Trigonometric Functions • Topic 2 of 3

Trigonometric Functions and Identities

The six trigonometric functions are defined for any 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 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$$-$$+$$-$

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 formulae, with three equivalent forms for $\cos 2A$:

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

The last two forms of $\cos 2A$ rearrange into the half-angle identities $\sin^2 A = \dfrac{1 - \cos 2A}{2}$ and $\cos^2 A = \dfrac{1 + \cos 2A}{2}$.

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 $(\cos\theta, \sin\theta)$. 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-angle formulae are not new either: they are the sum formulae with $B$ replaced by $A$, and the three faces of $\cos 2A$ come from substituting $\sin^2 = 1 - \cos^2$ into one another. Even the signs across quadrants are not arbitrary rules to cram — they simply 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 $\tan\theta$.

Key Points

  • On the unit circle $\cos\theta = x$, $\sin\theta = y$; hence $-1 \le \sin\theta, \cos\theta \le 1$.
  • Signs by quadrant follow "All Silver Tea Cups": all $+$ in Q1, only $\sin$ in Q2, only $\tan$ in Q3, only $\cos$ in Q4.
  • 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 $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$.
  • Double angle: $\sin 2A = 2\sin A\cos A$ and $\cos 2A = 1 - 2\sin^2 A = 2\cos^2 A - 1$.
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.