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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.

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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.

Test yourself — a quick 12-question Trigonometric Identities recall quiz, instant score.

Concept 1 of 5: The fundamental identities

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

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

The same idea in a real exam question:

NDA · Apr 2021 · Q17Easy

Example 2 · Trigonometric Identities · Specific Values and Quadrants

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

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

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

The same idea in a real exam question:

NDA · Sep 2019 · Q66Moderate

Example 3 · Trigonometric Identities · Specific Values and Quadrants

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

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 4 more on signs by quadrant (astc) and reductions

Concept 4 of 5: All ratios from one ratio and a quadrant

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

The same idea in a real exam question:

NDA · Sep 2019 · Q41Moderate

Example 4 · Trigonometric Identities · Specific Values and Quadrants

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 ?

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°)

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
sin 36°√(10 − 2√5)/4
cos 18°√(10 + 2√5)/4
sin 15°(√6 − √2)/4
cos 15°(√6 + √2)/4
cot 15°2 + √3
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

The same idea in a real exam question:

NDA · Apr 2017 · Q46Hard

Example 5 · Trigonometric Identities · Specific Values and Quadrants

What is the value of tan⁡18°\tan 18°?
Drill 10 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°)11 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
sin 36°√(10 − 2√5)/4
cos 18°√(10 + 2√5)/4
sin 15°(√6 − √2)/4
cos 15°(√6 + √2)/4
cot 15°2 + √3
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)

Test yourself on Trigonometric Identities

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