Trig identities and harder equations
This topic combines two skills that lean on each other constantly: using two standard identities to simplify expressions and prove others are true, and solving trig equations that are more involved than the single-step \(\sin\theta=k\) equations from Trigonometric Ratios.
Five skills make up this topic:
- The two key identities — \(\sin^2\theta+\cos^2\theta\equiv1\) and \(\tan\theta\equiv\dfrac{\sin\theta}{\cos\theta}\), used to simplify and prove
- Multiple-angle equations — solving equations like \(\sin(2\theta)=k\)
- Shifted-angle equations — solving equations like \(\cos(\theta-40°)=k\)
- Quadratic-form equations — using an identity to turn a mixed sin/cos equation into a solvable quadratic
- Equations solved by tan substitution — turning a mixed sin/cos equation into a single tan equation
The two key identities
Every trig identity in this topic comes back to two facts, both provable from a right-angled triangle with hypotenuse 1: \(\sin^2\theta+\cos^2\theta\equiv1\) (the Pythagorean identity) and \(\tan\theta\equiv\dfrac{\sin\theta}{\cos\theta}\). "Prove" and "show that" questions almost always start by substituting one of these in.
Worked Example 1
Prove that \((\sin\theta+\cos\theta)^2 \equiv 1+2\sin\theta\cos\theta\).
1
Expand the left-hand side: \((\sin\theta+\cos\theta)^2 = \sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta\)
2
Group the squared terms: \(= (\sin^2\theta+\cos^2\theta)+2\sin\theta\cos\theta\)
3
Substitute \(\sin^2\theta+\cos^2\theta\equiv1\): \(= 1+2\sin\theta\cos\theta\), which is the right-hand side
AnswerLHS = RHS, as required
Worked Example 2
Show that \(\cos\theta\tan\theta \equiv \sin\theta\).
1
Substitute \(\tan\theta\equiv\dfrac{\sin\theta}{\cos\theta}\): \(\cos\theta\tan\theta = \cos\theta \times \dfrac{\sin\theta}{\cos\theta}\)
2
Cancel the common factor of \(\cos\theta\): \(= \sin\theta\)
AnswerLHS = RHS, as required
Multiple-angle equations
For an equation like \(\sin(2\theta)=k\), substitute \(u=2\theta\), solve for \(u\) over the correspondingly widened range, then convert back to \(\theta\) at the very end. Doubling the angle inside the function doubles how many solutions land inside \(0°\) to \(360°\).
Worked Example 3
Solve \(\sin(2\theta) = 0.5\) for \(0° \leq \theta \leq 360°\).
1
Let \(u=2\theta\). Since \(0°\leq\theta\leq360°\), the range for \(u\) is \(0°\leq u\leq720°\)
2
Solve \(\sin u = 0.5\) over this doubled range: \(u = 30°, 150°, 390°, 510°\)
3
Divide every solution by 2 to return to \(\theta\): \(\theta = 15°, 75°, 195°, 255°\)
Answer\(\theta = 15°, 75°, 195°,\) or \(255°\)
Shifted-angle equations
For an equation like \(\cos(\theta - 40°)=k\), the same substitution idea applies, but this time the range shifts rather than stretches: substitute \(u=\theta-40°\), and adjust the bounds by the same shift.
Worked Example 4
Solve \(\cos(\theta - 40°) = 0.6\) for \(0° \leq \theta \leq 360°\).
1
Let \(u=\theta-40°\). Since \(0°\leq\theta\leq360°\), the range for \(u\) is \(-40°\leq u\leq320°\)
2
Solve \(\cos u = 0.6\) over this shifted range: principal value \(u=53.1°\); the other solutions of \(\cos u=0.6\) are \(-53.1°\) and \(306.9°\), and only \(u=53.1°\) and \(u=306.9°\) fall inside \([-40°, 320°]\)
3
Add 40° back to each: \(\theta = 93.1°, 346.9°\)
Answer\(\theta = 93.1°\) or \(346.9°\) (1 d.p.)
Quadratic-form equations
When an equation mixes \(\sin^2\theta\) (or \(\cos^2\theta\)) with a single power of the other, substitute the Pythagorean identity to rewrite everything in terms of one function — the result is a quadratic you can factorise or solve normally.
Worked Example 5
Solve \(2\sin^2\theta + \cos\theta = 2\) for \(0° \leq \theta \leq 360°\).
1
Substitute \(\sin^2\theta \equiv 1-\cos^2\theta\): \(2(1-\cos^2\theta)+\cos\theta = 2\)
2
Expand and simplify: \(2-2\cos^2\theta+\cos\theta = 2 \implies 2\cos^2\theta - \cos\theta = 0\)
3
Factorise: \(\cos\theta(2\cos\theta-1)=0 \implies \cos\theta=0\) or \(\cos\theta=0.5\)
4
Solve each: \(\cos\theta=0 \implies \theta=90°,270°\); \(\cos\theta=0.5 \implies \theta=60°,300°\)
Answer\(\theta = 60°, 90°, 270°,\) or \(300°\)
Equations solved by tan substitution
An equation with both \(\sin\theta\) and \(\cos\theta\) but no squares can usually be solved fastest by dividing every term by \(\cos\theta\), turning \(\dfrac{\sin\theta}{\cos\theta}\) into \(\tan\theta\) and collapsing the equation to a single trig function.
Worked Example 6
Solve \(2\sin\theta = 3\cos\theta\) for \(0° \leq \theta \leq 360°\).
1
Divide both sides by \(\cos\theta\): \(2\dfrac{\sin\theta}{\cos\theta} = 3 \implies 2\tan\theta = 3\)
2
Rearrange: \(\tan\theta = 1.5\)
3
Principal value: \(\theta=\tan^{-1}(1.5)=56.3°\); tan has period \(180°\), so the second solution is \(56.3°+180°=236.3°\)
Answer\(\theta = 56.3°\) or \(236.3°\) (1 d.p.)
Exam tips
💡 Tip 1
Substitute a variable for multiple/shifted angles rather than guessing
Writing \(u=2\theta\) or \(u=\theta-40°\) as an explicit first line, and correctly widening or shifting the range for \(u\), is what prevents missed solutions — do this every time, even when it feels unnecessary for simple cases.
💡 Tip 2
Convert back to \(\theta\) as the very last step
Find every solution for \(u\) first, then divide/shift each one back — converting partway through is a common source of losing a solution.
💡 Tip 3
Never divide an equation by \(\sin\theta\) or \(\cos\theta\) if it could be zero
Dividing away a factor that could legitimately be zero silently deletes valid solutions — factorise instead of dividing whenever the equation could equal zero (see Worked Example 5).
💡 Tip 4
Check which identity turns two functions into one
A mix of \(\sin^2\) and \(\cos\) (or vice versa) signals the Pythagorean identity; a mix of plain \(\sin\theta\) and \(\cos\theta\) with no squares signals dividing through for \(\tan\theta\).
💡 Tip 5
Reject any "solution" for cosine or sine outside \([-1, 1]\)
Quadratic-form equations sometimes produce an algebraic root like \(\cos\theta=1.4\) — this has no real angle solution and must be discarded, not forced into a calculator.
Common mistakes
Common Mistake 1
Forgetting to widen or shift the solution range for \(u\)
For \(\sin(2\theta)=k\), the range for \(u=2\theta\) must become \(0°\) to \(720°\), not stay at \(0°\) to \(360°\) — using the un-widened range misses half the solutions.
Common Mistake 2
Dividing by \(\cos\theta\) or \(\sin\theta\) without checking it isn't zero
This silently discards any solution where that function equals zero — always factorise a quadratic-form equation instead of dividing through.
Common Mistake 3
Substituting the wrong identity
Using \(\tan\theta\equiv\sin\theta/\cos\theta\) when the equation actually needs \(\sin^2\theta+\cos^2\theta\equiv1\) (or vice versa) leads nowhere — identify which functions are mixed in the equation before choosing.
Common Mistake 4
Stopping after the principal value only
Even inside a substitution, every equation still has (typically) two solutions per \(360°\) cycle of \(u\) — apply the same symmetry rules used for simple equations at every stage.
Common Mistake 5
Converting solutions for \(u\) back to \(\theta\) incorrectly
For \(u=2\theta\), \(\theta=\dfrac{u}{2}\) — a common slip is dividing only the first solution and forgetting to divide the rest, or adding/subtracting instead of dividing.
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1Prove that \(\sin^2\theta - \cos^2\theta \equiv 1 - 2\cos^2\theta\).Core Skills
Q2Show that \(\dfrac{\sin\theta}{\tan\theta} \equiv \cos\theta\).Core Skills
Q3Solve \(\sin\theta = 0.3\) for \(0° \leq \theta \leq 360°\), to 1 decimal place.Core Skills
Q4Solve \(\cos\theta = -0.4\) for \(0° \leq \theta \leq 360°\), to 1 decimal place.Core Skills
Q5Simplify \(\sin\theta\tan\theta + \cos\theta\).Core Skills
Exam-Style
Q6Solve \(\sin(2\theta) = 0.6\) for \(0° \leq \theta \leq 360°\), to 1 decimal place.Exam-Style
Q7Solve \(\cos(\theta+30°) = 0.5\) for \(0° \leq \theta \leq 360°\).Exam-Style
Q8Solve \(2\cos^2\theta + \cos\theta - 1 = 0\) for \(0° \leq \theta \leq 360°\).Exam-Style
Q9Solve \(3\sin\theta = 2\cos\theta\) for \(0° \leq \theta \leq 360°\), to 1 decimal place.Exam-Style
Q10Show that \((1-\sin\theta)(1+\sin\theta) \equiv \cos^2\theta\).Exam-Style
A* Challenge
Q11Solve \(\sin(3\theta) = -0.5\) for \(0° \leq \theta \leq 360°\).A* Challenge
Q12Solve \(2\sin^2\theta = 1 + \cos\theta\) for \(0° \leq \theta \leq 360°\).A* Challenge
Q13Prove that \(\tan\theta + \dfrac{1}{\tan\theta} \equiv \dfrac{1}{\sin\theta\cos\theta}\).A* Challenge
Q14Solve \(4\sin\theta\tan\theta = 3\) for \(0° \leq \theta \leq 360°\), to 1 decimal place.A* Challenge
Q15Solve \(\cos(2\theta - 20°) = -0.3\) for \(0° \leq \theta \leq 360°\), to 1 decimal place.A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — Identity proved
1
Substitute \(\sin^2\theta \equiv 1-\cos^2\theta\) into the LHS: \((1-\cos^2\theta)-\cos^2\theta\)
\(= 1-2\cos^2\theta\), which is the RHS, as required
Q2 — Identity proved
1
\(\dfrac{\sin\theta}{\tan\theta} = \sin\theta \div \dfrac{\sin\theta}{\cos\theta} = \sin\theta \times \dfrac{\cos\theta}{\sin\theta}\)
\(= \cos\theta\), as required
Q3 — \(\theta = 17.5°\) or \(162.5°\)
1
Principal value: \(\sin^{-1}(0.3) = 17.5°\)
Second solution: \(180°-17.5°=162.5°\)
Q4 — \(\theta = 113.6°\) or \(246.4°\)
1
Principal value: \(\cos^{-1}(-0.4) = 113.6°\)
Second solution: \(360°-113.6°=246.4°\)
Q5 — \(\dfrac{1}{\cos\theta}\)
1
\(\sin\theta\tan\theta = \sin\theta \times \dfrac{\sin\theta}{\cos\theta} = \dfrac{\sin^2\theta}{\cos\theta}\)
2
\(\dfrac{\sin^2\theta}{\cos\theta}+\cos\theta = \dfrac{\sin^2\theta+\cos^2\theta}{\cos\theta}\)
\(= \dfrac{1}{\cos\theta}\)
Exam-Style (Q6–Q10)
Q6 — \(\theta = 18.4°, 71.6°, 198.4°,\) or \(251.6°\)
1
Let \(u=2\theta\), range \(0°\) to \(720°\)
2
\(\sin u = 0.6 \implies u = 36.9°, 143.1°, 396.9°, 503.1°\)
\(\theta = \dfrac{u}{2} = 18.4°, 71.6°, 198.4°, 251.6°\)
Q7 — \(\theta = 30°\) or \(270°\)
1
Let \(u=\theta+30°\), range \(30°\) to \(390°\)
2
\(\cos u = 0.5 \implies u = 60°\) or \(300°\) (both within range)
\(\theta = u-30° = 30°\) or \(270°\)
Q8 — \(\theta = 60°, 180°,\) or \(300°\)
1
Factorise: \((2\cos\theta-1)(\cos\theta+1)=0\)
2
\(\cos\theta=0.5 \implies \theta=60°,300°\)
\(\cos\theta=-1 \implies \theta=180°\)
Q9 — \(\theta = 33.7°\) or \(213.7°\)
1
Divide by \(\cos\theta\): \(3\tan\theta = 2 \implies \tan\theta = 0.667\)
\(\theta = 33.7°\) or \(33.7°+180°=213.7°\)
Q10 — Identity proved
1
Expand the LHS (difference of two squares): \(1-\sin^2\theta\)
\(= \cos^2\theta\) by the Pythagorean identity, as required
A* Challenge (Q11–Q15)
Q11 — \(\theta = 70°, 110°, 190°, 230°, 310°,\) or \(350°\)
1
Let \(u=3\theta\), range \(0°\) to \(1080°\)
2
\(\sin u=-0.5\): reference angle \(30°\), solutions each cycle at \(210°, 330°\)
3
\(u = 210°, 330°, 570°, 690°, 930°, 1050°\)
\(\theta = \dfrac{u}{3} = 70°, 110°, 190°, 230°, 310°, 350°\)
Q12 — \(\theta = 60°, 180°,\) or \(300°\)
1
Substitute \(\sin^2\theta\equiv1-\cos^2\theta\): \(2-2\cos^2\theta=1+\cos\theta\)
2
Rearrange: \(2\cos^2\theta+\cos\theta-1=0 \implies (2\cos\theta-1)(\cos\theta+1)=0\)
\(\cos\theta=0.5 \implies \theta=60°,300°\); \(\cos\theta=-1 \implies \theta=180°\)
Q13 — Identity proved
1
\(\tan\theta+\dfrac{1}{\tan\theta} = \dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}\)
2
Combine over a common denominator: \(\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\)
\(= \dfrac{1}{\sin\theta\cos\theta}\) by the Pythagorean identity, as required
Q14 — \(\theta = 46.1°\) or \(313.9°\)
1
\(4\sin\theta\tan\theta = 4\dfrac{\sin^2\theta}{\cos\theta} = 4\dfrac{1-\cos^2\theta}{\cos\theta} = 3\)
2
\(4-4\cos^2\theta = 3\cos\theta \implies 4\cos^2\theta+3\cos\theta-4=0\)
3
Quadratic formula: \(\cos\theta = \dfrac{-3\pm\sqrt{73}}{8}\); only \(\cos\theta=0.693\) is valid (the other root is outside \([-1,1]\))
\(\theta = 46.1°\) or \(360°-46.1°=313.9°\)
Q15 — \(\theta = 63.7°, 136.3°, 243.7°,\) or \(316.3°\)
1
Let \(u=2\theta-20°\), range \(-20°\) to \(700°\)
2
\(\cos u=-0.3\): reference angle \(72.5°\), principal solution \(u=180°-72.5°=107.5°\)
3
Within range: \(u = 107.5°, 252.5°, 467.5°, 612.5°\)
\(\theta = \dfrac{u+20°}{2} = 63.7°, 136.3°, 243.7°, 316.3°\)
Summary
- Two identities underpin everything here: \(\sin^2\theta+\cos^2\theta\equiv1\) and \(\tan\theta\equiv\dfrac{\sin\theta}{\cos\theta}\) — recognise which one an equation or "prove" question needs from the functions mixed together in it.
- For multiple-angle and shifted-angle equations, substitute a new variable \(u\), correctly widen or shift the solution range, solve for \(u\), then convert back to \(\theta\) last.
- For quadratic-form equations, substitute the Pythagorean identity to leave a single function, then factorise — never divide by a function that could be zero.
- For mixed sin/cos equations with no squares, divide through by \(\cos\theta\) to reduce to a single \(\tan\theta\) equation.
- Always check every algebraic root of \(\sin\theta\) or \(\cos\theta\) lies within \([-1,1]\) before trying to find an angle from it.
This topic builds directly on Trigonometric Ratios — the graph symmetry and exact-value reasoning used there is exactly what identifies every second solution here — and reuses the factorising technique from Quadratics. Next in Pure Year 1: Exponentials & Logarithms.
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