Mastering Trigonometric Identities and Equations for Your Maths Exam

What the examiner is testing

The examiner is assessing your ability to manipulate trigonometric expressions using fundamental identities to prove equivalences or simplify, and to solve equations by transforming them into a solvable form. Marks are awarded for each correct application of an identity and for accurate algebraic steps in solving.

The method

1. Identify the type of problem: Determine if you need to prove an identity (show LHS = RHS) or solve an equation (find values for the variable).
2. For proving identities:
   • Start with the more complex side (usually the LHS) and aim to transform it into the simpler side.
   • Look for opportunities to use fundamental identities: \( \sin^2 \theta + \cos^2 \theta = 1 \), \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), \( \sec \theta = \frac{1}{\cos \theta} \), \( \csc \theta = \frac{1}{\sin \theta} \), \( \cot \theta = \frac{1}{\tan \theta} \).
   • Consider expressing everything in terms of \( \sin \theta \) and \( \cos \theta \).
   • Look for common denominators if fractions are involved.
   • Factorise or expand expressions where appropriate.
   • Apply double angle formulae (\( \sin 2A = 2 \sin A \cos A \), \( \cos 2A = \cos^2 A - \sin^2 A = 2 \cos^2 A - 1 = 1 - 2 \sin^2 A \), \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \)) or compound angle formulae (\( \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \), etc.) if angles like \( 2\theta \) or \( (x+30^\circ) \) are present.
3. For solving equations:
   • Rearrange the equation to isolate a single trigonometric function (e.g., \( \sin x = k \), \( \tan 2x = k \)).
   • If multiple trigonometric functions are present, try to express them in terms of a single function using identities (e.g., replace \( \sin^2 x \) with \( 1 - \cos^2 x \)).
   • If the equation is quadratic in a trigonometric function (e.g., \( 2 \cos^2 x + \cos x - 1 = 0 \)), substitute a dummy variable (e.g., \( y = \cos x \)) to solve the quadratic, then find the values of the trigonometric function.
   • Determine the principal value (the first solution in the specified range, often using the inverse trigonometric function on your calculator).
   • Use the symmetry of the trigonometric graphs and the given range to find all other solutions. Remember:
     • For \( \sin x = k \): \( x = \alpha \) and \( x = 180^\circ - \alpha \) (or \( \pi - \alpha \)).
     • For \( \cos x = k \): \( x = \alpha \) and \( x = 360^\circ - \alpha \) (or \( 2\pi - \alpha \)).
     • For \( \tan x = k \): \( x = \alpha \) and \( x = 180^\circ + \alpha \) (or \( \pi + \alpha \)).
   • If the angle is transformed (e.g., \( 2x \), \( x+30^\circ \)), adjust the range accordingly before finding solutions, then divide/subtract at the end.

Worked example

Prove the identity \( \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} \equiv 2 \csc x \).

$$ \begin{aligned} \text{LHS} &= \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} \\ &= \frac{\sin x (\sin x) + (1 + \cos x)(1 + \cos x)}{(1 + \cos x)(\sin x)} \\ &= \frac{\sin^2 x + (1 + 2 \cos x + \cos^2 x)}{(1 + \cos x)\sin x} \\ &= \frac{\sin^2 x + \cos^2 x + 1 + 2 \cos x}{(1 + \cos x)\sin x} \\ &= \frac{1 + 1 + 2 \cos x}{(1 + \cos x)\sin x} \quad (\text{since } \sin^2 x + \cos^2 x = 1) \\ &= \frac{2 + 2 \cos x}{(1 + \cos x)\sin x} \\ &= \frac{2(1 + \cos x)}{(1 + \cos x)\sin x} \\ &= \frac{2}{\sin x} \\ &= 2 \csc x \\ &= \text{RHS} \end{aligned} $$
The LHS transforms exactly into the RHS, thus the identity is proven.

Worked example: a harder one

Solve \( 2 \cos 2\theta + \sin \theta = 1 \) for \( 0^\circ \le \theta < 360^\circ \).

Initial thought: Replace \( \cos 2\theta \) with \( \cos^2 \theta - \sin^2 \theta \).
$$ 2(\cos^2 \theta - \sin^2 \theta) + \sin \theta = 1 $$
This leaves both \( \sin \theta \) and \( \cos \theta \) terms, which is hard to solve.

Better approach: Use the double angle identity for \( \cos 2\theta \) that only involves \( \sin \theta \), i.e., \( \cos 2\theta = 1 - 2 \sin^2 \theta \).
$$ \begin{aligned} 2 (1 - 2 \sin^2 \theta) + \sin \theta &= 1 \\ 2 - 4 \sin^2 \theta + \sin \theta &= 1 \\ 0 &= 4 \sin^2 \theta - \sin \theta - 1 \end{aligned} $$
This is a quadratic in \( \sin \theta \). Let \( y = \sin \theta \).
$$ 4y^2 - y - 1 = 0 $$
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
$$ \begin{aligned} y &= \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-1)}}{2(4)} \\ y &= \frac{1 \pm \sqrt{1 + 16}}{8} \\ y &= \frac{1 \pm \sqrt{17}}{8} \end{aligned} $$
So, \( \sin \theta = \frac{1 + \sqrt{17}}{8} \) or \( \sin \theta = \frac{1 - \sqrt{17}}{8} \).

For \( \sin \theta = \frac{1 + \sqrt{17}}{8} \approx \frac{1 + 4.123}{8} \approx 0.64038 \):
\( \theta = \arcsin(0.64038) \approx 39.83^\circ \) (to 2 decimal places).
Since \( \sin \theta \) is positive, solutions are in the 1st and 2nd quadrants.
\( \theta_1 = 39.83^\circ \)
\( \theta_2 = 180^\circ - 39.83^\circ = 140.17^\circ \)

For \( \sin \theta = \frac{1 - \sqrt{17}}{8} \approx \frac{1 - 4.123}{8} \approx -0.39038 \):
\( \theta = \arcsin(-0.39038) \approx -22.98^\circ \).
Since \( \sin \theta \) is negative, solutions are in the 3rd and 4th quadrants.
\( \theta_3 = 180^\circ - (-22.98^\circ) = 202.98^\circ \) (This is equivalent to \( 180^\circ + |\text{principal value}| \))
\( \theta_4 = 360^\circ + (-22.98^\circ) = 337.02^\circ \) (This is equivalent to \( 360^\circ - |\text{principal value}| \))

The solutions for \( 0^\circ \le \theta < 360^\circ \) are \( 39.83^\circ, 140.17^\circ, 202.98^\circ, 337.02^\circ \).

Practice

1. Prove the identity \( \tan \theta + \cot \theta \equiv \sec \theta \csc \theta \).
2. Solve \( 3 \tan x = 2 \cos x \) for \( 0 \le x < 2\pi \). Give your answers in radians to 3 significant figures.
3. Show that \( \frac{1 - \cos 2\theta}{\sin 2\theta} \equiv \tan \theta \).
4. Find all solutions to the equation \( 4 \sin^2 \theta + 5 \cos \theta = 5 \) for \( -180^\circ \le \theta < 180^\circ \).

Answers:
1. \( \text{LHS} = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta \sin \theta} = \frac{1}{\cos \theta \sin \theta} = \frac{1}{\cos \theta} \times \frac{1}{\sin \theta} = \sec \theta \csc \theta = \text{RHS} \)
2. \( x = 0.536, 2.61 \) radians
3. \( \text{LHS} = \frac{1 - (1 - 2\sin^2 \theta)}{2\sin \theta \cos \theta} = \frac{2\sin^2 \theta}{2\sin \theta \cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{RHS} \)
4. Working for Q4:
$$ \begin{aligned} 4 \sin^2 \theta + 5 \cos \theta &= 5 \\ 4(1 - \cos^2 \theta) + 5 \cos \theta &= 5 \\ 4 - 4 \cos^2 \theta + 5 \cos \theta &= 5 \\ 0 &= 4 \cos^2 \theta - 5 \cos \theta + 1 \end{aligned} $$
Let \( y = \cos \theta \).
\( 4y^2 - 5y + 1 = 0 \)
\( (4y - 1)(y - 1) = 0 \)
So, \( y = 1 \) or \( y = \frac{1}{4} \).
Case 1: \( \cos \theta = 1 \)
For \( -180^\circ \le \theta < 180^\circ \), \( \theta = 0^\circ \).
Case 2: \( \cos \theta = \frac{1}{4} \)
\( \theta = \arccos(0.25) \approx 75.52^\circ \) (to 2 decimal places).
Since \( \cos \theta \) is positive, solutions are in the 1st and 4th quadrants.
\( \theta_1 = 75.52^\circ \)
\( \theta_2 = -75.52^\circ \) (This is \( 360^\circ - 75.52^\circ \) within the negative range)
Solutions are \( -75.5^\circ, 0^\circ, 75.5^\circ \) (to 1 decimal place).

The three mistakes that lose marks

1. Dividing by a trigonometric function that could be zero: Forgetting that \( \sin x = 0 \) or \( \cos x = 0 \) can lead to lost solutions. For example, in \( \sin x \cos x = \sin x \), if you divide by \( \sin x \) you get \( \cos x = 1 \). This misses the solutions where \( \sin x = 0 \). The correct approach is to rearrange to \( \sin x \cos x - \sin x = 0 \), then factorise: \( \sin x (\cos x - 1) = 0 \), leading to \( \sin x = 0 \) or \( \cos x = 1 \).
2. Incorrectly applying inverse trigonometric functions for all solutions: Only using the principal value from a calculator and not considering the symmetry of the sine, cosine, and tangent graphs. For example, \( \sin x = 0.5 \) gives \( x = 30^\circ \) from a calculator. Forgetting the second solution \( x = 180^\circ - 30^\circ = 150^\circ \) (within \( 0^\circ \le x < 360^\circ \)) is a common error.
3. Mixing angle units (degrees and radians): Using \( \pi \) for \( 180^\circ \) in one part of a calculation and \( 360^\circ \) in another, or setting calculator to the wrong mode. For example, if the range is \( 0 \le x < 2\pi \) but you use \( 180^\circ - \alpha \) for sine solutions, you will get incorrect numerical values. Stick to one unit throughout the problem.

30-second recap

Identities prove equivalence, equations find specific values. Use fundamental identities to simplify or transform expressions. For equations, isolate a single trig function, find the principal value, then use graph symmetry and the given range to find all solutions. Avoid dividing by variables that could be zero.