Chapter 3 · Trigonometric Functions

3.4 Multiple-Angle, Product and Sum Formulae

Put $y=x$ in the sum formulae and out come $\sin2x$, $\cos2x$ and $\tan2x$; one more step gives $\sin3x$ and $\cos3x$; adding and subtracting them turns products into sums and sums into products — the toolkit for proving identities and for understanding beats in sound.

A cricket ball thrown at speed $u$ at an angle $\theta$ to the ground (ignoring air resistance) lands at a distance $R=\dfrac{u^2\sin2\theta}{g}$. Why $\sin2\theta$, and why does that tell us that $45^\circ$ gives the longest throw and that $30^\circ$ and $60^\circ$ give the same one? Two guitar strings that are almost but not quite in tune produce a sound that swells and fades — "beats" — a few times a second. Why? Both answers come from formulae that rewrite trigonometric expressions in a more useful shape.

In this section we derive them all from Section 3.3. Setting $y=x$ gives the double-angle formulae, and with them the values of $\sin$ and $\cos$ at half angles such as $22\tfrac12^\circ$. Applying the sum formula once more gives the triple-angle formulae. Finally, adding and subtracting the sum and difference formulae gives the product–sum formulae, which convert a product such as $2\sin x\cos y$ into a sum and back. With these four families you can prove almost every identity a Board or JEE paper will ask.

Double-Angle Formulae

📐 Theorem 3.10 — Double-angle formulae

For every real $x$:

$$\sin2x=2\sin x\cos x,$$ $$\cos2x=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x .$$

If $x$ is not an odd multiple of $\tfrac{\pi}{2}$:

$$\sin2x=\frac{2\tan x}{1+\tan^2x},\qquad \cos2x=\frac{1-\tan^2x}{1+\tan^2x};$$

and if, in addition, $2x$ is not an odd multiple of $\tfrac{\pi}{2}$:

$$\tan2x=\frac{2\tan x}{1-\tan^2x}.$$

Proof. Put $y=x$ in Theorem 3.7: $\sin2x=\sin x\cos x+\cos x\sin x=2\sin x\cos x$ and $\cos2x=\cos^2x-\sin^2x$. Replacing $\sin^2x$ by $1-\cos^2x$ gives $2\cos^2x-1$; replacing $\cos^2x$ by $1-\sin^2x$ gives $1-2\sin^2x$. For the tangent forms, divide by $1=\cos^2x+\sin^2x$ and then divide the numerator and the denominator by $\cos^2x$: $\sin2x=\dfrac{2\sin x\cos x}{\cos^2x+\sin^2x}=\dfrac{2\tan x}{1+\tan^2x}$, and similarly for $\cos2x$. Finally put $y=x$ in Theorem 3.9 for $\tan2x$. $\blacksquare$

Turning the cosine formulae round gives the power-reduction (or half-angle) formulae, which replace a square by a first power of a double angle:

$$\sin^2x=\frac{1-\cos2x}{2},\qquad \cos^2x=\frac{1+\cos2x}{2},\qquad \tan^2x=\frac{1-\cos2x}{1+\cos2x}.$$

Writing $\tfrac{x}{2}$ for $x$: $\ \sin\tfrac{x}{2}=\pm\sqrt{\dfrac{1-\cos x}{2}}$ and $\cos\tfrac{x}{2}=\pm\sqrt{\dfrac{1+\cos x}{2}}$, where the sign is decided by the quadrant of $\tfrac{x}{2}$ — not of $x$.

✏️ Example 1 — Double angles from one value
If $\tan x=-\tfrac43$ and $x$ lies in the second quadrant, find $\sin2x$, $\cos2x$ and $\tan2x$.
sin, cos $\sec^2x=1+\tfrac{16}{9}=\tfrac{25}{9}$; in Quadrant II cosine is negative, so $\cos x=-\tfrac35$ and $\sin x=\tan x\cos x=\tfrac45$.
Double $\sin2x=2\cdot\tfrac45\cdot\left(-\tfrac35\right)=-\tfrac{24}{25}$; $\ \cos2x=\cos^2x-\sin^2x=\tfrac{9}{25}-\tfrac{16}{25}=-\tfrac{7}{25}$; $\ \tan2x=\dfrac{-24/25}{-7/25}=\tfrac{24}{7}$.
Check $\dfrac{2\tan x}{1-\tan^2x}=\dfrac{-8/3}{1-16/9}=\dfrac{-8/3}{-7/9}=\tfrac{24}{7}$ ✓, and $\left(\tfrac{24}{25}\right)^2+\left(\tfrac{7}{25}\right)^2=1$ ✓.
✏️ Example 2 — Half angles
(a) Find the exact values of $\sin22\tfrac12^\circ$, $\cos22\tfrac12^\circ$ and $\tan22\tfrac12^\circ$.
(b) If $\cos x=-\tfrac13$ and $\pi\lt x\lt\tfrac{3\pi}{2}$, find $\sin\tfrac{x}{2}$, $\cos\tfrac{x}{2}$ and $\tan\tfrac{x}{2}$.
(a) With $x=22\tfrac12^\circ$, $2x=45^\circ$: $\sin^2x=\dfrac{1-\tfrac{1}{\sqrt2}}{2}=\dfrac{2-\sqrt2}{4}$ and $\cos^2x=\dfrac{2+\sqrt2}{4}$. The angle is acute, so $\sin22\tfrac12^\circ=\tfrac12\sqrt{2-\sqrt2}$ and $\cos22\tfrac12^\circ=\tfrac12\sqrt{2+\sqrt2}$.
tan $\tan x=\dfrac{1-\cos2x}{\sin2x}=\dfrac{1-\tfrac1{\sqrt2}}{\tfrac1{\sqrt2}}=\sqrt2-1$ (using $\tfrac{1-\cos2x}{\sin2x}=\tfrac{2\sin^2x}{2\sin x\cos x}$).
(b) Quadrant $\pi\lt x\lt\tfrac{3\pi}{2}$ gives $\tfrac{\pi}{2}\lt\tfrac x2\lt\tfrac{3\pi}{4}$: Quadrant II, where sine is positive and cosine negative.
(b) Values $\sin\tfrac x2=\sqrt{\dfrac{1+\tfrac13}{2}}=\sqrt{\tfrac23}=\tfrac{\sqrt6}{3}$; $\ \cos\tfrac x2=-\sqrt{\dfrac{1-\tfrac13}{2}}=-\tfrac{1}{\sqrt3}$; $\ \tan\tfrac x2=-\sqrt2$. Check: $\cos x=1-2\sin^2\tfrac x2=1-\tfrac43=-\tfrac13$ ✓.

Triple-Angle Formulae

📐 Theorem 3.11 — Triple-angle formulae

For every real $x$:

$$\sin3x=3\sin x-4\sin^3x,\qquad \cos3x=4\cos^3x-3\cos x .$$

If neither $x$ nor $3x$ is an odd multiple of $\tfrac{\pi}{2}$:

$$\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.$$

Proof. $\sin3x=\sin(2x+x)=\sin2x\cos x+\cos2x\sin x=2\sin x\cos^2x+(1-2\sin^2x)\sin x$. Using $\cos^2x=1-\sin^2x$: $\ 2\sin x-2\sin^3x+\sin x-2\sin^3x=3\sin x-4\sin^3x$.
$\cos3x=\cos2x\cos x-\sin2x\sin x=(2\cos^2x-1)\cos x-2\sin^2x\cos x=2\cos^3x-\cos x-2(1-\cos^2x)\cos x=4\cos^3x-3\cos x$.
$\tan3x=\dfrac{\tan2x+\tan x}{1-\tan2x\tan x}$; substitute $\tan2x=\dfrac{2\tan x}{1-\tan^2x}$ and multiply the numerator and the denominator by $1-\tan^2x$: $\ \dfrac{2\tan x+\tan x-\tan^3x}{1-\tan^2x-2\tan^2x}$. $\blacksquare$

Interactive 1 — See an Identity Hold Everywhere

The wide translucent band is the left-hand side, the thin dashed curve the right-hand side. If they lie on top of each other for every $x$, the identity holds. The faint curves are the ingredients — for example $\sin x$ and $\cos x$ inside $2\sin x\cos x$. The last option is a false "identity": watch the curves separate.

$x$—
left side—
right side—
difference—
✏️ Example 3 — The exact value of $\sin18^\circ$
(a) Evaluate $3\sin10^\circ-4\sin^310^\circ$ and $4\cos^320^\circ-3\cos20^\circ$.
(b) Use $2\times18^\circ=90^\circ-3\times18^\circ$ to find the exact value of $\sin18^\circ$.
(a) These are $\sin(3\times10^\circ)=\sin30^\circ=\tfrac12$ and $\cos(3\times20^\circ)=\cos60^\circ=\tfrac12$.
(b) Set up Let $\theta=18^\circ$, so $2\theta=90^\circ-3\theta$ and $\sin2\theta=\cos3\theta$: $\ 2\sin\theta\cos\theta=4\cos^3\theta-3\cos\theta$.
Solve Divide by $\cos\theta\ne0$: $2\sin\theta=4\cos^2\theta-3=4(1-\sin^2\theta)-3=1-4\sin^2\theta$. So $s=\sin\theta$ satisfies $4s^2+2s-1=0$, giving $s=\dfrac{-2\pm\sqrt{4+16}}{8}=\dfrac{-1\pm\sqrt5}{4}$.
Choose $\sin18^\circ\gt0$, so $\sin18^\circ=\dfrac{\sqrt5-1}{4}\approx0.3090$ ✓ (calculator: $0.30902$).

Product and Sum Formulae

📐 Theorem 3.12 — Products as sums

For all real $x$ and $y$:

$$2\sin x\cos y=\sin(x+y)+\sin(x-y),$$ $$2\cos x\sin y=\sin(x+y)-\sin(x-y),$$ $$2\cos x\cos y=\cos(x+y)+\cos(x-y),$$ $$2\sin x\sin y=\cos(x-y)-\cos(x+y).$$

Proof. Add the formulae for $\sin(x+y)$ and $\sin(x-y)$: the $\cos x\sin y$ terms cancel, leaving $2\sin x\cos y$. Subtracting instead leaves $2\cos x\sin y$. Adding $\cos(x+y)$ and $\cos(x-y)$ leaves $2\cos x\cos y$; subtracting $\cos(x+y)$ from $\cos(x-y)$ leaves $2\sin x\sin y$. $\blacksquare$

📐 Theorem 3.13 — Sums as products

For all real $x$ and $y$:

$$\sin x+\sin y=2\sin\tfrac{x+y}{2}\cos\tfrac{x-y}{2},$$ $$\sin x-\sin y=2\cos\tfrac{x+y}{2}\sin\tfrac{x-y}{2},$$ $$\cos x+\cos y=2\cos\tfrac{x+y}{2}\cos\tfrac{x-y}{2},$$ $$\cos x-\cos y=-2\sin\tfrac{x+y}{2}\sin\tfrac{x-y}{2}.$$

Proof. In Theorem 3.12 write $A=\tfrac{x+y}{2}$ and $B=\tfrac{x-y}{2}$, so that $A+B=x$ and $A-B=y$. Then $2\sin A\cos B=\sin x+\sin y$, $\ 2\cos A\sin B=\sin x-\sin y$, $\ 2\cos A\cos B=\cos x+\cos y$, and $2\sin A\sin B=\cos y-\cos x$, i.e. $\cos x-\cos y=-2\sin A\sin B$. $\blacksquare$

🧭 Problem-solving strategy — Proving identities
  1. Start from the more complicated side and work towards the other; never "cross-multiply both sides" of the identity you are trying to prove.
  2. Different angles ($x$, $2x$, $3x$)? Bring them to one angle with the multiple-angle formulae. Sums of sines or cosines, especially with angles in arithmetic progression ($3x,5x,7x$)? Pair the outer two with Theorem 3.13 — a common factor usually appears.
  3. Products of sines and cosines? Turn them into sums with Theorem 3.12, or multiply by $2\sin(\text{smallest angle})$ to start a double-angle chain (Example 6).
  4. Check the finished identity numerically at one angle such as $x=10^\circ$.
✏️ Example 4 — Converting both ways
(a) Write $2\sin3x\cos x$ as a sum. (b) Find $\sin75^\circ\cos15^\circ$. (c) Find $\cos75^\circ+\cos15^\circ$. (d) Factorise $\sin7x-\sin3x$.
(a) $2\sin3x\cos x=\sin(3x+x)+\sin(3x-x)=\sin4x+\sin2x$.
(b) $\sin75^\circ\cos15^\circ=\tfrac12(\sin90^\circ+\sin60^\circ)=\tfrac12\left(1+\tfrac{\sqrt3}{2}\right)=\dfrac{2+\sqrt3}{4}$.
(c) $2\cos\tfrac{75^\circ+15^\circ}{2}\cos\tfrac{75^\circ-15^\circ}{2}=2\cos45^\circ\cos30^\circ=2\cdot\tfrac1{\sqrt2}\cdot\tfrac{\sqrt3}{2}=\dfrac{\sqrt6}{2}$.
(d) $\sin7x-\sin3x=2\cos\tfrac{7x+3x}{2}\sin\tfrac{7x-3x}{2}=2\cos5x\sin2x$.
Interactive 2 — Beats: a Sum Becomes a Product

Two waves $\cos at$ and $\cos bt$ (faint) are added. When $a$ and $b$ are close, the sum (thick) is a fast wave whose size swells and fades inside the dashed envelope $\pm2\cos\tfrac{(a-b)t}{2}$ — exactly what $\cos at+\cos bt=2\cos\tfrac{(a+b)t}{2}\cos\tfrac{(a-b)t}{2}$ predicts. Each swell is one "beat"; move $b$ towards $a$ and the beats slow down.

1210
fast wave—
envelope—
time between beats—
sum = product at cursor—
✏️ Example 5 — Identities from sums of sines and cosines
Prove that (a) $\dfrac{\sin5x+\sin3x}{\cos5x+\cos3x}=\tan4x$ (b) $\dfrac{\cos7x+\cos5x}{\sin7x-\sin5x}=\cot x$, wherever the denominators are non-zero.
(a) $\dfrac{2\sin4x\cos x}{2\cos4x\cos x}=\dfrac{\sin4x}{\cos4x}=\tan4x$ (we may cancel $\cos x$, since the denominator $2\cos4x\cos x$ is non-zero).
(b) $\dfrac{2\cos6x\cos x}{2\cos6x\sin x}=\dfrac{\cos x}{\sin x}=\cot x$.
Check (a) at $x=10^\circ$: $\dfrac{\sin50^\circ+\sin30^\circ}{\cos50^\circ+\cos30^\circ}=\dfrac{1.2660}{1.5088}\approx0.8391=\tan40^\circ$ ✓.
✏️ Example 6 — A product that collapses
Prove that $\cos20^\circ\cos40^\circ\cos80^\circ=\tfrac18$.
Idea Each angle is double the one before, so multiply by $8\sin20^\circ$ and use $2\sin\theta\cos\theta=\sin2\theta$ three times. Call the product $p$.
Chain $8\sin20^\circ\,p=4(2\sin20^\circ\cos20^\circ)\cos40^\circ\cos80^\circ=2(2\sin40^\circ\cos40^\circ)\cos80^\circ=2\sin80^\circ\cos80^\circ=\sin160^\circ$.
Finish $\sin160^\circ=\sin(180^\circ-20^\circ)=\sin20^\circ\ne0$, so $8p=1$ and $p=\tfrac18$. $\blacksquare$
Check $0.9397\times0.7660\times0.1736\approx0.1250$ ✓.
⚠️ Common mistakes with multiple angles and products

1. $\sin2x\ne2\sin x$. At $x=\tfrac{\pi}{2}$: $\sin\pi=0$ but $2\sin\tfrac{\pi}{2}=2$. The "2" goes inside the bracket, and the formula is $2\sin x\cos x$.

2. Choosing the wrong form of $\cos2x$. All three are correct; pick the one that matches the rest of the problem — $1-2\sin^2x$ if you know $\sin x$, $2\cos^2x-1$ if you know $\cos x$, $1\pm\cos2x$ to clear a "$1$".

3. Half-angle signs. The sign of $\sin\tfrac x2$ depends on the quadrant of $\tfrac x2$, which can differ from that of $x$ (Example 2(b)).

4. Losing the factor 2 or the minus sign. $\sin x\cos y=\tfrac12[\sin(x+y)+\sin(x-y)]$ — the $\tfrac12$ is easily forgotten. And $\cos x-\cos y=-2\sin\tfrac{x+y}{2}\sin\tfrac{x-y}{2}$ has a minus sign: $\cos0-\cos\pi=2$, and $-2\sin\tfrac{\pi}{2}\sin\left(-\tfrac{\pi}{2}\right)=2$ ✓.

📘 In the CBSE syllabus — school assessment only (not in the year-end board exam)

Taught and tested in school, but not in the year-end board exam; JEE Main and JEE Advanced both list trigonometric equations and ask for general solutions — all real $x$ satisfying a trigonometric equation. The basic facts (with $n\in\mathbb{Z}$):

$\sin x=\sin\alpha\iff x=n\pi+(-1)^n\alpha$; $\quad\cos x=\cos\alpha\iff x=2n\pi\pm\alpha$; $\quad\tan x=\tan\alpha\iff x=n\pi+\alpha$.

(For example, $\cos x=\cos\alpha$ means $P$ for $x$ is $P$ for $\alpha$ or its mirror image in the $x$-axis, plus any number of whole turns.) Product–sum formulae are the usual way to reach these basic forms.

Worked example. Find the general solution of $\cos3x=\cos x$.
Solution. $3x=2n\pi\pm x$. With "$+$": $2x=2n\pi$, $x=n\pi$. With "$-$": $4x=2n\pi$, $x=\tfrac{n\pi}{2}$. The second family contains the first, so the general solution is $x=\tfrac{n\pi}{2}$, $n\in\mathbb{Z}$. (Check: $\cos x-\cos3x=2\sin2x\sin x=0$ exactly when $\sin2x=0$ or $\sin x=0$, i.e. $x=\tfrac{n\pi}{2}$ ✓.)

Exercises

I
Double-angle formulae.
  • 1 $\sin x=\tfrac35$, $x$ acute. Find $\sin2x$, $\cos2x$, $\tan2x$.
  • 2 $\cos x=-\tfrac{12}{13}$, $x$ in Quadrant III. Find $\sin2x$, $\cos2x$, $\tan2x$.
  • 3 $\tan x=\tfrac12$. Find $\tan2x$, and (using the tangent forms) $\sin2x$ and $\cos2x$.
  • 4 Evaluate $2\sin15^\circ\cos15^\circ$.
  • 5 Evaluate $\cos^215^\circ-\sin^215^\circ$.
  • 6 Evaluate $\dfrac{2\tan22\tfrac12^\circ}{1-\tan^222\tfrac12^\circ}$.
  • 7 Prove that $\dfrac{1-\cos2x}{\sin2x}=\tan x$, where defined.
  • 8 Prove that $\dfrac{\sin2x}{1+\cos2x}=\tan x$, where defined.
  • 9 Prove that $\cos^4x-\sin^4x=\cos2x$.
1 $\cos x=\tfrac45$: $\sin2x=\tfrac{24}{25}$, $\cos2x=\tfrac{16}{25}-\tfrac9{25}=\tfrac7{25}$, $\tan2x=\tfrac{24}{7}$.
2 $\sin x=-\tfrac5{13}$: $\sin2x=2\left(-\tfrac5{13}\right)\left(-\tfrac{12}{13}\right)=\tfrac{120}{169}$, $\cos2x=\tfrac{144-25}{169}=\tfrac{119}{169}$, $\tan2x=\tfrac{120}{119}$.
3 $\tan2x=\dfrac{1}{1-\tfrac14}=\tfrac43$; $\sin2x=\dfrac{1}{1+\tfrac14}=\tfrac45$; $\cos2x=\dfrac{1-\tfrac14}{1+\tfrac14}=\tfrac35$.
4 $\sin30^\circ=\tfrac12$.
5 $\cos30^\circ=\tfrac{\sqrt3}{2}$.
6 $\tan45^\circ=1$.
7 $\dfrac{2\sin^2x}{2\sin x\cos x}=\dfrac{\sin x}{\cos x}=\tan x$.
8 $\dfrac{2\sin x\cos x}{2\cos^2x}=\tan x$.
9 $(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)=\cos2x\cdot1$.
II
Half angles and triple angles.
  • 1 $\cos x=\tfrac13$ with $\tfrac{3\pi}{2}\lt x\lt2\pi$. Find $\sin\tfrac x2$, $\cos\tfrac x2$, $\tan\tfrac x2$.
  • 2 Use $\cos^2x=\tfrac{1+\cos2x}{2}$ to find $\cos15^\circ$, and show your answer equals $\tfrac{\sqrt3+1}{2\sqrt2}$.
  • 3 Find $\tan67\tfrac12^\circ$.
  • 4 Evaluate $3\sin20^\circ-4\sin^320^\circ$.
  • 5 Evaluate $4\cos^340^\circ-3\cos40^\circ$.
  • 6 Prove that $\dfrac{\sin3x}{\sin x}-\dfrac{\cos3x}{\cos x}=2$, where defined.
  • 7 If $\tan x=\tfrac12$, find $\tan3x$.
  • 8 Using $\sin18^\circ=\tfrac{\sqrt5-1}{4}$, find $\cos36^\circ$.
1 $\tfrac{3\pi}{4}\lt\tfrac x2\lt\pi$: Quadrant II. $\sin\tfrac x2=\sqrt{\tfrac{1-1/3}{2}}=\tfrac{1}{\sqrt3}$, $\cos\tfrac x2=-\sqrt{\tfrac{1+1/3}{2}}=-\sqrt{\tfrac23}$, $\tan\tfrac x2=-\tfrac{1}{\sqrt2}$.
2 $\cos^215^\circ=\tfrac{1+\sqrt3/2}{2}=\tfrac{2+\sqrt3}{4}$, so $\cos15^\circ=\tfrac12\sqrt{2+\sqrt3}$. And $\left(\tfrac{\sqrt3+1}{2\sqrt2}\right)^2=\tfrac{4+2\sqrt3}{8}=\tfrac{2+\sqrt3}{4}$; both are positive, so they are equal.
3 $\tan67\tfrac12^\circ=\dfrac{1-\cos135^\circ}{\sin135^\circ}=\dfrac{1+\tfrac1{\sqrt2}}{\tfrac1{\sqrt2}}=\sqrt2+1$.
4 $\sin60^\circ=\tfrac{\sqrt3}{2}$.
5 $\cos120^\circ=-\tfrac12$.
6 $\dfrac{\sin3x\cos x-\cos3x\sin x}{\sin x\cos x}=\dfrac{\sin2x}{\sin x\cos x}=\dfrac{2\sin x\cos x}{\sin x\cos x}=2$.
7 $\dfrac{\tfrac32-\tfrac18}{1-\tfrac34}=\dfrac{11/8}{1/4}=\tfrac{11}{2}$.
8 $\cos36^\circ=1-2\sin^218^\circ=1-2\cdot\tfrac{6-2\sqrt5}{16}=\tfrac{1+\sqrt5}{4}$.
III
Products and sums.
  • 1 Write $2\sin5x\cos3x$ as a sum.
  • 2 Write $2\cos4x\cos x$ as a sum.
  • 3 Write $2\sin3x\sin x$ as a difference.
  • 4 Factorise $\sin7x+\sin3x$.
  • 5 Factorise $\cos6x-\cos2x$.
  • 6 Evaluate $\sin105^\circ+\sin15^\circ$.
  • 7 Evaluate $\cos105^\circ-\cos15^\circ$.
  • 8 Prove that $\dfrac{\sin x+\sin3x}{\cos x+\cos3x}=\tan2x$, where defined.
  • 9 Prove that $\dfrac{\sin3x+\sin5x+\sin7x}{\cos3x+\cos5x+\cos7x}=\tan5x$, where defined.
Check: product ↔ sum at any pair of angles
left side—
right side—
$x$, $y$—
1 $\sin8x+\sin2x$.
2 $\cos5x+\cos3x$.
3 $\cos2x-\cos4x$.
4 $2\sin5x\cos2x$.
5 $-2\sin4x\sin2x$.
6 $2\sin60^\circ\cos45^\circ=2\cdot\tfrac{\sqrt3}{2}\cdot\tfrac{1}{\sqrt2}=\tfrac{\sqrt6}{2}$.
7 $-2\sin60^\circ\sin45^\circ=-\tfrac{\sqrt6}{2}$.
8 $\dfrac{2\sin2x\cos x}{2\cos2x\cos x}=\tan2x$.
9 $\sin3x+\sin7x=2\sin5x\cos2x$ and $\cos3x+\cos7x=2\cos5x\cos2x$. The quotient is $\dfrac{\sin5x(2\cos2x+1)}{\cos5x(2\cos2x+1)}=\tan5x$.
IV
Applications and harder problems.
  • 1 Prove that $\sin20^\circ\sin40^\circ\sin80^\circ=\tfrac{\sqrt3}{8}$.
  • 2 Projectiles. The range of a ball thrown at speed $u$ and angle $\theta$ is $R=\dfrac{u^2\sin2\theta}{g}$. Show that $\theta$ and $90^\circ-\theta$ give the same range. For $u=20$ m/s and $g=9.8\ \text{m/s}^2$, find $R$ at $\theta=30^\circ$ and the greatest possible range.
  • 3 Beats. Two tuning forks produce $y=\sin(512\pi t)+\sin(520\pi t)$ ($t$ in seconds). Write $y$ as a product and find the number of beats (swells of loudness) per second.
  • 4 Prove that $\cos x\cos\left(\tfrac{\pi}{3}-x\right)\cos\left(\tfrac{\pi}{3}+x\right)=\tfrac14\cos3x$.
  • 5 If $\sin x+\sin y=a$ and $\cos x+\cos y=b$ with $b\ne0$, show that $\tan\tfrac{x+y}{2}=\tfrac ab$.
  • 6 Express $\sin^4x$ in terms of $\cos2x$ and $\cos4x$.
1 $2\sin40^\circ\sin80^\circ=\cos40^\circ-\cos120^\circ=\cos40^\circ+\tfrac12$. Multiply by $\sin20^\circ$: $\sin20^\circ\cos40^\circ+\tfrac12\sin20^\circ=\tfrac12(\sin60^\circ-\sin20^\circ)+\tfrac12\sin20^\circ=\tfrac12\sin60^\circ$. So $2\times$ (product) $=\tfrac{\sqrt3}{4}$, and the product is $\tfrac{\sqrt3}{8}$.
2 $\sin2(90^\circ-\theta)=\sin(180^\circ-2\theta)=\sin2\theta$. At $30^\circ$: $R=\tfrac{400\times\sqrt3/2}{9.8}\approx35.3$ m. Greatest when $\sin2\theta=1$, $\theta=45^\circ$: $R=\tfrac{400}{9.8}\approx40.8$ m.
3 $y=2\sin(516\pi t)\cos(4\pi t)$. The loudness follows $|2\cos4\pi t|$, which peaks when $4\pi t$ is a multiple of $\pi$, i.e. every $\tfrac14$ s: $4$ beats per second (the difference of the frequencies $260$ Hz and $256$ Hz).
4 $\cos\left(\tfrac{\pi}{3}-x\right)\cos\left(\tfrac{\pi}{3}+x\right)=\tfrac12\left(\cos\tfrac{2\pi}{3}+\cos2x\right)=\tfrac12\left(\cos2x-\tfrac12\right)$. Times $\cos x$: $\tfrac12\cos2x\cos x-\tfrac14\cos x=\tfrac14(\cos3x+\cos x)-\tfrac14\cos x=\tfrac14\cos3x$.
5 $a=2\sin\tfrac{x+y}{2}\cos\tfrac{x-y}{2}$, $b=2\cos\tfrac{x+y}{2}\cos\tfrac{x-y}{2}$. Since $b\ne0$, $\cos\tfrac{x-y}{2}\ne0$ and we may divide: $\tfrac ab=\tan\tfrac{x+y}{2}$.
6 $\sin^4x=\left(\tfrac{1-\cos2x}{2}\right)^2=\tfrac14\left(1-2\cos2x+\tfrac{1+\cos4x}{2}\right)=\tfrac18(3-4\cos2x+\cos4x)$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: this is the most heavily tested part of the chapter. Board papers carry 2–5-mark identity proofs using $\sin2x$, $\cos2x$, and sum–product conversions, plus half-angle values. JEE favours products of cosines with doubling angles, values like $\sin18^\circ$, and equations reduced with product–sum formulae.

E1. Board · 1 mark If $\cos x=\tfrac13$, then $\cos2x$ equals (A) $-\tfrac79$   (B) $\tfrac79$   (C) $\tfrac23$   (D) $-\tfrac13$

Idea Use the form with $\cos x$: $\cos2x=2\cos^2x-1=\tfrac29-1=-\tfrac79$.
Trap (C) is "$2\cos x$" — the classic $\cos2x=2\cos x$ error. (B) has the sign of $1-2\cos^2x$, which is $-\cos2x$.
Answer (A) $-\tfrac79$.

E2. Board · 2 marks Prove that $\dfrac{\sin2x}{1-\cos2x}=\cot x$, where defined.

Rewrite $\sin2x=2\sin x\cos x$ and $1-\cos2x=2\sin^2x$. (1 mark)
Answer $\dfrac{2\sin x\cos x}{2\sin^2x}=\dfrac{\cos x}{\sin x}=\cot x$. $\blacksquare$ (1 mark)

E3. Board · 3 marks Prove that $1+\cos2x+\cos4x+\cos6x=4\cos x\cos2x\cos3x$.

Group $(1+\cos6x)+(\cos2x+\cos4x)=2\cos^23x+2\cos3x\cos x$, using $1+\cos2\theta=2\cos^2\theta$ and Theorem 3.13. (1 mark)
Factor $=2\cos3x(\cos3x+\cos x)$. (1 mark)
Answer $\cos3x+\cos x=2\cos2x\cos x$, so the expression is $2\cos3x\cdot2\cos2x\cos x=4\cos x\cos2x\cos3x$. $\blacksquare$ (1 mark)

E4. Board · Case study In a school sports meet a ball is kicked at $20$ m/s at an angle $\theta$ to the ground; its range is $R=\dfrac{u^2\sin2\theta}{g}$ with $g=10\ \text{m/s}^2$. (i) Find $R$ for $\theta=15^\circ$ and for $\theta=75^\circ$. (2 marks) (ii) What is the greatest range, and at what angle? (1 mark) (iii) Find the two angles that give a range of $20\sqrt3$ m. (1 mark)

(i) $\tfrac{u^2}{g}=\tfrac{400}{10}=40$ m. $\theta=15^\circ$: $40\sin30^\circ=20$ m. $\theta=75^\circ$: $40\sin150^\circ=20$ m — the same, since $\sin(180^\circ-30^\circ)=\sin30^\circ$. (2 marks)
(ii) $\sin2\theta\le1$, with equality at $2\theta=90^\circ$: greatest range $40$ m at $\theta=45^\circ$. (1 mark)
Answer (iii) $40\sin2\theta=20\sqrt3$ gives $\sin2\theta=\tfrac{\sqrt3}{2}$; with $0^\circ\lt2\theta\lt180^\circ$, $2\theta=60^\circ$ or $120^\circ$: $\theta=30^\circ$ or $60^\circ$. (1 mark)

E5. JEE Main The value of $\cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ$ is (A) $\tfrac1{16}$   (B) $\tfrac18$   (C) $\tfrac{\sqrt3}{16}$   (D) $\tfrac14$

Idea By Example 6, $\cos20^\circ\cos40^\circ\cos80^\circ=\tfrac18$; and $\cos60^\circ=\tfrac12$.
Trap (B) is the product without $\cos60^\circ$. (C) uses $\sin60^\circ=\tfrac{\sqrt3}{2}$ in place of $\cos60^\circ=\tfrac12$.
Answer (A) $\tfrac18\cdot\tfrac12=\tfrac1{16}$. ⏱ Fast route: $\cos\theta\cos(60^\circ-\theta)\cos(60^\circ+\theta)=\tfrac14\cos3\theta$ (Exercise IV 4); at $\theta=20^\circ$ it gives $\cos20^\circ\cos40^\circ\cos80^\circ=\tfrac14\cos60^\circ=\tfrac18$.

E6. JEE Advanced Numerical answer. Find the number of solutions of $\sin x+\sin3x=0$ in the interval $[0,2\pi]$.

Factorise $\sin3x+\sin x=2\sin2x\cos x=0$, so $\sin2x=0$ or $\cos x=0$.
Solve $\sin2x=0$: $2x=n\pi$, $x=\tfrac{n\pi}{2}$, giving $0,\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2},2\pi$ in $[0,2\pi]$. $\cos x=0$: $x=\tfrac{\pi}{2},\tfrac{3\pi}{2}$ — already counted.
Answer $5$. Trap: counting $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}$ twice gives $7$.
🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. Can you hear a product–sum formula? How does the beat rate of two notes depend on how far apart they are?

Why it matters. Musicians tune a guitar or a tanpura by listening for beats and adjusting until they disappear; piano tuners do the same, and engineers use the same principle in radio receivers ("heterodyning") and in ultrasound Doppler scanners that measure blood flow.

What to do.

  1. Use a free tone-generator app (or two phones) to play two pure tones together: $440$ Hz with $441$, $442$, $444$ and $448$ Hz.
  2. For each pair, count the swells of loudness in $10$ seconds (record a video and count from it if the beats are fast).
  3. Write $\sin(2\pi f_1t)+\sin(2\pi f_2t)$ as a product using Theorem 3.13 and predict the number of beats per second.
  4. Tabulate predicted against counted beats and plot one against the other.

Tools: Tone-generator app, phone stopwatch or video, spreadsheet

Hand in: The table and graph, the product form of the sum, and a sentence explaining why the beat rate is $|f_1-f_2|$ and not $\tfrac12|f_1-f_2|$.

Going further. What happens when $f_2$ is $20$ Hz or more away from $f_1$? Why does the ear stop hearing "beats" then?

Instructor note: the envelope $2\cos\left(\pi(f_1-f_2)t\right)$ has period $\tfrac{2}{|f_1-f_2|}$, but loudness depends on its absolute value, which peaks twice per period — hence $|f_1-f_2|$ beats per second. Beyond roughly $15$–$20$ Hz difference the beats merge into a rough "two-note" sound.

🧪 Self-check quiz

0/5 correct

Q1. $\sin2x$ equals

Q2. Which of these is not equal to $\cos2x$?

Q3. $\sin3x$ equals

Q4. $\sin5x+\sin3x$ equals

Q5. $2\cos x\cos y$ equals

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