Exercises — Double angle formulas — sin 2A, cos 2A (three forms), tan 2A

WHAT: apply $\sin 2A=2\sin A\cos A$ directly. WHY: both $\sin A$ and $\cos A$ are handed to us, so the double-sine formula needs no extra work. $\sin 2A=2\cdot \frac{3}{5}\cdot \frac{4}{5}=\frac{24}{25}=0.96$

WHAT: use Form 2, $\cos 2A=2{\cos }^{2}A-1$. WHY this form: we were given only $\cos A$, so the "cosine-only" form lets us finish without hunting for $\sin A$. $\cos 2A=2{\left(\frac{4}{5}\right)}^2-1=2\cdot \frac{16}{25}-1=\frac{32}{25}-\frac{25}{25}=\frac{7}{25}$

WHAT: apply $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$. WHY: we have $\tan A$ alone, and this formula is written purely in $\tan A$. $\tan 2A=\frac{2\cdot \frac{1}{3}}{1-{\left(\frac{1}{3}\right)}^2}=\frac{\frac{2}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{2}{3}\cdot \frac{9}{8}=\frac{3}{4}$

Step 1 — find $\cos A$. We only have $\sin A$, so use Pythagoras: $\cos A=\pm \sqrt{1-{\sin }^{2}A}$. Since $A$ is acute (Quadrant I), cosine is positive, so we take the $+$ root. $\cos A=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$ Step 2 — $\sin 2A$: $2\sin A\cos A=2\cdot \frac{5}{13}\cdot \frac{12}{13}=\frac{120}{169}$. Step 3 — $\cos 2A$: use Form 3 (only sine, so no re-using $\cos A$): $1-2{\sin }^{2}A=1-2\cdot \frac{25}{169}=1-\frac{50}{169}=\frac{119}{169}$.

Step 1 — sign of $\cos A$. In Quadrant II, cosine is negative, so we take the minus root. $\cos A=-\sqrt{1-\frac{9}{25}}=-\frac{4}{5}$ Look at Figure s01: the terminal arm points up-and-left, so its horizontal ($\cos$) reach is negative while its vertical ($\sin$) reach stays positive. Step 2 — $\sin 2A$: $2\cdot \frac{3}{5}\cdot \left(-\frac{4}{5}\right)=-\frac{24}{25}$. Step 3 — $\cos 2A$: Form 3 (only sine): $1-2\cdot \frac{9}{25}=1-\frac{18}{25}=\frac{7}{25}$. Step 4 — locate $2A$: $\sin 2A<0$ and $\cos 2A>0$ ⇒ $2A$ is in Quadrant IV. (Check: $A\approx 143^{\circ }$, so $2A\approx 287^{\circ }$, which is indeed Q4.)

Step 1 — find $\sin A$. In Quadrant III both sine and cosine are negative. $\sin A=-\sqrt{1-\frac{64}{289}}=-\sqrt{\frac{225}{289}}=-\frac{15}{17}$ Step 2 — $\tan A$: a negative divided by a negative is positive: $\tan A=\frac{\sin A}{\cos A}=\frac{-15/17}{-8/17}=\frac{15}{8}$ Step 3 — $\tan 2A$: $\tan 2A=\frac{2\cdot \frac{15}{8}}{1-{\left(\frac{15}{8}\right)}^2}=\frac{\frac{30}{8}}{1-\frac{225}{64}}=\frac{\frac{30}{8}}{\frac{64-225}{64}}=\frac{\frac{30}{8}}{-\frac{161}{64}}=\frac{30}{8}\cdot \left(-\frac{64}{161}\right)=-\frac{240}{161}$

Numerator: we want a clean single term, so pick the $\cos 2A$ form whose constant cancels the $1$. Form 3 gives $1-\cos 2A=1-(1-2{\sin }^{2}A)=2{\sin }^{2}A$. Denominator: $\sin 2A=2\sin A\cos A$. $\frac{2{\sin }^{2}A}{2\sin A\cos A}=\frac{\sin A}{\cos A}=\tan A■$ WHY Form 3 here: it turns $1-\cos 2A$ into $2{\sin }^{2}A$, matching the $\sin$ in the denominator so a factor cancels cleanly.

Idea: the right side is built from $\tan A$, so divide top and bottom of a $\cos 2A$ form by ${\cos }^{2}A$ to manufacture ${\tan }^{2}A={\sin }^{2}A/{\cos }^{2}A$. Start from Form 1: $\cos 2A={\cos }^{2}A-{\sin }^{2}A$. Write it over the Pythagorean $1={\cos }^{2}A+{\sin }^{2}A$ (multiplying by $1$ changes nothing): $\cos 2A=\frac{{\cos }^{2}A-{\sin }^{2}A}{{\cos }^{2}A+{\sin }^{2}A}$ Divide every term by ${\cos }^{2}A$: $=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}■$ WHY divide by ${\cos }^{2}A$: it is the standard move to convert an expression in $\sin ,\cos$ into one purely in $\tan$.

Same trick: write $\sin 2A=2\sin A\cos A$ over $1={\cos }^{2}A+{\sin }^{2}A$: $\sin 2A=\frac{2\sin A\cos A}{{\cos }^{2}A+{\sin }^{2}A}$ Divide top and bottom by ${\cos }^{2}A$. Top: $\frac{2\sin A\cos A}{{\cos }^{2}A}=2\tan A$. Bottom: $1+{\tan }^{2}A$. $\sin 2A=\frac{2\tan A}{1+{\tan }^{2}A}■$ These last two results are the seeds of the Weierstrass t = tan(A/2) Substitution.

Key insight: $4A=2(2A)$, so $\sin 4A=2\sin 2A\cos 2A$ — the double-angle formula applied to the angle $2A$. Step 1 — build the $2A$ pieces. With $A$ acute, $\cos A=\frac{4}{5}$. $\sin 2A=2\cdot \frac{3}{5}\cdot \frac{4}{5}=\frac{24}{25},\cos 2A=1-2\cdot \frac{9}{25}=\frac{7}{25}$ Step 2 — apply again: $\sin 4A=2\sin 2A\cos 2A=2\cdot \frac{24}{25}\cdot \frac{7}{25}=\frac{336}{625}$

Split: $3A=2A+A$, then use the addition formula: $\sin 3A=\sin 2A\cos A+\cos 2A\sin A$ Substitute the double-angle forms — choose Form 3 for $\cos 2A$ so everything ends in $\sin A$: $=(2\sin A\cos A)\cos A+(1-2{\sin }^{2}A)\sin A$ $=2\sin A{\cos }^{2}A+\sin A-2{\sin }^{3}A$ Kill the ${\cos }^{2}A$ with Pythagoras, ${\cos }^{2}A=1-{\sin }^{2}A$: $=2\sin A(1-{\sin }^{2}A)+\sin A-2{\sin }^{3}A=2\sin A-2{\sin }^{3}A+\sin A-2{\sin }^{3}A$ $=3\sin A-4{\sin }^{3}A■$

Step 1 — $\tan 2A$: $\tan 2A=\frac{2\cdot 2}{1-2^2}=\frac{4}{1-4}=\frac{4}{-3}=-\frac{4}{3}$ Step 2 — double again, treating $2A$ as the new "single" angle: $\tan 4A=\frac{2\tan 2A}{1-{\tan }^{2}2A}=\frac{2\cdot \left(-\frac{4}{3}\right)}{1-{\left(-\frac{4}{3}\right)}^2}=\frac{-\frac{8}{3}}{1-\frac{16}{9}}=\frac{-\frac{8}{3}}{-\frac{7}{9}}=\frac{8}{3}\cdot \frac{9}{7}=\frac{24}{7}$

WHY Form 3: the right side is $\sin A$, so choose the $\cos 2A$ form written in $\sin A$ — that makes one variable throughout. $1-2{\sin }^{2}A=\sin A$ Rearrange into a quadratic in $s=\sin A$: $2s^2+s-1=0⟹(2s-1)(s+1)=0⟹s=\frac{1}{2}\text{ or }s=-1$ Case $\sin A=\frac{1}{2}$: $A=30^{\circ }$ or $A=150^{\circ }$. Case $\sin A=-1$: $A=270^{\circ }$. $\boxed{A=30^{\circ },150^{\circ },270^{\circ }}$ Every case of the sine values is covered, and all lie in $[0^{\circ },360^{\circ })$.

WHY power-reduction: ${\sin }^{2}x$ has no elementary antiderivative as written, but rearranging Form 3 gives ${\sin }^{2}x=\frac{1-\cos 2x}{2}$, a sum of things we can integrate. This is the whole point of Power-Reduction / Half-Angle Formulas and Integration of sin²x and cos²x. ${\int }_0^{\pi /2}{\sin }^{2}xdx={\int }_0^{\pi /2}\frac{1-\cos 2x}{2}dx=\frac{1}{2}{\left[x-\frac{1}{2}\sin 2x\right]}_0^{\pi /2}$ At $x=\frac{\pi }{2}$: $\frac{\pi }{2}-\frac{1}{2}\sin \pi =\frac{\pi }{2}-0$. At $x=0$: $0-0=0$. $=\frac{1}{2}\cdot \frac{\pi }{2}=\frac{\pi }{4}$

Read the triangle (Figure s02): with equal legs, $\tan A=\frac{A_y}{A_x}=1$, and the hypotenuse has length $\sqrt{1^2+1^2}=\sqrt{2}$, so $\sin A=\cos A=\frac{1}{\sqrt{2}}$. Double-angle values: $\sin 2A=2\cdot \frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=2\cdot \frac{1}{2}=1$ $\cos 2A={\cos }^{2}A-{\sin }^{2}A=\frac{1}{2}-\frac{1}{2}=0$ Confirm the angle: $\sin 2A=1$ and $\cos 2A=0$ pinpoint $2A=90^{\circ }$ (so $A=45^{\circ }$ ✓). $\tan 2A$: the formula $\frac{2\tan A}{1-{\tan }^{2}A}=\frac{2}{1-1}=\frac{2}{0}$ is undefined — exactly matching $\cos 2A=0$, since $\tan 2A=\sin 2A/\cos 2A$ divides by zero. This is the degenerate case flagged in the parent note (${\tan }^{2}A=1\Rightarrow 2A=90^{\circ }$).

Only two ideas, over and over: (1) $2A=A+A$ turns every double angle into the addition formula with the same number twice; (2) ${\sin }^2+{\cos }^2=1$ lets you swap a square, giving three $\cos 2A$ costumes. Levels 1–2 just plug in (watching quadrant signs); Level 3 chooses the right costume to cancel a $1$; Level 4 reuses the template on $2A$ to reach $3A$ and $4A$; Level 5 connects the swap-trick to solving equations and integrating.