Trigonometric Functions • Topic 2 of 3

Trigonometric Functions and Identities

The six trigonometric functions are defined for any real number (angle) using the unit circle (radius $1$, centred at the origin). If the terminal side of angle $\theta$ meets the circle at the point $P(x, y)$, then:

$$\cos\theta = x, \qquad \sin\theta = y, \qquad \tan\theta = \dfrac{y}{x}\ (x \ne 0)$$

The remaining three are reciprocals: $\csc\theta = \dfrac{1}{\sin\theta}$, $\sec\theta = \dfrac{1}{\cos\theta}$, $\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$. Because $x$ and $y$ never exceed $1$ in magnitude on the unit circle, $-1 \le \sin\theta \le 1$ and $-1 \le \cos\theta \le 1$ always. The functions are periodic: $\sin$, $\cos$, $\csc$, $\sec$ repeat every $2\pi$, while $\tan$ and $\cot$ repeat every $\pi$.

Domain and range of all six functions (where $n \in \mathbb{Z}$):

FunctionDomainRange
$\sin\theta$$\mathbb{R}$$[-1, 1]$
$\cos\theta$$\mathbb{R}$$[-1, 1]$
$\tan\theta$$\mathbb{R} \setminus \{(2n+1)\tfrac{\pi}{2}\}$$\mathbb{R}$
$\cot\theta$$\mathbb{R} \setminus \{n\pi\}$$\mathbb{R}$
$\sec\theta$$\mathbb{R} \setminus \{(2n+1)\tfrac{\pi}{2}\}$$(-\infty,-1] \cup [1,\infty)$
$\csc\theta$$\mathbb{R} \setminus \{n\pi\}$$(-\infty,-1] \cup [1,\infty)$

The sign of each function depends only on the signs of $x$ and $y$ in that quadrant — remembered as "All Silver Tea Cups" (All positive in Q1, Sin in Q2, Tan in Q3, Cos in Q4):

QuadrantPositive functions$\sin$$\cos$$\tan$
I ($0$ to $90^\circ$)all$+$$+$$+$
II ($90$ to $180^\circ$)$\sin,\csc$$+$$-$$-$
III ($180$ to $270^\circ$)$\tan,\cot$$-$$-$$+$
IV ($270$ to $360^\circ$)$\cos,\sec$$-$$+$$-$

Values at standard angles are the backbone of every calculation. Memorise the $\sin$ row; the $\cos$ row is the same read backwards, and $\tan = \sin/\cos$:

$\theta$$0$$\dfrac{\pi}{6}$$\dfrac{\pi}{4}$$\dfrac{\pi}{3}$$\dfrac{\pi}{2}$
$\sin\theta$$0$$\dfrac{1}{2}$$\dfrac{1}{\sqrt2}$$\dfrac{\sqrt3}{2}$$1$
$\cos\theta$$1$$\dfrac{\sqrt3}{2}$$\dfrac{1}{\sqrt2}$$\dfrac{1}{2}$$0$
$\tan\theta$$0$$\dfrac{1}{\sqrt3}$$1$$\sqrt3$$\infty$

Allied angles. Angles such as $-\theta$, $\dfrac{\pi}{2}\pm\theta$, $\pi\pm\theta$, $2\pi-\theta$ are "allied" to $\theta$ and their function values follow two rules. For odd multiples of $\dfrac{\pi}{2}$ (i.e. $\dfrac{\pi}{2}\pm\theta$, $\dfrac{3\pi}{2}\pm\theta$) the function changes to its co-function ($\sin\leftrightarrow\cos$, $\tan\leftrightarrow\cot$, $\sec\leftrightarrow\csc$); for even multiples ($\pi\pm\theta$, $2\pi\pm\theta$) the function stays the same. The sign is then fixed by the ASTC quadrant rule. Two consequences used constantly: $\sin(-\theta) = -\sin\theta$ (odd), $\cos(-\theta) = \cos\theta$ (even), and $\sin\!\left(\dfrac{\pi}{2}-\theta\right) = \cos\theta$.

From $x^2 + y^2 = 1$ on the unit circle come the three Pythagorean identities:

$$\sin^2\theta + \cos^2\theta = 1, \qquad 1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \csc^2\theta$$

The sum and difference formulae let you break a compound angle apart:

$$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$$
$$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$$
$$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$$

Setting $B = A$ gives the double-angle (multiple-angle) formulae, with three equivalent forms for $\cos 2A$:

$$\sin 2A = 2\sin A\cos A = \dfrac{2\tan A}{1+\tan^2 A}$$
$$\cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A = 2\cos^2 A - 1 = \dfrac{1-\tan^2 A}{1+\tan^2 A}$$
$$\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$$

The triple-angle formulae extend the pattern to $3A$:

$$\sin 3A = 3\sin A - 4\sin^3 A, \qquad \cos 3A = 4\cos^3 A - 3\cos A, \qquad \tan 3A = \dfrac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$$

The two faces of $\cos 2A$ rearrange into the sub-multiple (half-angle) identities, writing $A$ in terms of $\dfrac{A}{2}$:

$$\sin^2\dfrac{A}{2} = \dfrac{1 - \cos A}{2}, \qquad \cos^2\dfrac{A}{2} = \dfrac{1 + \cos A}{2}, \qquad \sin A = \dfrac{2\tan\frac{A}{2}}{1+\tan^2\frac{A}{2}}, \qquad \cos A = \dfrac{1-\tan^2\frac{A}{2}}{1+\tan^2\frac{A}{2}}$$

Product-to-sum formulae turn a product of two functions into a sum, which is essential before integrating or simplifying:

$$2\sin A\cos B = \sin(A+B) + \sin(A-B), \qquad 2\cos A\sin B = \sin(A+B) - \sin(A-B)$$
$$2\cos A\cos B = \cos(A+B) + \cos(A-B), \qquad 2\sin A\sin B = \cos(A-B) - \cos(A+B)$$

Reading them backwards gives the sum-to-product formulae (let $C = A+B$, $D = A-B$):

$$\sin C + \sin D = 2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}, \qquad \sin C - \sin D = 2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$$
$$\cos C + \cos D = 2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2}, \qquad \cos C - \cos D = -2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$$

Deeper Insight — one definition generates the entire formula sheet: Students often try to memorise dozens of identities as separate facts, but they all descend from a single source — the unit-circle point $(\cos\theta, \sin\theta)$ together with the two sum formulae. The Pythagorean identities are literally just $x^2 + y^2 = 1$ rewritten, and dividing that one equation by $\cos^2\theta$ or $\sin^2\theta$ produces the other two for free. The double-, triple- and half-angle formulae are not new either: they are the sum formula with $B$ chosen as $A$, as $2A$, or read in reverse. Product-to-sum and sum-to-product are simply $\sin(A+B)$ and $\sin(A-B)$ added or subtracted. Even the signs across quadrants and the allied-angle rules are not arbitrary — they read off whether $x$ and $y$ are positive or negative where the terminal side lands. If you internalise the unit circle and the two sum formulae, you can reconstruct the rest in seconds under exam pressure, which is far safer than recalling a memorised list and hoping you got a sign right.

Unit circle defining cosine and sine as coordinates Unit Circle: P = (cosθ, sinθ) θ P cosθ sinθ ASTC Q1: all +Q2: sin +Q3: tan +Q4: cos + Graphs of sine and cosine over one period y = sin x (orange) and y = cos x (blue) π1−1
1
Worked Example
Given $\sin\theta = \dfrac{3}{5}$ and $\theta$ lies in the second quadrant, find $\cos\theta$ and $\tan\theta$.
Solution
  1. Use $\cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$, so $\cos\theta = \pm\dfrac{4}{5}$.
  2. In Quadrant II cosine is negative, so $\cos\theta = -\dfrac{4}{5}$.
  3. $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}$.

Answer: $\cos\theta = -\dfrac{4}{5}$, $\tan\theta = -\dfrac{3}{4}$.

2
Worked Example
Find the exact value of $\cos 15^\circ$.
Solution
  1. Write $15^\circ = 45^\circ - 30^\circ$ and use $\cos(A - B) = \cos A\cos B + \sin A\sin B$.
  2. $\cos 15^\circ = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ$.
  3. $= \dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}}\cdot\dfrac{1}{2} = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}$.
  4. Rationalise: $\dfrac{\sqrt{3} + 1}{2\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.

Answer: $\cos 15^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.

3
Worked Example
Prove that $\dfrac{1 + \tan^2\theta}{1 + \cot^2\theta} = \tan^2\theta$.
Solution
  1. Use the Pythagorean identities: numerator $= \sec^2\theta$, denominator $= \csc^2\theta$.
  2. $\dfrac{\sec^2\theta}{\csc^2\theta} = \dfrac{1/\cos^2\theta}{1/\sin^2\theta} = \dfrac{\sin^2\theta}{\cos^2\theta}$.
  3. $= \tan^2\theta$, which equals the right-hand side.

Answer: Identity proved: both sides equal $\tan^2\theta$.

4
Worked Example
If $\tan A = \dfrac{1}{2}$ and $\tan B = \dfrac{1}{3}$, find $\tan(A + B)$.
Solution
  1. Apply $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$.
  2. Numerator: $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$.
  3. Denominator: $1 - \dfrac{1}{2}\cdot\dfrac{1}{3} = 1 - \dfrac{1}{6} = \dfrac{5}{6}$.
  4. $\tan(A + B) = \dfrac{5/6}{5/6} = 1$.

Answer: $\tan(A + B) = 1$ (so $A + B = 45^\circ$).

5
Worked Example
If $\cos\theta = \dfrac{3}{5}$ with $\theta$ in the first quadrant, find $\sin 2\theta$ and $\cos 2\theta$.
Solution
  1. In Q1, $\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$.
  2. $\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\dfrac{4}{5}\cdot\dfrac{3}{5} = \dfrac{24}{25}$.
  3. $\cos 2\theta = 2\cos^2\theta - 1 = 2\cdot\dfrac{9}{25} - 1 = \dfrac{18}{25} - 1 = -\dfrac{7}{25}$.

Answer: $\sin 2\theta = \dfrac{24}{25}$, $\cos 2\theta = -\dfrac{7}{25}$.

6
Worked Example
Prove that $\dfrac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$.
Solution
  1. Rewrite the numerator with the double-angle formula: $\sin 2\theta = 2\sin\theta\cos\theta$.
  2. Rewrite the denominator using $\cos 2\theta = 2\cos^2\theta - 1$, so $1 + \cos 2\theta = 2\cos^2\theta$.
  3. $\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \dfrac{\sin\theta}{\cos\theta}$.
  4. $= \tan\theta$, matching the right-hand side.

Answer: Identity proved: the expression simplifies to $\tan\theta$.

7
Worked Example
Using allied angles, evaluate $\sin\dfrac{31\pi}{3}$.
Solution
  1. Sine has period $2\pi$, so subtract whole multiples of $2\pi = \dfrac{6\pi}{3}$: $\dfrac{31\pi}{3} - 5\times 2\pi = \dfrac{31\pi}{3} - \dfrac{30\pi}{3} = \dfrac{\pi}{3}$.
  2. Thus $\sin\dfrac{31\pi}{3} = \sin\dfrac{\pi}{3}$.
  3. $\sin\dfrac{\pi}{3} = \dfrac{\sqrt3}{2}$.

Answer: $\sin\dfrac{31\pi}{3} = \dfrac{\sqrt3}{2}$.

8
Worked Example
Find the exact value of $\tan 75^\circ$ using a sum formula.
Solution
  1. Write $75^\circ = 45^\circ + 30^\circ$ and use $\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$.
  2. $\tan 75^\circ = \dfrac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ\tan 30^\circ} = \dfrac{1 + \frac{1}{\sqrt3}}{1 - \frac{1}{\sqrt3}}$.
  3. Multiply top and bottom by $\sqrt3$: $\dfrac{\sqrt3 + 1}{\sqrt3 - 1}$.
  4. Rationalise by $(\sqrt3+1)$: $\dfrac{(\sqrt3+1)^2}{(\sqrt3)^2 - 1^2} = \dfrac{3 + 2\sqrt3 + 1}{2} = \dfrac{4 + 2\sqrt3}{2} = 2 + \sqrt3$.

Answer: $\tan 75^\circ = 2 + \sqrt3$.

9
Worked Example
Express $2\sin 5\theta\cos 3\theta$ as a sum of two sines.
Solution
  1. Use the product-to-sum rule $2\sin A\cos B = \sin(A+B) + \sin(A-B)$ with $A = 5\theta$, $B = 3\theta$.
  2. $A + B = 8\theta$ and $A - B = 2\theta$.
  3. So the product becomes $\sin 8\theta + \sin 2\theta$.

Answer: $2\sin 5\theta\cos 3\theta = \sin 8\theta + \sin 2\theta$.

10
Worked Example
Prove that $\dfrac{\sin 5\theta + \sin 3\theta}{\cos 5\theta + \cos 3\theta} = \tan 4\theta$.
Solution
  1. Apply sum-to-product. Numerator: $\sin 5\theta + \sin 3\theta = 2\sin\dfrac{5\theta+3\theta}{2}\cos\dfrac{5\theta-3\theta}{2} = 2\sin 4\theta\cos\theta$.
  2. Denominator: $\cos 5\theta + \cos 3\theta = 2\cos\dfrac{5\theta+3\theta}{2}\cos\dfrac{5\theta-3\theta}{2} = 2\cos 4\theta\cos\theta$.
  3. Divide: $\dfrac{2\sin 4\theta\cos\theta}{2\cos 4\theta\cos\theta} = \dfrac{\sin 4\theta}{\cos 4\theta} = \tan 4\theta$.

Answer: Identity proved: the ratio simplifies to $\tan 4\theta$.

11
Worked Example
If $\sin A = \dfrac{3}{5}$ and $\cos B = \dfrac{9}{41}$, both $A$ and $B$ acute, find $\sin(A+B)$.
Solution
  1. Since $A$ is acute, $\cos A = \sqrt{1 - \tfrac{9}{25}} = \dfrac{4}{5}$.
  2. Since $B$ is acute, $\sin B = \sqrt{1 - \tfrac{81}{1681}} = \sqrt{\tfrac{1600}{1681}} = \dfrac{40}{41}$.
  3. $\sin(A+B) = \sin A\cos B + \cos A\sin B = \dfrac{3}{5}\cdot\dfrac{9}{41} + \dfrac{4}{5}\cdot\dfrac{40}{41}$.
  4. $= \dfrac{27}{205} + \dfrac{160}{205} = \dfrac{187}{205}$.

Answer: $\sin(A+B) = \dfrac{187}{205}$.

12
Worked Example
Using the triple-angle formula, find $\sin 3\theta$ when $\sin\theta = \dfrac{1}{2}$.
Solution
  1. Use $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$.
  2. Substitute $\sin\theta = \dfrac{1}{2}$: $\sin^3\theta = \dfrac{1}{8}$.
  3. $\sin 3\theta = 3\cdot\dfrac{1}{2} - 4\cdot\dfrac{1}{8} = \dfrac{3}{2} - \dfrac{1}{2} = 1$.
  4. (Consistent: $\sin\theta = \tfrac12$ at $\theta = 30^\circ$ gives $\sin 90^\circ = 1$.)

Answer: $\sin 3\theta = 1$.

13
Worked Example
Prove the half-angle result $\tan\dfrac{A}{2} = \dfrac{1 - \cos A}{\sin A}$.
Solution
  1. Use the sub-multiple forms $1 - \cos A = 2\sin^2\dfrac{A}{2}$ and $\sin A = 2\sin\dfrac{A}{2}\cos\dfrac{A}{2}$.
  2. Substitute: $\dfrac{1 - \cos A}{\sin A} = \dfrac{2\sin^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}$.
  3. Cancel $2\sin\dfrac{A}{2}$: $= \dfrac{\sin\frac{A}{2}}{\cos\frac{A}{2}} = \tan\dfrac{A}{2}$.

Answer: Proved: $\tan\dfrac{A}{2} = \dfrac{1 - \cos A}{\sin A}$.

Key Points

  • On the unit circle $\cos\theta = x$, $\sin\theta = y$; hence $-1 \le \sin\theta, \cos\theta \le 1$, and $\tan,\cot$ range over all of $\mathbb{R}$.
  • Signs by quadrant follow "All Silver Tea Cups": all $+$ in Q1, only $\sin$ in Q2, only $\tan$ in Q3, only $\cos$ in Q4.
  • Allied angles: at $\dfrac{\pi}{2}\pm\theta$ the function turns into its co-function; at $\pi\pm\theta$ it stays the same; the sign comes from ASTC.
  • Pythagorean identities: $\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$, $1 + \cot^2\theta = \csc^2\theta$.
  • Sum/difference: $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$ and $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$.
  • Multiple angles: $\sin 2A = 2\sin A\cos A$, $\cos 2A = 2\cos^2 A - 1$, $\sin 3A = 3\sin A - 4\sin^3 A$, $\cos 3A = 4\cos^3 A - 3\cos A$.
  • Half-angle: $\sin^2\dfrac{A}{2} = \dfrac{1-\cos A}{2}$, $\cos^2\dfrac{A}{2} = \dfrac{1+\cos A}{2}$.
  • Product-to-sum and sum-to-product convert between $2\sin A\cos B = \sin(A+B)+\sin(A-B)$ and $\sin C+\sin D = 2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$.
Tap an option to check your answer0 / 4
Q1.$\sin^2\theta+\cos^2\theta=$
Explanation: The fundamental Pythagorean identity.
Q2.$\sin\tfrac{\pi}{6}=$
Explanation: $\sin30^\circ=\tfrac12$.
Q3.$\cos 0=$
Explanation: $\cos0^\circ=1$.
Q4.$1+\tan^2\theta=$
Explanation: A Pythagorean identity.