IB Maths AA SLTopic 3 — Trig Equations & IdentitiesPaper 1 & 2~9 min read
Simple Trigonometric Identities
An identity is an equation that’s true for every value of the angle. The two simple identities in this note — the tan identity and the Pythagorean identity — unlock almost every “prove that…” and “solve the equation…” question on Paper 1 and Paper 2.
📘 What you need to know
An identity holds for every angle θ — unlike an equation, you don’t “solve” it, you use it as a tool.
The tan identity: tan θ = sin θcos θ — use this to turn a tan into a sin and cos (or vice-versa).
The Pythagorean identity: sin2θ + cos2θ = 1 — use this to swap between sin2 and cos2.
Both identities are in your formula booklet — you don’t need to memorise them, but you do need to know when to reach for them.
The notation sin2θ means (sin θ)2, not sin(θ2).
What is a trigonometric identity?
An equation like sin x = 0.5 is only true for some values of x — those are the things you go solve for. An identity is different: it’s a statement that’s true for every angle you can plug in.
You don’t solve an identity. You use it as a tool — to swap one piece of an expression for an equivalent piece that’s easier to work with.
Think of identities as your “swap cards” in trig. When you’re stuck on an equation that mixes sin, cos, and tan, an identity lets you rewrite the messy bits so everything is in the same language.
The two identities you need at SL
For IB Maths AA SL, two identities do almost all the heavy lifting. Both are listed in your formula booklet:
The tan identity
tan θ = sin θcos θ
The Pythagorean identity
sin2θ + cos2θ = 1
Notice the notation. When you see sin2θ, that’s a shorthand for (sin θ)2 — square the value of sin θ. It does not mean sin(θ2), and it doesn’t mean apply sin twice.
Where do these identities come from?
You don’t need to derive these for the exam, but seeing where they come from makes them stick — and it helps for your IA too.
Draw a right-angled triangle and label one of the non-right angles θ. Choose the hypotenuse to be 1 unit long. Then by basic SOHCAHTOA:
The side oppositeθ has length sin θ (since sin = opp/hyp = opp/1).
The side adjacent to θ has length cos θ (since cos = adj/hyp = adj/1).
Right-triangle derivation (hypotenuse = 1)
🤔 Why does sin²θ + cos²θ = 1?
Look at the triangle: the two legs have lengths sin θ and cos θ, and the hypotenuse is 1. Apply Pythagoras’ theorem (a2 + b2 = c2): (sin θ)2 + (cos θ)2 = 12.
Tidy up the squared notation and you get sin2θ + cos2θ = 1. That’s literally Pythagoras’ theorem applied to a unit-hypotenuse triangle — which is why it’s called the Pythagorean identity.
🤔 Why does tan θ = sin θ ÷ cos θ?
By SOHCAHTOA, tan θ = opposite ÷ adjacent. Substitute the lengths from the unit-hypotenuse triangle: tan θ = sin θ ÷ cos θ. Done.
The three faces of the Pythagorean identity
The Pythagorean identity is most useful when you rearrange it. The same identity can be written three ways depending on what you want to swap out:
Form
→
Use it when…
sin2θ + cos2θ = 1
→
You want to recognise that something simplifies to 1.
sin2θ = 1 − cos2θ
→
You want to get rid of sin2 and write everything in cos.
cos2θ = 1 − sin2θ
→
You want to get rid of cos2 and write everything in sin.
The trick when you’re stuck: ask yourself which letter is “in the way”. If the question only has cos in the linear terms but a stray sin2, swap the sin2 for 1 − cos2. Same idea the other way.
When do I use which identity?
Both identities serve the same goal — getting an expression into a single language so you can simplify or solve. Here’s the quick decision tree:
Reach for the tan identity
tan θ = sin θcos θ
…when you have tan mixed with sin or cos. Rewrite the tan to get everything in sin and cos only.
Reach for the Pythagorean identity
sin2θ + cos2θ = 1
…when you have sin2 or cos2 alongside other terms. Swap one for the other to get a single ratio.
Spotting the swap quickly
Most “show that…” questions tell you the target form. Compare the start and the end — whatever’s missing at the end is the thing you need to substitute out.
2sin2x − cos x = 0
→
target uses cos only
3 − 4cos2θ = sin θ
→
target uses sin only
2 sin x − tan x = 0
→
target needs tan replaced
🧠
Memory trick: “Match the linear, swap the squared”
If a quadratic term and a linear term don’t match (sin2 with cos, or cos2 with sin), change the squared one using the Pythagorean identity. Linear terms are harder to manipulate — leave them alone.
Step-by-step method for “show that” questions
You’ll see the same pattern in almost every Paper 1 identity question. Follow these four steps and you’ll rarely get stuck:
Read the target. See which trig function the answer is in (sin only? cos only? a particular form?).
Pick the swap. Decide which identity to use — usually the one that removes whatever’s missing from the target.
Substitute carefully. Be ruthless with brackets — a sign error here is the most common mark-loss.
Tidy up. Expand, collect like terms, and rearrange so it matches the requested form exactly.
📍
Don’t just stop at “looks right” — match the target form exactly
If the question asks for the form a cos2x + b cos x + c = 0, your answer must be in that order with that sign convention. Examiners only award the “show that” mark when your final line genuinely matches the requested form.
Worked examples
WE 1
Rewrite an equation as a quadratic in cos x
Show that the equation 2sin2x − cos x = 0 can be written in the form a cos2x + b cos x + c = 0, where a, b, c are integers to be found.
2sin2 x − cos x = 0Equation has both sin x and cos x, so it needs changing before it can be solved.Use the identitysin2 x = 1 − cos2 xSubstitute:2(1 − cos2 x) − cos x = 0Expand:2 − 2cos2 x − cos x = 0Rearrange (× −1):2cos2 x + cos x − 2 = 0a = 2, b = 1, c = −2
WE 2
Rewrite an equation in terms of sin only
Show that 3cos2θ + 5 sin θ = 1 can be written as 3sin2θ − 5 sin θ − 2 = 0.
3cos2 θ + 5 sin θ = 1Has cos2 and sin — target is in sin only, so swap out the cos2.Use the identitycos2 θ = 1 − sin2 θSubstitute:3(1 − sin2 θ) + 5 sin θ = 1Expand:3 − 3sin2 θ + 5 sin θ = 1Move to one side:2 − 3sin2 θ + 5 sin θ = 0× −1:3sin2 θ − 5 sin θ − 2 = 03sin2 θ − 5 sin θ − 2 = 0 ✓
WE 3
Use the tan identity to simplify and factorise
The equation 2 sin x − tan x = 0 holds for some angle x. Use the tan identity to rewrite it in terms of sin and cos only, then factorise the result.
2 sin x − tan x = 0Mixing sin and tan — replace tan to get one language.Use the identitytan x = sin xcos xSubstitute:2 sin x − sin xcos x = 0× cos x:2 sin x cos x − sin x = 0Factor sin x:sin x (2 cos x − 1) = 0sin x (2 cos x − 1) = 0then set each factor = 0 → sin x = 0 or cos x = ½
WE 4
Find the exact value of cos θ given sin θ
Given that sin θ = 35 and that θ is acute, find the exact value of cos θ using the Pythagorean identity.
sin θ = 35, θ acuteUse the identitysin2 θ + cos2 θ = 1Substitute:(35)2 + cos2 θ = 1Simplify:925 + cos2 θ = 1Rearrange:cos2 θ = 1625Square root:cos θ = ± 45θ is acute → cos θ > 0, so take the +cos θ = 45always check the quadrant before picking + or −!
WE 5
Show an expression simplifies to a single ratio
Show that 1 − cos2xcos x = sin x · tan x.
LHS = 1 − cos2 xcos xSpot 1 − cos2 x in the numerator — that’s the Pythagorean identity rearranged.Use the identity1 − cos2 x = sin2 xSubstitute:LHS = sin2 xcos xSplit sin² x:LHS = sin x · sin xcos xUse tan = sin / cos:LHS = sin x · tan xLHS = sin x · tan x = RHS ✓always finish a “show that” proof by writing LHS = RHS
💡 Top tips
Both identities are in the formula booklet — check it for the exact form before you start substituting. Don’t waste exam time worrying you’ve remembered them wrong.
Match the squared, leave the linear. If the equation has a sin2 and a linear cos, swap the sin2 using 1 − cos2. The reverse if it’s cos2 with linear sin.
For “show that…” questions, always finish by writing the target form exactly. The final mark is for matching the form, not just for getting somewhere close.
Bracket the substitution. Write 1 − cos2x as (1 − cos2x) when you sub it in, so the minus sign distributes correctly.
If you see a tan term mixed with sin or cos, the very first move is to rewrite tan as sin/cos and clear the fraction.
The identity sin2θ + cos2θ = 1 also works in reverse — sometimes you’ll spot a 1 in an expression that should be replaced by sin2 + cos2.
When asked to find an exact ratio (like cos θ from sin θ), always check which quadrant θ is in before deciding on the sign.
⚠ Common mistakes
Reading sin2θ as sin(θ²). The 2 is on sin, not on θ. It means (sin θ)2 — square the value of sin θ, don’t square the angle.
Forgetting brackets when substituting. Writing 2 · 1 − cos2x instead of 2(1 − cos2x) loses you the distribution and the answer falls apart.
Swapping the wrong term. Replacing the linear cos when you have sin2 floating around — it’s the squared one you swap, not the linear one.
Sign errors when distributing the minus. 2(1 − cos2x) = 2 − 2cos2x, not 2 − 2cos2x with a stray + somewhere.
Picking the wrong sign for the square root. When cos2θ = 1625, you get cos θ = ± 45. The quadrant tells you which one — don’t just take the positive answer by default.
Stopping too early. If the “show that” target is 2cos2x + cos x − 2 = 0 and you’ve reached −2cos2x − cos x + 2 = 0, you still need to multiply by −1 to match.
Confusing the tan identity direction. tan = sin/cos, not cos/sin. Mix this up and every step from here is wrong.
Once these two identities feel automatic, double-angle formulas and the rest of the trig chapter become much easier — they’re all built on top of these foundations.
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