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NDA Maths · Trigonometric Identities

Standard Values, Signs & Special Angles

The bedrock: the fundamental identities, the standard-angle table, which ratios are positive in which quadrant, how to recover every ratio from one, and the exact values of the special angles.

All Trigonometric Identities notesNDA Maths strategy

Why this matters

Almost every other identity question silently assumes you can read off a standard value, fix a sign by quadrant, or know that tan 15° = 2 − √3. These are the cheapest marks in the chapter — and the most common silent error is a sign wrong for the quadrant.

Concept 1 of 5

The fundamental identities

Intuition

Three families generate everything: the Pythagorean identities, the reciprocal pairs, and the quotient relations. Every simplification eventually reduces to one of these.

Definition

  • Pythagorean: sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta, 1+cot⁡2θ=csc⁡2θ1+\cot^2\theta=\csc^2\theta.
  • Reciprocal: csc⁡θ=1sin⁡θ\csc\theta=\tfrac{1}{\sin\theta}, sec⁡θ=1cos⁡θ\sec\theta=\tfrac{1}{\cos\theta}, cot⁡θ=1tan⁡θ\cot\theta=\tfrac{1}{\tan\theta}.
  • Quotient: tan⁡θ=sin⁡θcos⁡θ\tan\theta=\tfrac{\sin\theta}{\cos\theta}, cot⁡θ=cos⁡θsin⁡θ\cot\theta=\tfrac{\cos\theta}{\sin\theta}.

The three Pythagorean identities

sin⁡2θ+cos⁡2θ=1,sec⁡2θ−tan⁡2θ=1,csc⁡2θ−cot⁡2θ=1\sin^2\theta+\cos^2\theta=1,\quad \sec^2\theta-\tan^2\theta=1,\quad \csc^2\theta-\cot^2\theta=1

Worked example

Simplify (1−cos⁡2θ)csc⁡2θ(1-\cos^2\theta)\csc^2\theta.
Practice this conceptself-check · 4 quick reps

It's 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta, never 1−tan⁡2θ1-\tan^2\theta

The classic slip is writing sec⁡2θ−tan⁡2θ\sec^2\theta-\tan^2\theta as anything other than 11, or "remembering" 1−tan⁡2θ=sec⁡2θ1-\tan^2\theta=\sec^2\theta. The Pythagorean form is 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta (so sec⁡2θ−tan⁡2θ=1\sec^2\theta-\tan^2\theta=1) and 1+cot⁡2θ=csc⁡2θ1+\cot^2\theta=\csc^2\theta. The plus sign is what makes them work — sec⁡θ≥1\sec\theta\ge 1 means sec⁡2θ\sec^2\theta must be the *larger* term.
Drill 1 more on the fundamental identities

Concept 2 of 5

Standard-angle values and allied reductions

Intuition

The values at 0°, 30°, 45°, 60°, 90° must be instant. Everything outside the first quadrant reduces to these via the allied-angle rules — add or subtract a multiple of 90° and fix the sign.

Definition

Read the table left-to-right. For angles beyond 90°, reduce with allied rules: sin⁡(180°−θ)=sin⁡θ\sin(180°-\theta)=\sin\theta, cos⁡(180°−θ)=−cos⁡θ\cos(180°-\theta)=-\cos\theta, sin⁡(360°+θ)=sin⁡θ\sin(360°+\theta)=\sin\theta (periodicity), and 90°±θ90°\pm\theta swaps sin↔\leftrightarrowcos.

Anglesincostan
0°010
30°1/2√3/21/√3
45°1/√21/√21
60°√3/21/2√3
90°10∞ (undefined)
Beyond 90°, reduce by allied angles and fix the sign from the quadrant.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Trigonometric IdentitiesEASY
What is the value of csc⁡(7π6)sec⁡(5π3)\csc\left(\frac{7\pi}{6}\right)\sec\left(\frac{5\pi}{3}\right)?

[Q17 · Apr · 2021]

Don't swap sin and cos at 30° and 60°

Under exam pressure students write sin⁡30°=32\sin 30°=\tfrac{\sqrt3}{2} — that's actually cos⁡30°\cos 30°. The correct pair is sin⁡30°=12, cos⁡30°=32\sin 30°=\tfrac12,\ \cos 30°=\tfrac{\sqrt3}{2} and sin⁡60°=32, cos⁡60°=12\sin 60°=\tfrac{\sqrt3}{2},\ \cos 60°=\tfrac12. Anchor on the small angle: the *smaller* angle has the *smaller* sine, so sin⁡30°=12\sin 30°=\tfrac12 is the small one.
Drill 1 more on standard-angle values and allied reductions

Concept 3 of 5

Signs by quadrant (ASTC) and reductions

Intuition

Which ratios are positive depends only on the quadrant: All in I, Sine in II, Tangent in III, Cosine in IV ("All Students Take Calculus"). Pair this with allied reductions to collapse angles like 80° + 40° − 20°.

Definition

  • Quadrant I: all positive. II: sin (and csc) positive. III: tan (and cot) positive. IV: cos (and sec) positive.
  • A square root like 1+sin⁡A=±(sin⁡A2+cos⁡A2)\sqrt{1+\sin A}=\pm(\sin\tfrac{A}{2}+\cos\tfrac{A}{2}) — the sign is decided by the quadrant of A/2A/2, never assumed positive.
  • Allied reductions (cos⁡80°+cos⁡40°=cos⁡20°\cos 80°+\cos 40°=\cos 20° etc.) shrink an awkward combination to a standard value.
Aall +Ssin, csc +Ttan, cot +Ccos, sec +All Students Take Calculus

Worked example

If sin⁡θ>0\sin\theta>0 and tan⁡θ<0\tan\theta<0, which quadrant is θ\theta in?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Trigonometric IdentitiesMODERATE
What is cos⁡80°+cos⁡40°−cos⁡20°\cos80° + \cos40° - \cos20° equal to ?

[Q66 · Sep · 2019]

An allied angle can FLIP the sign — don't assume the ratio stays positive

Students apply cos⁡(180°−θ)=cos⁡θ\cos(180°-\theta)=\cos\theta (dropping the minus) or treat sin⁡(180°+θ)\sin(180°+\theta) as +sin⁡θ+\sin\theta. Use ASTC: in quadrant II only sine/cosec are positive, in III only tan/cot, in IV only cos/sec — so cos⁡(180°−θ)=−cos⁡θ\cos(180°-\theta)=-\cos\theta and sin⁡(180°+θ)=−sin⁡θ\sin(180°+\theta)=-\sin\theta. Decide the sign from the quadrant of the whole angle first, then attach it.
Drill 3 more on signs by quadrant (astc) and reductions

Concept 4 of 5

All ratios from one ratio and a quadrant

Intuition

Given one ratio plus the quadrant, every other ratio is fixed. Build a right triangle from the given ratio for the magnitudes, then attach signs from the quadrant.

Definition

From a single ratio: get the third side by Pythagoras (e.g. sin⁡θ=pr⇒cos⁡θ=±r2−p2r\sin\theta=\tfrac{p}{r}\Rightarrow\cos\theta=\pm\tfrac{\sqrt{r^2-p^2}}{r}), then the quadrant fixes each sign. The quadrant is essential — without it the signs are ambiguous.

Worked example

If cos⁡θ=35\cos\theta=\tfrac{3}{5} and θ\theta is in quadrant IV, find sin⁡θ\sin\theta and tan⁡θ\tan\theta.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Trigonometric IdentitiesMODERATE
If cosec⁡θ=2921\cosec\theta = \dfrac{29}{21} where 0<θ<90°0 < \theta < 90°, then what is the value of 4sec⁡θ+4tan⁡θ4\sec\theta + 4\tan\theta ?

[Q41 · Sep · 2019]

One ratio fixes magnitudes only — the QUADRANT fixes the sign

Given cos⁡θ=35\cos\theta=\tfrac35, students write sin⁡θ=45\sin\theta=\tfrac45 and stop — but sin⁡θ=±45\sin\theta=\pm\tfrac45 until the quadrant is used. If θ\theta is in quadrant IV then sin⁡θ=−45\sin\theta=-\tfrac45 and tan⁡θ=−43\tan\theta=-\tfrac43. Never default to the positive root: from a single ratio every other ratio is determined only up to sign, and the quadrant supplies that sign.
Drill 3 more on all ratios from one ratio and a quadrant

Concept 5 of 5

Special-angle exact values (15°, 18°, 36°, 22.5°, 75°)

Intuition

A handful of non-standard angles recur with exact surd values. Either memorise them or derive on the spot — 15° and 75° from compound angles, 18° and 36° from the pentagon relations, 22.5° from the half-angle of 45°.

Definition

Derive when unsure: tan⁡15°=tan⁡(45°−30°)=2−3\tan 15°=\tan(45°-30°)=2-\sqrt3; tan⁡75°=2+3\tan 75°=2+\sqrt3; tan⁡22.5°=2−1\tan 22.5°=\sqrt2-1; sin⁡18°=5−14\sin 18°=\tfrac{\sqrt5-1}{4}; cos⁡36°=5+14\cos 36°=\tfrac{\sqrt5+1}{4}. Note tan⁡15°\tan 15° and cot⁡15°\cot 15° are conjugate surds, so tan⁡15°+cot⁡15°=4\tan 15°+\cot 15°=4.

AngleExact value
tan 15°2 − √3
tan 75°2 + √3
tan 22.5°√2 − 1
sin 18°(√5 − 1)/4
cos 36°(√5 + 1)/4
tan 18°√(25 − 10√5)/5
15°/75° via compound angle; 18°/36° via the pentagon; 22.5° via half-angle of 45°.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Trigonometric IdentitiesHARD
What is the value of tan⁡18°\tan 18°?

[Q46 · Apr · 2017]

Drill 9 more on special-angle exact values (15°, 18°, 36°, 22.5°, 75°)

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

  • The fundamental identities

    The three Pythagorean identities

    sin⁡2θ+cos⁡2θ=1,sec⁡2θ−tan⁡2θ=1,csc⁡2θ−cot⁡2θ=1\sin^2\theta+\cos^2\theta=1,\quad \sec^2\theta-\tan^2\theta=1,\quad \csc^2\theta-\cot^2\theta=1

Reference tables (2)

Standard-angle values and allied reductions5 rows
Anglesincostan
0°010
30°1/2√3/21/√3
45°1/√21/√21
60°√3/21/2√3
90°10∞ (undefined)
Beyond 90°, reduce by allied angles and fix the sign from the quadrant.
Special-angle exact values (15°, 18°, 36°, 22.5°, 75°)6 rows
AngleExact value
tan 15°2 − √3
tan 75°2 + √3
tan 22.5°√2 − 1
sin 18°(√5 − 1)/4
cos 36°(√5 + 1)/4
tan 18°√(25 − 10√5)/5
15°/75° via compound angle; 18°/36° via the pentagon; 22.5° via half-angle of 45°.

Watch out for (4)

Drill every past-year question on this subtopic

21 questions from the bank — paginated, with cart and Word-export support.

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