value of $\sin 15^\circ$

Solving $\sin(2x) = \sin x$ on $[0, 2\pi)$