Chapter 3 · Trigonometric Functions

3.3 Sum and Difference Formulae

Is $\sin(x+y)$ equal to $\sin x+\sin y$? No — and the correct formulae, proved here from one distance calculation on the unit circle, let you find exact values such as $\sin75^\circ$, reduce any angle to an acute one, and prove a whole family of identities.

You know $\sin30^\circ=\tfrac12$ and $\sin45^\circ=\tfrac{1}{\sqrt2}$. What is $\sin75^\circ$? It is tempting to add — but $\tfrac12+\tfrac1{\sqrt2}\approx1.21$ is bigger than $1$, which no sine can be. Sine is not "additive", and finding the right rule for $\sin(x+y)$ and $\cos(x+y)$ is the goal of this section. These sum and difference formulae are the engine of trigonometry: every identity in Section 3.4, the derivatives of $\sin x$ and $\cos x$ in Chapter 12, and the way engineers combine two alternating currents or two sound waves all come from them.

We prove one formula — for $\cos(x-y)$ — directly from the unit circle using the distance formula, and then obtain all the others from it by clever substitutions. Along the way we get the allied-angle rules, such as $\sin(\pi-x)=\sin x$ and $\cos\left(\tfrac{\pi}{2}+x\right)=-\sin x$, which reduce the sine or cosine of any angle to that of an acute angle.

The Cosine of a Difference

📐 Theorem 3.6 — Cosine of a difference

For all real numbers $x$ and $y$,

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

Proof. On the unit circle take $P_1=(\cos x,\sin x)$ and $P_2=(\cos y,\sin y)$, and also $A=(1,0)$ and $P_3=(\cos(x-y),\sin(x-y))$. The arc from $P_2$ to $P_1$ and the arc from $A$ to $P_3$ both correspond to the angle $x-y$: turning the whole picture about $O$ through the angle $-y$ carries $P_2$ to $A$ and $P_1$ to $P_3$. A rotation does not change distances, so $P_1P_2=AP_3$. By the distance formula,

$$P_1P_2^{\,2}=(\cos x-\cos y)^2+(\sin x-\sin y)^2=2-2(\cos x\cos y+\sin x\sin y),$$

using $\cos^2x+\sin^2x=1$ and $\cos^2y+\sin^2y=1$. Similarly

$$AP_3^{\,2}=(\cos(x-y)-1)^2+\sin^2(x-y)=2-2\cos(x-y).$$

Equating the two: $2-2\cos(x-y)=2-2(\cos x\cos y+\sin x\sin y)$, which is the formula. $\blacksquare$

Everything else follows by substitution. Write $-y$ for $y$ and use $\cos(-y)=\cos y$, $\sin(-y)=-\sin y$; put $x=\tfrac{\pi}{2}$; and so on:

📐 Theorem 3.7 — Sum and difference formulae for sine and cosine

For all real $x$ and $y$:

$$\cos(x+y)=\cos x\cos y-\sin x\sin y,$$ $$\cos\left(\tfrac{\pi}{2}-x\right)=\sin x,\qquad \sin\left(\tfrac{\pi}{2}-x\right)=\cos x,$$ $$\sin(x+y)=\sin x\cos y+\cos x\sin y,$$ $$\sin(x-y)=\sin x\cos y-\cos x\sin y .$$

Proof. (1) Replace $y$ by $-y$ in Theorem 3.6: $\cos(x+y)=\cos x\cos(-y)+\sin x\sin(-y)=\cos x\cos y-\sin x\sin y$.
(2) Put $\tfrac{\pi}{2}$ for $x$ and $x$ for $y$ in Theorem 3.6: $\cos\left(\tfrac{\pi}{2}-x\right)=0\cdot\cos x+1\cdot\sin x=\sin x$. Applying this to $\tfrac{\pi}{2}-x$ in place of $x$: $\sin\left(\tfrac{\pi}{2}-x\right)=\cos\left(\tfrac{\pi}{2}-\left(\tfrac{\pi}{2}-x\right)\right)=\cos x$.
(3) $\sin(x+y)=\cos\left(\left(\tfrac{\pi}{2}-x\right)-y\right)=\cos\left(\tfrac{\pi}{2}-x\right)\cos y+\sin\left(\tfrac{\pi}{2}-x\right)\sin y=\sin x\cos y+\cos x\sin y$.
(4) Replace $y$ by $-y$ in (3). $\blacksquare$

✏️ Example 1 — Exact values from $30^\circ$ and $45^\circ$

Find the exact values of (a) $\cos15^\circ$ (b) $\sin75^\circ$ (c) $\sin15^\circ$.

💡 Show solution▶

(a) $\cos15^\circ=\cos(45^\circ-30^\circ)=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ=\tfrac{1}{\sqrt2}\cdot\tfrac{\sqrt3}{2}+\tfrac{1}{\sqrt2}\cdot\tfrac12=\dfrac{\sqrt3+1}{2\sqrt2}$.

(b) $\sin75^\circ=\sin(45^\circ+30^\circ)=\tfrac{1}{\sqrt2}\cdot\tfrac{\sqrt3}{2}+\tfrac{1}{\sqrt2}\cdot\tfrac12=\dfrac{\sqrt3+1}{2\sqrt2}$ — the same as $\cos15^\circ$, as it must be, since $\sin(90^\circ-x)=\cos x$.

(c) $\sin15^\circ=\sin(45^\circ-30^\circ)=\tfrac{1}{\sqrt2}\cdot\tfrac{\sqrt3}{2}-\tfrac{1}{\sqrt2}\cdot\tfrac12=\dfrac{\sqrt3-1}{2\sqrt2}$.

Check $\dfrac{\sqrt3+1}{2\sqrt2}\approx0.9659$ and $\dfrac{\sqrt3-1}{2\sqrt2}\approx0.2588$, matching a calculator; and $0.9659^2+0.2588^2\approx1$ ✓. (Rationalised: $\dfrac{\sqrt6\pm\sqrt2}{4}$.)

The black segment $OP$ has length $1$ and makes the angle $A+B$ with the $x$-axis, so the height of $P$ is $\sin(A+B)$. Build it in two steps: walk $\cos B$ along a ray at angle $A$ to $Q$, then $\sin B$ at right angles to $P$. The height of $Q$ is $\cos B\cdot\sin A$ (red), and the climb from $Q$ to $P$ is $\sin B\cdot\cos A$ (blue) — because the angle at $P$ is again $A$. Switch to cosine to see the horizontal version, where the second piece is subtracted.

Show:A:B:

red piece—

blue piece—

formula—

direct value—

✏️ Example 2 — Sum and difference from given values

If $\sin x=\tfrac35$ and $\cos y=-\tfrac{12}{13}$, where $x$ and $y$ both lie in the second quadrant, find $\sin(x+y)$, $\cos(x+y)$ and $\cos(x-y)$.

💡 Show solution▶

Missing values In Quadrant II cosine is negative and sine positive: $\cos x=-\sqrt{1-\tfrac{9}{25}}=-\tfrac45$ and $\sin y=\sqrt{1-\tfrac{144}{169}}=\tfrac{5}{13}$.

$\sin(x+y)$ $\sin x\cos y+\cos x\sin y=\tfrac35\left(-\tfrac{12}{13}\right)+\left(-\tfrac45\right)\tfrac{5}{13}=-\tfrac{36}{65}-\tfrac{20}{65}=-\tfrac{56}{65}$.

$\cos(x\pm y)$ $\cos(x+y)=\cos x\cos y-\sin x\sin y=\tfrac{48}{65}-\tfrac{15}{65}=\tfrac{33}{65}$; $\ \cos(x-y)=\tfrac{48}{65}+\tfrac{15}{65}=\tfrac{63}{65}$.

Check $\left(\tfrac{56}{65}\right)^2+\left(\tfrac{33}{65}\right)^2=\tfrac{3136+1089}{4225}=1$ ✓. Also $x+y$ lies between $\pi$ and $2\pi$, and $\sin(x+y)\lt0$ agrees.

Allied Angles

Angles such as $\tfrac{\pi}{2}\pm x$, $\pi\pm x$, $\tfrac{3\pi}{2}\pm x$ and $2\pi-x$ are called allied angles of $x$. Putting $\tfrac{\pi}{2}$, $\pi$ or $\tfrac{3\pi}{2}$ into the sum and difference formulae (where $\sin$ and $\cos$ of these angles are $0$ or $\pm1$) gives:

📐 Theorem 3.8 — Allied-angle formulae

For every real $x$ (and wherever the tangent is defined):

angle	$\sin$	$\cos$	$\tan$
$-x$	$-\sin x$	$\cos x$	$-\tan x$
$\tfrac{\pi}{2}-x$	$\cos x$	$\sin x$	$\cot x$
$\tfrac{\pi}{2}+x$	$\cos x$	$-\sin x$	$-\cot x$
$\pi-x$	$\sin x$	$-\cos x$	$-\tan x$
$\pi+x$	$-\sin x$	$-\cos x$	$\tan x$
$\tfrac{3\pi}{2}-x$	$-\cos x$	$-\sin x$	$\cot x$
$\tfrac{3\pi}{2}+x$	$-\cos x$	$\sin x$	$-\cot x$
$2\pi-x$	$-\sin x$	$\cos x$	$-\tan x$

Proof (two rows; the rest are similar). $\sin(\pi-x)=\sin\pi\cos x-\cos\pi\sin x=0-(-1)\sin x=\sin x$. $\ \cos\left(\tfrac{\pi}{2}+x\right)=\cos\tfrac{\pi}{2}\cos x-\sin\tfrac{\pi}{2}\sin x=-\sin x$. The tangent column is the sine column divided by the cosine column. $\blacksquare$

🧭 Problem-solving strategy — Reducing any angle to an acute one

Remove whole turns ($2n\pi$ or $n\cdot360^\circ$); for a negative angle use $\sin(-x)=-\sin x$, $\cos(-x)=\cos x$.
Write the angle as $\pi\pm x$ or $2\pi-x$ (the function name stays the same), or as $\tfrac{\pi}{2}\pm x$ or $\tfrac{3\pi}{2}\pm x$ (the name swaps: $\sin\leftrightarrow\cos$, $\tan\leftrightarrow\cot$), with $x$ acute.
Fix the sign by the quadrant of the original angle (ASTC), pretending $x$ is acute.

Example: $\cos210^\circ=\cos(180^\circ+30^\circ)$; the name stays; $210^\circ$ is in Quadrant III where cosine is negative: $-\cos30^\circ=-\tfrac{\sqrt3}{2}$.

$P=(\cos x,\sin x)$ is the blue point. Choose an allied angle: its point $P'$ (red) is a reflection of $P$ in a line (for $-x$, $\tfrac{\pi}{2}-x$, $\pi-x$, $\tfrac{3\pi}{2}-x$, $2\pi-x$) or a rotation of $P$ about $O$ (for $\tfrac{\pi}{2}+x$, $\pi+x$, $\tfrac{3\pi}{2}+x$). Read the coordinates of $P'$ to get its cosine and sine in terms of those of $P$.

Allied angle:x (°):

$x$—

allied angle—

its cosine—

its sine—

✏️ Example 3 — Values at any angle

Find (a) $\sin150^\circ$ (b) $\cos210^\circ$ (c) $\tan315^\circ$ (d) $\cos(-1710^\circ)$ (e) $\sin\tfrac{29\pi}{6}$.

💡 Show solution▶

(a), (b) $\sin150^\circ=\sin(180^\circ-30^\circ)=\sin30^\circ=\tfrac12$. $\ \cos210^\circ=\cos(180^\circ+30^\circ)=-\cos30^\circ=-\tfrac{\sqrt3}{2}$.

(d) $\cos(-1710^\circ)=\cos1710^\circ$ and $1710^\circ=4\times360^\circ+270^\circ$, so the value is $\cos270^\circ=0$.

(e) $\tfrac{29\pi}{6}=4\pi+\tfrac{5\pi}{6}$, and $\sin\tfrac{5\pi}{6}=\sin\left(\pi-\tfrac{\pi}{6}\right)=\sin\tfrac{\pi}{6}=\tfrac12$.

Tangent and Cotangent of a Sum

📐 Theorem 3.9 — Sum and difference formulae for tangent and cotangent

If none of $x$, $y$, $x+y$ is an odd multiple of $\tfrac{\pi}{2}$, then

$$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y};$$

if none of $x$, $y$, $x-y$ is an odd multiple of $\tfrac{\pi}{2}$, then

$$\tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}.$$

If none of $x$, $y$, $x\pm y$ is a multiple of $\pi$, then

$$\cot(x\pm y)=\frac{\cot x\cot y\mp1}{\cot y\pm\cot x}.$$

Proof. $\tan(x+y)=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$. The conditions guarantee $\cos x\cos y\ne0$ and $\cos(x+y)\ne0$; divide the numerator and the denominator by $\cos x\cos y$ to get the formula. Replacing $y$ by $-y$ (and using $\tan(-y)=-\tan y$) gives $\tan(x-y)$. For the cotangent, divide $\dfrac{\cos x\cos y\mp\sin x\sin y}{\sin x\cos y\pm\cos x\sin y}$ by $\sin x\sin y$. $\blacksquare$

A useful special case, with $x=\tfrac{\pi}{4}$ and $\tan\tfrac{\pi}{4}=1$: $\ \tan\left(\tfrac{\pi}{4}+x\right)=\dfrac{1+\tan x}{1-\tan x}$ and $\tan\left(\tfrac{\pi}{4}-x\right)=\dfrac{1-\tan x}{1+\tan x}$.

✏️ Example 4 — Exact tangents

Find (a) $\tan15^\circ$ (b) $\tan75^\circ$ (c) $\tan\tfrac{13\pi}{12}$.

💡 Show solution▶

(a) $\tan(45^\circ-30^\circ)=\dfrac{1-\tfrac{1}{\sqrt3}}{1+\tfrac{1}{\sqrt3}}=\dfrac{\sqrt3-1}{\sqrt3+1}=\dfrac{(\sqrt3-1)^2}{3-1}=\dfrac{4-2\sqrt3}{2}=2-\sqrt3$.

(b) $\tan(45^\circ+30^\circ)=\dfrac{\sqrt3+1}{\sqrt3-1}=\dfrac{(\sqrt3+1)^2}{2}=2+\sqrt3$.

Check $\tan15^\circ\cdot\tan75^\circ=(2-\sqrt3)(2+\sqrt3)=1$, as it must be since $\tan(90^\circ-x)=\cot x$ ✓.

✏️ Example 5 — Proving identities

Prove that (a) $\cos\left(\tfrac{\pi}{4}+x\right)+\cos\left(\tfrac{\pi}{4}-x\right)=\sqrt2\cos x$ (b) $\sin(x+y)\sin(x-y)=\sin^2x-\sin^2y$ (c) $\tan3x\tan2x\tan x=\tan3x-\tan2x-\tan x$, wherever all the tangents are defined.

💡 Show solution▶

(a) Expand both: $\left(\cos\tfrac{\pi}{4}\cos x-\sin\tfrac{\pi}{4}\sin x\right)+\left(\cos\tfrac{\pi}{4}\cos x+\sin\tfrac{\pi}{4}\sin x\right)=2\cdot\tfrac{1}{\sqrt2}\cos x=\sqrt2\cos x$.

(b) $(\sin x\cos y+\cos x\sin y)(\sin x\cos y-\cos x\sin y)=\sin^2x\cos^2y-\cos^2x\sin^2y$. Replace $\cos^2y=1-\sin^2y$ and $\cos^2x=1-\sin^2x$: $\ \sin^2x-\sin^2x\sin^2y-\sin^2y+\sin^2x\sin^2y=\sin^2x-\sin^2y$.

(c) $3x=2x+x$, so $\tan3x=\dfrac{\tan2x+\tan x}{1-\tan2x\tan x}$. Cross-multiply: $\tan3x-\tan3x\tan2x\tan x=\tan2x+\tan x$. Rearranging: $\tan3x\tan2x\tan x=\tan3x-\tan2x-\tan x$.

Check (c) at $x=10^\circ$: $\tan30^\circ\tan20^\circ\tan10^\circ\approx0.03705$, and $\tan30^\circ-\tan20^\circ-\tan10^\circ\approx0.03705$ ✓.

✏️ Example 6 — Two angles that add up to $45^\circ$

If $\tan A=\tfrac12$ and $\tan B=\tfrac13$, where $A$ and $B$ are acute, show that $A+B=\tfrac{\pi}{4}$.

💡 Show solution▶

Tangent of the sum $\tan(A+B)=\dfrac{\tfrac12+\tfrac13}{1-\tfrac12\cdot\tfrac13}=\dfrac{5/6}{5/6}=1$.

Locate the angle $A$, $B$ acute gives $0\lt A+B\lt\pi$. Moreover $\tan A\lt1$ and $\tan B\lt1$ mean $A,B\lt\tfrac{\pi}{4}$, so $0\lt A+B\lt\tfrac{\pi}{2}$. The only angle there with tangent $1$ is $\tfrac{\pi}{4}$.

Answer $A+B=\tfrac{\pi}{4}$. (A neat consequence: $\tfrac{\pi}{4}$ splits into two angles with "nice" tangents $\tfrac12$ and $\tfrac13$.)

⚠️ Common mistakes with the sum formulae

1. "Distributing" sine. $\sin(x+y)\ne\sin x+\sin y$. Test at $x=y=\tfrac{\pi}{2}$: the left side is $\sin\pi=0$, the right side is $2$.

2. The sign in $\cos(x+y)$. The cosine formulae have the opposite sign: $\cos(x+y)=\cos x\cos y\,\mathbf{-}\,\sin x\sin y$.

3. Mixed-up allied angles. $\cos\left(\tfrac{\pi}{2}+x\right)=-\sin x$, not $\sin x$. With $\tfrac{\pi}{2}$ or $\tfrac{3\pi}{2}$ the name changes; with $\pi$ or $2\pi$ it does not. The sign comes from the quadrant of the whole angle.

4. Ignoring the conditions on $\tan(x+y)$. At $x=y=\tfrac{\pi}{4}$ the formula gives $\tfrac{2}{0}$ — correct, because $\tan\tfrac{\pi}{2}$ is undefined. Never use the formula when $x$ or $y$ is an odd multiple of $\tfrac{\pi}{2}$.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026 (an application of the trigonometric identities and addition formulae both list); not in the CBSE year-end exam. JEE Main in particular frequently asks for the greatest and least values of $a\sin x+b\cos x$.

Method. Let $R=\sqrt{a^2+b^2}\gt0$ and choose $\alpha$ with $\cos\alpha=\tfrac{a}{R}$, $\sin\alpha=\tfrac{b}{R}$ (possible because $\left(\tfrac aR\right)^2+\left(\tfrac bR\right)^2=1$). Then $a\sin x+b\cos x=R(\sin x\cos\alpha+\cos x\sin\alpha)=R\sin(x+\alpha)$. Since $\sin(x+\alpha)$ takes every value in $[-1,1]$, the expression takes every value in $[-R,R]$.

Worked example. Find the range of $f(x)=3\sin x+4\cos x+2$.
Solution. $R=\sqrt{9+16}=5$, so $3\sin x+4\cos x=5\sin(x+\alpha)$ with $\cos\alpha=\tfrac35$, $\sin\alpha=\tfrac45$, and it ranges over $[-5,5]$. Hence $f$ ranges over $[-3,7]$: greatest value $7$ (when $x+\alpha=\tfrac{\pi}{2}$, i.e. $\sin x=\tfrac35$, $\cos x=\tfrac45$), least value $-3$.

Exercises

Exact values with the sum and difference formulae.

1 $\cos105^\circ$
2 $\tan105^\circ$
3 $\sin\tfrac{\pi}{12}$
4 $\cos\tfrac{5\pi}{12}$
5 $\cos75^\circ\cos15^\circ-\sin75^\circ\sin15^\circ$
6 $\sin40^\circ\cos20^\circ+\cos40^\circ\sin20^\circ$
7 $\dfrac{\tan66^\circ+\tan69^\circ}{1-\tan66^\circ\tan69^\circ}$
8 $\sin\tfrac{7\pi}{12}\cos\tfrac{\pi}{4}-\cos\tfrac{7\pi}{12}\sin\tfrac{\pi}{4}$

💡 Show solutions▶

1 $\cos(60^\circ+45^\circ)=\tfrac12\cdot\tfrac1{\sqrt2}-\tfrac{\sqrt3}{2}\cdot\tfrac1{\sqrt2}=\dfrac{1-\sqrt3}{2\sqrt2}$ ($\approx-0.2588$).

2 $\tan(60^\circ+45^\circ)=\dfrac{\sqrt3+1}{1-\sqrt3}=-\dfrac{(\sqrt3+1)^2}{2}=-(2+\sqrt3)$.

3 $\sin15^\circ=\dfrac{\sqrt3-1}{2\sqrt2}$ (Example 1).

4 $\cos75^\circ=\cos(45^\circ+30^\circ)=\dfrac{\sqrt3-1}{2\sqrt2}$.

5 $\cos(75^\circ+15^\circ)=\cos90^\circ=0$.

6 $\sin(40^\circ+20^\circ)=\sin60^\circ=\tfrac{\sqrt3}{2}$.

7 $\tan(66^\circ+69^\circ)=\tan135^\circ=-1$.

8 $\sin\left(\tfrac{7\pi}{12}-\tfrac{\pi}{4}\right)=\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$.

Allied angles.

1 $\sin120^\circ$
2 $\cos135^\circ$
3 $\tan240^\circ$
4 $\sec(-300^\circ)$
5 $\sin\tfrac{11\pi}{6}$
6 $\cot\left(-\tfrac{3\pi}{4}\right)$
7 $\sin(-690^\circ)$
8 Simplify $\dfrac{\sin(\pi-x)\cos\left(\tfrac{\pi}{2}+x\right)}{\cos(\pi+x)\sin\left(\tfrac{3\pi}{2}-x\right)}$.
9 Use the sum formula to show that $\cos\left(\tfrac{3\pi}{2}+x\right)=\sin x$.

💡 Show solutions▶

1 $\sin(180^\circ-60^\circ)=\sin60^\circ=\tfrac{\sqrt3}{2}$.

2 $\cos(180^\circ-45^\circ)=-\cos45^\circ=-\tfrac{1}{\sqrt2}$.

3 $\tan(180^\circ+60^\circ)=\tan60^\circ=\sqrt3$.

4 $\sec(-300^\circ)=\sec300^\circ=\dfrac{1}{\cos(360^\circ-60^\circ)}=\dfrac{1}{\cos60^\circ}=2$.

5 $\sin\left(2\pi-\tfrac{\pi}{6}\right)=-\sin\tfrac{\pi}{6}=-\tfrac12$.

6 $\cot\left(-\tfrac{3\pi}{4}\right)=-\cot\tfrac{3\pi}{4}=-\cot\left(\pi-\tfrac{\pi}{4}\right)=-\left(-\cot\tfrac{\pi}{4}\right)=1$.

7 $\sin(-690^\circ)=-\sin690^\circ=-\sin(720^\circ-30^\circ)=-(-\sin30^\circ)=\tfrac12$.

8 $\dfrac{\sin x\cdot(-\sin x)}{(-\cos x)(-\cos x)}=-\tan^2x$.

9 $\cos\tfrac{3\pi}{2}\cos x-\sin\tfrac{3\pi}{2}\sin x=0\cdot\cos x-(-1)\sin x=\sin x$.

III

Given values and identities.

1 $\sin x=\tfrac45$ and $\cos y=\tfrac{5}{13}$, with $x$, $y$ acute. Find $\sin(x+y)$ and $\cos(x-y)$.
2 $\tan x=\tfrac34$ and $\tan y=\tfrac17$. Find $\tan(x+y)$ and $\tan(x-y)$.
3 Prove that $\cos(x+y)\cos(x-y)=\cos^2x-\sin^2y$, and hence find $\cos^245^\circ-\sin^215^\circ$.
4 Prove that $\sin5x\sin4x+\cos5x\cos4x=\cos x$.
5 Prove that $\tan\left(\tfrac{\pi}{4}+x\right)\tan\left(\tfrac{\pi}{4}-x\right)=1$, where defined.
6 Prove that $\dfrac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\left(\tfrac{\pi}{2}+x\right)}=\cot^2x$.
7 Prove that $\dfrac{\tan\left(\tfrac{\pi}{4}+x\right)}{\tan\left(\tfrac{\pi}{4}-x\right)}=\left(\dfrac{1+\tan x}{1-\tan x}\right)^2$.
8 Prove that $\dfrac{\sin(x-y)}{\cos x\cos y}+\dfrac{\sin(y-z)}{\cos y\cos z}+\dfrac{\sin(z-x)}{\cos z\cos x}=0$.
9 If $\tan A=\tfrac56$ and $\tan B=\tfrac1{11}$ with $A$, $B$ acute, show that $A+B=\tfrac{\pi}{4}$.

Check: test an identity at any angle

Identity:x (°):

left side—

right side—

difference—

💡 Show solutions▶

1 $\cos x=\tfrac35$, $\sin y=\tfrac{12}{13}$. $\sin(x+y)=\tfrac45\cdot\tfrac5{13}+\tfrac35\cdot\tfrac{12}{13}=\tfrac{56}{65}$; $\cos(x-y)=\tfrac35\cdot\tfrac5{13}+\tfrac45\cdot\tfrac{12}{13}=\tfrac{63}{65}$.

2 $\tan(x+y)=\dfrac{\tfrac34+\tfrac17}{1-\tfrac{3}{28}}=\dfrac{25/28}{25/28}=1$; $\ \tan(x-y)=\dfrac{\tfrac34-\tfrac17}{1+\tfrac3{28}}=\dfrac{17/28}{31/28}=\tfrac{17}{31}$.

3 $(\cos x\cos y)^2-(\sin x\sin y)^2=\cos^2x(1-\sin^2y)-(1-\cos^2x)\sin^2y=\cos^2x-\sin^2y$. With $x=45^\circ$, $y=15^\circ$: $\cos60^\circ\cos30^\circ=\tfrac12\cdot\tfrac{\sqrt3}{2}=\tfrac{\sqrt3}{4}$.

4 The left side is $\cos5x\cos4x+\sin5x\sin4x=\cos(5x-4x)=\cos x$.

5 $\dfrac{1+\tan x}{1-\tan x}\cdot\dfrac{1-\tan x}{1+\tan x}=1$.

6 $\dfrac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}=\dfrac{\cos^2x}{\sin^2x}=\cot^2x$.

7 $\dfrac{(1+\tan x)/(1-\tan x)}{(1-\tan x)/(1+\tan x)}=\left(\dfrac{1+\tan x}{1-\tan x}\right)^2$.

8 $\dfrac{\sin(x-y)}{\cos x\cos y}=\dfrac{\sin x\cos y-\cos x\sin y}{\cos x\cos y}=\tan x-\tan y$. Likewise the others are $\tan y-\tan z$ and $\tan z-\tan x$; the three add to $0$.

9 $\tan(A+B)=\dfrac{\tfrac56+\tfrac1{11}}{1-\tfrac{5}{66}}=\dfrac{61/66}{61/66}=1$; both tangents are less than $1$, so $0\lt A+B\lt\tfrac{\pi}{2}$, and $A+B=\tfrac{\pi}{4}$.

Applications and harder problems.

1 $\cos x=-\tfrac35$ with $x$ in Quadrant II, and $\sin y=-\tfrac{12}{13}$ with $y$ in Quadrant III. Find $\sin(x+y)$ and $\cos(x+y)$, and decide in which quadrant $x+y$ lies (up to whole turns).
2 Hill roads. Two straight roads climb from the same point with gradients $2$ and $\tfrac13$ (so $\tan\theta_1=2$, $\tan\theta_2=\tfrac13$). Find the angle between them.
3 Flagpole. A building $10$ m tall has a $10$ m flagpole on its roof. From a point on level ground $20$ m from the foot of the building, find the tangent of the angle subtended by the flagpole.
4 Show that $\cos\left(\tfrac{\pi}{4}-x\right)\cos\left(\tfrac{\pi}{4}-y\right)-\sin\left(\tfrac{\pi}{4}-x\right)\sin\left(\tfrac{\pi}{4}-y\right)=\sin(x+y)$.
5 Show that $\sin x+\sqrt3\cos x=2\sin\left(x+\tfrac{\pi}{3}\right)$ for every real $x$.

💡 Show solutions▶

1 $\sin x=\tfrac45$, $\cos y=-\tfrac5{13}$. $\sin(x+y)=\tfrac45\left(-\tfrac5{13}\right)+\left(-\tfrac35\right)\left(-\tfrac{12}{13}\right)=\tfrac{16}{65}$; $\ \cos(x+y)=\left(-\tfrac35\right)\left(-\tfrac5{13}\right)-\tfrac45\left(-\tfrac{12}{13}\right)=\tfrac{63}{65}$. Both positive: $x+y$ lies in Quadrant I (up to whole turns).

2 $\tan(\theta_1-\theta_2)=\dfrac{2-\tfrac13}{1+\tfrac23}=1$, and $0\lt\theta_1-\theta_2\lt\tfrac{\pi}{2}$, so the roads meet at $45^\circ$.

3 The top of the pole is seen at angle $\alpha$ with $\tan\alpha=\tfrac{20}{20}=1$; the roof at angle $\beta$ with $\tan\beta=\tfrac{10}{20}=\tfrac12$. $\tan(\alpha-\beta)=\dfrac{1-\tfrac12}{1+\tfrac12}=\tfrac13$.

4 The left side is $\cos\left(\left(\tfrac{\pi}{4}-x\right)+\left(\tfrac{\pi}{4}-y\right)\right)=\cos\left(\tfrac{\pi}{2}-(x+y)\right)=\sin(x+y)$.

5 $2\sin\left(x+\tfrac{\pi}{3}\right)=2\left(\sin x\cdot\tfrac12+\cos x\cdot\tfrac{\sqrt3}{2}\right)=\sin x+\sqrt3\cos x$.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask for exact values ($\sin75^\circ$, $\tan15^\circ$), allied-angle simplifications, and 3–5-mark identity proofs built on the sum formulae. JEE tests fluency: recognising $\cos A\cos B+\sin A\sin B$ instantly, splitting an angle as a sum, and using $\tan(A+B)$ with a condition such as $A+B=\tfrac{\pi}{4}$.

E1. Board · 1 mark The value of $\cos68^\circ\cos8^\circ+\sin68^\circ\sin8^\circ$ is (A) $\tfrac12$ (B) $\tfrac{\sqrt3}{2}$ (C) $\tfrac1{\sqrt2}$ (D) $0$

💡 Show solution▶

Idea This is $\cos(68^\circ-8^\circ)=\cos60^\circ$.

Trap (B) is $\sin60^\circ$ — it comes from mixing up sine and cosine at the last step. Remember that "$\cos\cos+\sin\sin$" is the cosine of the difference, not of the sum $76^\circ$.

Answer (A) $\tfrac12$.

E2. Board · 2 marks Prove that $\tan\left(\tfrac{\pi}{4}+x\right)=\dfrac{1+\tan x}{1-\tan x}$, and hence find $\tan75^\circ$.

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Proof $\tan\left(\tfrac{\pi}{4}+x\right)=\dfrac{\tan\tfrac{\pi}{4}+\tan x}{1-\tan\tfrac{\pi}{4}\tan x}=\dfrac{1+\tan x}{1-\tan x}$. (1 mark)

Answer With $x=30^\circ$: $\tan75^\circ=\dfrac{1+\tfrac1{\sqrt3}}{1-\tfrac1{\sqrt3}}=\dfrac{\sqrt3+1}{\sqrt3-1}=2+\sqrt3$. (1 mark)

E3. Board · 3 marks Prove that $\dfrac{\sin(x+y)-2\sin x\cos y}{2\sin x\sin y+\cos(x+y)}=\tan(y-x)$.

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Numerator $\sin x\cos y+\cos x\sin y-2\sin x\cos y=\cos x\sin y-\sin x\cos y=\sin(y-x)$. (1 mark)

Denominator $2\sin x\sin y+\cos x\cos y-\sin x\sin y=\cos x\cos y+\sin x\sin y=\cos(x-y)=\cos(y-x)$. (1 mark)

Answer The quotient is $\dfrac{\sin(y-x)}{\cos(y-x)}=\tan(y-x)$. $\blacksquare$ (1 mark)

E4. Board · Case study A $15$ m advertising hoarding stands on the roof of a $30$ m building. A surveyor stands on level ground $30$ m from the foot of the building and sees the top of the hoarding at an angle of elevation $\alpha$ and its bottom at an angle $\beta$. (i) Find $\tan\alpha$ and $\tan\beta$. (2 marks) (ii) Find the tangent of the angle $\alpha-\beta$ subtended by the hoarding. (1 mark) (iii) Is this angle more or less than $15^\circ$? Justify. (1 mark)

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(i) $\tan\alpha=\tfrac{45}{30}=\tfrac32$, $\ \tan\beta=\tfrac{30}{30}=1$. (2 marks)

(ii) $\tan(\alpha-\beta)=\dfrac{\tfrac32-1}{1+\tfrac32}=\dfrac{1/2}{5/2}=\tfrac15$. (1 mark)

Answer (iii) $\tan15^\circ=2-\sqrt3\approx0.268\gt0.2$, and tangent increases on $\left(0,\tfrac{\pi}{2}\right)$, so $\alpha-\beta\lt15^\circ$ (it is about $11.3^\circ$). (1 mark)

E5. JEE Main If $A+B=\tfrac{\pi}{4}$ (with $\tan A$, $\tan B$ defined), then $(1+\tan A)(1+\tan B)$ equals (A) $1$ (B) $2$ (C) $0$ (D) $\sqrt2$

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Idea $\tan(A+B)=1$ gives $\tan A+\tan B=1-\tan A\tan B$, i.e. $\tan A+\tan B+\tan A\tan B=1$.

Expand $(1+\tan A)(1+\tan B)=1+(\tan A+\tan B+\tan A\tan B)=1+1=2$.

Answer (B). ⏱ Fast route: try $A=0$, $B=\tfrac{\pi}{4}$: $(1+0)(1+1)=2$. Trap: (D) comes from guessing with $\sin\tfrac{\pi}{4}$-type numbers.

E6. JEE Advanced One or more options correct. Let $0\lt\beta\lt\alpha\lt\tfrac{\pi}{4}$ with $\cos(\alpha+\beta)=\tfrac35$ and $\sin(\alpha-\beta)=\tfrac{5}{13}$. Which are true? (A) $\tan2\alpha=\tfrac{63}{16}$ (B) $\tan2\beta=\tfrac{33}{56}$ (C) $\sin2\alpha=\tfrac{63}{65}$ (D) $\cos2\beta=\tfrac{33}{65}$

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Idea $2\alpha=(\alpha+\beta)+(\alpha-\beta)$ and $2\beta=(\alpha+\beta)-(\alpha-\beta)$. Since $0\lt\alpha+\beta\lt\tfrac{\pi}{2}$ and $0\lt\alpha-\beta\lt\tfrac{\pi}{4}$: $\sin(\alpha+\beta)=\tfrac45$, $\cos(\alpha-\beta)=\tfrac{12}{13}$, so $\tan(\alpha+\beta)=\tfrac43$, $\tan(\alpha-\beta)=\tfrac5{12}$.

(A), (B) $\tan2\alpha=\dfrac{\tfrac43+\tfrac5{12}}{1-\tfrac{5}{9}}=\dfrac{21/12}{4/9}=\tfrac{63}{16}$ ✓. $\ \tan2\beta=\dfrac{\tfrac43-\tfrac5{12}}{1+\tfrac59}=\dfrac{11/12}{14/9}=\tfrac{33}{56}$ ✓.

(C), (D) $\sin2\alpha=\tfrac45\cdot\tfrac{12}{13}+\tfrac35\cdot\tfrac5{13}=\tfrac{63}{65}$ ✓. $\ \cos2\beta=\tfrac35\cdot\tfrac{12}{13}+\tfrac45\cdot\tfrac5{13}=\tfrac{56}{65}$, so (D) is false — $\tfrac{33}{65}$ is $\sin2\beta$.

Answer (A), (B), (C).

🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. Where should you stand to see a hoarding, a statue on a pedestal, or a cinema screen at the largest angle?

Why it matters. Architects place statues and screens, and advertisers place hoardings, so that viewers at the intended distance see them as large as possible. The angle subtended is $\alpha-\beta$, and its tangent comes straight from Theorem 3.9.

What to do.

Choose a real object above eye level: a notice board on a wall, a clock on a tower, a projector screen. Measure (or estimate) the heights $a$ and $b$ of its bottom and top above your eyes.
Standing at distance $d$, the angle subtended has $\tan(\alpha-\beta)=\dfrac{\tfrac bd-\tfrac ad}{1+\tfrac{ab}{d^2}}=\dfrac{(b-a)d}{d^2+ab}$. Derive this formula yourself.
In a spreadsheet, tabulate this for $d=0.5,1,1.5,\dots$ metres and find the $d$ that makes it largest.
Compare your best $d$ with $\sqrt{ab}$. Stand there and check that the object does look biggest.

Tools: Measuring tape, phone clinometer app (optional), spreadsheet

Hand in: Your measurements, the table and graph of the angle against $d$, and a comparison of your best distance with $\sqrt{ab}$.

Going further. Why does standing very close or very far make the angle tiny? Explain using the formula.

Instructor note: $\dfrac{(b-a)d}{d^2+ab}$ is largest at $d=\sqrt{ab}$ (by A.M. ≥ G.M., Chapter 8, applied to $d+\tfrac{ab}{d}$). For $a=1$ m, $b=4$ m the best distance is $2$ m and $\tan(\alpha-\beta)=\tfrac34$.

🧪 Self-check quiz

0/5 correct

Q1. $\sin(x+y)$ equals

Q2. $\cos15^\circ$ equals

Q3. $\cos\left(\tfrac{\pi}{2}+x\right)$ equals

Q4. $\tan\left(\tfrac{\pi}{4}-x\right)$ equals

Q5. $\sin50^\circ\cos10^\circ+\cos50^\circ\sin10^\circ$ equals

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