Pythagorean and Reciprocal Identities (AA SL)
Knowing just one trigonometric ratio - say \(\sin\theta\) - is enough to pin down the other two, as long as you know which quadrant \(\theta\) sits in. The Pythagorean identity does the heavy lifting, and this page shows how to use it cleanly, without ever needing to find \(\theta\) itself. It's part of the broader Identities & Exact Values topic.
17 questions on this sub-topic.
The Pythagorean identity
Covered under IB syllabus reference SL3.6: the Pythagorean identity, and using it to find one trigonometric ratio from another without finding \(\theta\) itself.
Pythagorean identity
\[\sin^2\theta+\cos^2\theta=1\]
True for every angle \(\theta\). Rearrange to \(\cos^2\theta=1-\sin^2\theta\) (or the other way round) whenever you're given one ratio and need another.
Choosing the sign
\[\cos\theta=\pm\sqrt{1-\sin^2\theta}\]
Squaring loses the sign, so you have to put it back by hand: check which quadrant \(\theta\) is in (acute means both sin and cos are positive) before you decide \(+\) or \(-\).
Need double-angle identities too? See Double and Compound Angle Formulae, or the full syllabus table at Identities & Exact Values.
Worked examples
Given \(\sin\theta = \tfrac35\) and \(\theta\) is acute:
(a)(i) Find \(\cos\theta\).
(a)(ii) Find \(\tan\theta\).
Worked solution
\(\cos^2\theta=1-\tfrac{9}{25}=\tfrac{16}{25}\Rightarrow \cos\theta=\tfrac45\) (acute). M1 A1
\(\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\tfrac34.\) M1 A1
Show that \((1 + \cos\theta)(1 - \cos\theta) = \sin^2\theta\).
Worked solution
The left-hand side is a difference of two squares: \((1+\cos\theta)(1-\cos\theta)=1-\cos^2\theta.\) M1
By the Pythagorean identity, \(\sin^2\theta+\cos^2\theta=1\), so \(1-\cos^2\theta=\sin^2\theta\), which is the right-hand side. R1
Prove \(\dfrac{1}{1-\sin\theta}+\dfrac{1}{1+\sin\theta}=\dfrac{2}{\cos^2\theta}.\)
Worked solution
\(\dfrac{(1+\sin\theta) + (1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}\) M1 \(= \dfrac{2}{1 - \sin^2\theta}.\) A1
\(1 - \sin^2\theta = \cos^2\theta\) M1 so the expression \(= \dfrac{2}{\cos^2\theta}.\) A1 AG ∎
Given \(\sin\theta = \dfrac{3}{5}\) and \(\cos\theta = \dfrac{4}{5}\), find \(\tan\theta\).
Worked solution
\(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) M1
\(=\dfrac{3/5}{4/5}\) A1
\(=\tfrac34.\) A1
Prove that \(\dfrac{1-\cos^2\theta}{\sin\theta}=\sin\theta.\)
Worked solution
\(1 - \cos^2\theta = \sin^2\theta.\) M1 A1
LHS \(= \dfrac{\sin^2\theta}{\sin\theta}\) M1 \(= \sin\theta.\) A1 AG ∎
Common mistakes
- Forgetting the \(\pm\) when square-rooting. \(\cos^2\theta=\tfrac{16}{25}\) gives \(\cos\theta=\pm\tfrac45\) - always check the quadrant before dropping one sign, rather than defaulting to positive.
- Rearranging the identity incorrectly. \(\sin^2\theta+\cos^2\theta=1\) rearranges to \(\cos^2\theta=1-\sin^2\theta\), not \(\cos\theta=1-\sin\theta\) - the squares must stay in place until the square root is taken.
- Trying to find \(\theta\) first. These questions are designed to be solved from the ratio directly - inverse trig functions are slower and can introduce rounding error that the exact-value approach avoids.
Ready to practise properly?
17 Pythagorean-identity questions, marked instantly like the real exam.
Quick answers
What is the Pythagorean identity in trigonometry?
\(\sin^2\theta+\cos^2\theta=1\), true for every value of \(\theta\). Rearranged, it lets you find \(\cos\theta\) from \(\sin\theta\), or vice versa, without knowing \(\theta\) itself.
How do I know whether the square root should be positive or negative?
Use the quadrant \(\theta\) is in. If \(\theta\) is acute or in the first quadrant, both sin and cos are positive; in other quadrants, check the sign of each ratio (ASTC) before choosing the root.