Trig Equations Using Identities (AA SL)
Once an equation mixes ratios - or the angle itself is doubled or tripled - the basic method from a plain \(\sin x = k\) equation isn't enough on its own. You need an identity to rewrite the equation in a single ratio, or a domain adjustment to handle the compressed angle. This page covers both, with worked examples and where marks tend to go missing. It's part of the broader Trig Equations & Graphs topic.
19 questions on this sub-topic.
The identities you'll need
Covered under IB syllabus reference SL3.8. The Pythagorean identity is in the formula booklet; the quotient identity is not - it's prior knowledge from the definition of \(\tan\theta\). The skill is spotting when an equation needs one.
Pythagorean identity
\(\sin^2\theta + \cos^2\theta = 1\)
Use it to swap between \(\sin^2\theta\) and \(\cos^2\theta\) - essential when an equation has both, or when it needs turning into a quadratic in one ratio.
Quotient identity
\(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)
Use it to eliminate \(\tan\theta\) in favour of \(\sin\theta\) and \(\cos\theta\), or to combine it with the Pythagorean identity when an equation mixes all three ratios.
Need the full syllabus wording and worked equations-and-graphs overview? See Trig Equations & Graphs.
Worked examples
Given \(\cos\theta=\tfrac35\) and \(\theta\) is acute, find \(\sin\theta.\)
Worked solution
Apply the Pythagorean identity: \(\sin^2\theta = 1 - \tfrac{9}{25}\) M1 \(= \tfrac{16}{25}.\) A1 As \(\theta\) is acute, \(\sin\theta = \tfrac45.\) A1
Solve \(\sin(2x) = \dfrac{\sqrt3}{2}\) for \(0 \le x \le \pi.\)
(a)(i) Give the value with \(x<0.785\).
(a)(ii) Give the value with \(x>0.785.\)
Worked solution
\(3\sin(2x)+1=2\Rightarrow \sin(2x)=\tfrac13.\) M1
\(0\le 2x\le 4\pi\): \(2x=0.3398,2.802,6.623,9.085.\) A1
\(x\approx 0.170,1.40,3.31,4.54.\) A1 A1 A1 A1
Common mistakes
- Forgetting to widen the domain for \(\sin(bx)\). When you substitute \(u=bx\), the domain for \(u\) is \(b\) times wider than the domain for \(x\) - skip this and you'll miss whole solutions before you even divide back down.
- Dropping the negative square root. Taking a square root of \(\sin^2\theta\) or \(\cos^2\theta\) gives \(\pm\), and both signs are valid unless the question restricts \(\theta\) to a particular quadrant (e.g. "acute").
- Using the wrong identity to eliminate a ratio. An equation with both \(\sin\theta\) and \(\cos\theta\) needs the Pythagorean identity; one with \(\tan\theta\) alongside \(\sin\theta\) or \(\cos\theta\) needs the quotient identity - picking the wrong one leaves the equation just as tangled.
Ready to practise properly?
19 identity-based trig equation questions, marked instantly like the real exam.
Quick answers
What identity do I need to solve trig equations with sin and cos in the same equation?
The Pythagorean identity \(\sin^2\theta+\cos^2\theta=1\) lets you rewrite the equation in a single ratio, usually turning it into a quadratic in \(\sin\theta\) or \(\cos\theta\) that you can factorise or solve with the quadratic formula.
How do you solve sin(2x) = k for x?
Widen the domain to match the doubled angle (so \(0\) to \(2\pi\) becomes \(0\) to \(4\pi\) for \(u=2x\)), solve for \(u\) using the usual method, then divide every solution by 2 to get \(x\).