Double Angle and Half Angle Identities

🔁 Double-Angle Identities

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

Topics in This Part

🔑 Key Concept: A "double angle" formula rewrites $\sin(2\theta)$ or $\cos(2\theta)$ in terms of $\sin\theta$ and $\cos\theta$ alone. Each one is just an angle-sum formula with both angles equal.

We Start From the Sum Formulas

You already know the angle-sum identities:

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

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.

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 three equivalent ways.

💡 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.

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 $\cos\theta = \tfrac{4}{5}$.

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)

What You've Got So Far

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

In Part 2 we start using these to compute exact values from a single given ratio. The whole rest of the lesson rests on the boxed formulas above — make sure they feel automatic.

🔁 Double-Angle Identities

Part 2 of 7 — Evaluating Double Angles From a Given Ratio

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

Half-angle identities and the tangent double-angle identity are not assessed on the AP Precalculus exam.

Step Zero: Find the Missing Ratio

Double-angle formulas need both $\sin\theta$ and $\cos\theta$ (unless you choose a one-function form of $\cos(2\theta)$). When only one is given, use the Pythagorean identity and the quadrant:

$\sin^2\theta + \cos^2\theta = 1 \;\Rightarrow\; \cos\theta = \pm\sqrt{1 - \sin^2\theta}$

Quick Example: $\sin\theta = \tfrac{5}{13}$, $\theta$ in Quadrant II

$\cos\theta = -\sqrt{1 - \tfrac{25}{169}} = -\sqrt{\tfrac{144}{169}} = -\tfrac{12}{13}$

The negative sign comes from the quadrant, not from the arithmetic.

Where Does $2\theta$ Land?

Once you have $\sin(2\theta)$ and $\cos(2\theta)$, their signs locate $2\theta$. With the values above, $\sin(2\theta) = 2\cdot\tfrac{5}{13}\cdot\left(-\tfrac{12}{13}\right) = -\tfrac{120}{169}$ and $\cos(2\theta) = 1 - 2\cdot\tfrac{25}{169} = \tfrac{119}{169}$. Sine negative and cosine positive puts $2\theta$ in Quadrant IV.

⚠️ Doubling an angle can move it to a different quadrant. Never assume $2\theta$ shares the quadrant of $\theta$.

Missing Ratio, Then Double 🧮

Given $\cos\theta = \tfrac{8}{17}$ with $\theta$ in Quadrant IV. Enter each answer as a fraction like 8/15.

1) $\sin\theta = \,?$ 2) $\sin(2\theta) = \,?$ 3) $\cos(2\theta) = \,?$

Worked Example: Build Everything From One Ratio

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

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. $\sin(2\theta) = 2\sin\theta\cos\theta = 2\cdot\tfrac{3}{5}\cdot\tfrac{4}{5} = \frac{24}{25}$ $\cos(2\theta) = 1 - 2\sin^2\theta = 1 - 2\cdot\tfrac{9}{25} = 1 - \tfrac{18}{25} = \frac{7}{25}$

Step 3 — Locate $2\theta$. Both values are positive, so $2\theta$ is in Quadrant I. (Check: $\theta \approx 0.644$, so $2\theta \approx 1.287$, which is less than $\frac{\pi}{2} \approx 1.571$.)

✅ 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)

🔁 Double-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.

Example

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

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 not — the $-1$ 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:

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

🔑 Reading identities backwards turns a product into a single sinusoid. For example, $y = \sin x\cos x = \tfrac{1}{2}\sin(2x)$ has amplitude $\tfrac{1}{2}$ and period $\pi$, facts that are hidden in the product form.

🔁 Double-Angle Identities

Part 4 of 7 — Verifying Identities with Double Angles

Topics in This Part

🔑 Key Concept: To verify an identity, you transform one side, step by step, until it matches the other side. Double-angle identities are often the first move, because they turn every angle into the same angle.

A Verification Game Plan

1. Start with the more complicated side. It has more to simplify.
2. Make every angle the same. Expand $\sin(2\theta)$ and $\cos(2\theta)$ into functions of $\theta$.
3. Pick the $\cos(2\theta)$ form that cancels a constant. $1 + \cos(2\theta) = 2\cos^2\theta$ and $1 - \cos(2\theta) = 2\sin^2\theta$.
4. Use the Pythagorean identity whenever $\sin^2\theta + \cos^2\theta$ appears.
5. Never move terms across the equals sign. You are showing the two sides are equal, so you may not assume it.

Worked Example 1: Verify $(\sin\theta + \cos\theta)^2 = 1 + \sin(2\theta)$

Start with the left side and expand the square:

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

Group the squares and recognize the double angle:

$= (\sin^2\theta + \cos^2\theta) + 2\sin\theta\cos\theta = 1 + \sin(2\theta) \;\checkmark$

Concept Check 🎯

Worked Example 2: Verify $\cos^4\theta - \sin^4\theta = \cos(2\theta)$

The left side is a difference of squares:

$\cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)$

The second factor is $1$ by the Pythagorean identity, and the first factor is a form of $\cos(2\theta)$:

$= \cos(2\theta)\cdot 1 = \cos(2\theta) \;\checkmark$

Worked Example 3: Verify $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$

Write $3\theta = 2\theta + \theta$ and use the sum identity for sine:

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

Expand both double angles, choosing the sine-only form of $\cos(2\theta)$:

$= (2\sin\theta\cos\theta)\cos\theta + (1 - 2\sin^2\theta)\sin\theta = 2\sin\theta\cos^2\theta + \sin\theta - 2\sin^3\theta$

Replace $\cos^2\theta$ with $1 - \sin^2\theta$:

$= 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta = 3\sin\theta - 4\sin^3\theta \;\checkmark$

💡 Check with a value: at $\theta = \frac{\pi}{6}$, the left side is $\sin \frac{\pi}{2} = 1$ and the right side is $3\cdot\tfrac{1}{2} - 4\cdot\tfrac{1}{8} = 1$. A numerical check cannot prove an identity, but it catches sign errors fast.

Choose the Next Step 🔽

Pick the move that makes progress in each verification.

Use the Verified Identities 🧮

Each value follows from an identity verified above.

1) If $\sin\theta + \cos\theta = 1.2$, then $\sin(2\theta) = \,?$ (enter a decimal) 2) If $\cos\theta = 0.6$, then $\cos^4\theta - \sin^4\theta = \,?$ (enter a decimal) 3) If $\sin\theta = \tfrac{1}{2}$, then $\sin(3\theta) = \,?$

🔁 Double-Angle Identities

Part 5 of 7 — Squared Sines and Cosines as Sinusoids

🔑 The Idea: Solving the cosine double-angle identity for the squared term trades a squared trig function for a first-power one. The result, sometimes called a power-reduction form, shows that $\sin^2\theta$ and $\cos^2\theta$ are sinusoids in disguise.

The Power-Reduction Formulas

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

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

Divide by $2$:

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

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

Why this is useful

A squared function is hard to read, but $\cos(2\theta)$ is a single, clean wave. Written as $\sin^2\theta = \tfrac{1}{2} - \tfrac{1}{2}\cos(2\theta)$, the function is a sinusoid with midline $y = \tfrac{1}{2}$, so its average value over a full period is $\tfrac{1}{2}$ (the $\cos(2\theta)$ term averages to $0$).

💡 Both forms describe the same function. The squared form is compact; the reduced form shows the midline, amplitude, and period directly.

Concept Check 🎯

Reading Sinusoid Features From the Reduced Form

Rewrite, then read the features as you would for any $y = a\cos(bx) + d$.

Example: $y = \sin^2 x$

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

The period is $\pi$, half the period of $\sin x$: squaring makes every negative value positive, so the graph repeats twice as often.

Example: $y = 6\cos^2(3x)$

$6\cos^2(3x) = 6\cdot\frac{1 + \cos(6x)}{2} = 3 + 3\cos(6x)$

Midline $y = 3$, amplitude $3$, period $\tfrac{2\pi}{6} = \tfrac{\pi}{3}$, range $0 \le y \le 6$.

⚠️ The inside angle doubles ($3x$ becomes $6x$). Forgetting the doubling gives the wrong period, which 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 midline of $y = \cos^2\theta$ is $y = \,?$ (enter as a fraction or decimal) 2) In $\sin^2(5\theta) = \dfrac{1 - \cos(k\theta)}{2}$, the value of $k$ is $\,?$ 3) In the reduced form of $\cos^2(2\theta)$, the angle inside the cosine term is $k\theta$. Find $k$.

Reduce the Power 🔽

Choose the correct power-reduction rewrite for each expression.

Recap

🔑 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 all built from the double-angle identity you started with in Part 1.

🔁 Double-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 every term the same angle. Then it factors.

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 $2\theta = \frac{\pi}{2}$, i.e. $\theta = \frac{\pi}{4}$. That's why $\frac{\pi}{4}$ gives maximum range.
• Two angles that add to $\frac{\pi}{2}$ (like $\frac{\pi}{6}$ and $\frac{\pi}{3}$) give the same range, because $\sin(2\cdot \frac{\pi}{6}) = \sin \frac{\pi}{3} = \sin \frac{2\pi}{3} = \sin(2\cdot \frac{\pi}{3})$.

💡 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 $k$.

🔁 Double-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) verify identities with them, (4) rewrite squared sines and cosines as sinusoids, and (5) solve equations. Let's pull it all together.

Master Reference Card

⚠️ Three habits that save points: match the $\cos(2\theta)$ form to what you're given; double the inside angle when you rewrite a square; 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.