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NDA Mathematics · Formula sheet

Trigonometric Identities formulas

10 formulas, 2 reference tables and 14 common traps for NDA Mathematics Trigonometric Identities, grouped by subtopic.

Full notes with worked examples

Standard Values, Signs & Special Angles

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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

Standard-angle values and allied reductions

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°)

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°.

Common traps

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.

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.

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.

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.

Compound Angles — sin/cos/tan(A ± B)

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sin(A ± B) and cos(A ± B)

Sum and difference

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B,cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,\quad \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B

tan(A ± B) and cot(A ± B)

Tangent of a sum/difference

tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}

Common traps

sin⁡(A+B)≠sin⁡A+sin⁡B\sin(A+B)\neq\sin A+\sin B — sine is NOT linear

The single most common error in the chapter: treating sin⁡(A+B)\sin(A+B) as sin⁡A+sin⁡B\sin A+\sin B (or cos⁡(A+B)\cos(A+B) as cos⁡A+cos⁡B\cos A+\cos B). A quick disproof: sin⁡(30°+60°)=sin⁡90°=1\sin(30°+60°)=\sin 90°=1, but sin⁡30°+sin⁡60°=12+32≈1.37\sin 30°+\sin 60°=\tfrac12+\tfrac{\sqrt3}{2}\approx1.37. You must expand: sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)=\sin A\cos B+\cos A\sin B.

The tan⁡(A+B)\tan(A+B) denominator is 1−tan⁡Atan⁡B1-\tan A\tan B, and the signs OPPOSE

Two errors cluster here: forgetting the denominator entirely (writing tan⁡(A+B)=tan⁡A+tan⁡B\tan(A+B)=\tan A+\tan B), and getting the denominator sign wrong. The rule is tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} — the bottom sign is opposite the top, so a sum on top means a minus on the bottom. (And tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}.)

cos⁡(A+B)cos⁡(A−B)=cos⁡2A−sin⁡2B\cos(A+B)\cos(A-B)=\cos^2 A-\sin^2 B, not cos⁡2A−cos⁡2B\cos^2 A-\cos^2 B

When collapsing a product like cos⁡(A+B)cos⁡(A−B)\cos(A+B)\cos(A-B), students misremember the result as cos⁡2A−cos⁡2B\cos^2 A-\cos^2 B. Expanding gives cos⁡2A−sin⁡2B\cos^2 A-\sin^2 B (equivalently cos⁡2B−sin⁡2A\cos^2 B-\sin^2 A). The companion is sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin(A+B)\sin(A-B)=\sin^2 A-\sin^2 B. The leftover square is sin⁡2B\sin^2 B, a different function from the cos⁡2A\cos^2 A it sits beside.

Double, Triple & Half-Angle

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Double-angle formulas

sin⁡2A=2sin⁡Acos⁡A,cos⁡2A=cos⁡2A−sin⁡2A=1−2sin⁡2A=2cos⁡2A−1,tan⁡2A=2tan⁡A1−tan⁡2A\sin 2A=2\sin A\cos A,\qquad \cos 2A=\cos^2 A-\sin^2 A=1-2\sin^2 A=2\cos^2 A-1,\qquad \tan 2A=\dfrac{2\tan A}{1-\tan^2 A}

Triple-angle formulas

sin⁡3A=3sin⁡A−4sin⁡3A,cos⁡3A=4cos⁡3A−3cos⁡A,tan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A\sin 3A=3\sin A-4\sin^3 A,\qquad \cos 3A=4\cos^3 A-3\cos A,\qquad \tan 3A=\dfrac{3\tan A-\tan^3 A}{1-3\tan^2 A}

Half-angle formulas and 1 ± cos A / 1 ± sin A

Half-angle formulas

sin⁡A2=±1−cos⁡A2,cos⁡A2=±1+cos⁡A2,tan⁡A2=1−cos⁡Asin⁡A=sin⁡A1+cos⁡A\sin\tfrac A2=\pm\sqrt{\tfrac{1-\cos A}{2}},\qquad \cos\tfrac A2=\pm\sqrt{\tfrac{1+\cos A}{2}},\qquad \tan\tfrac A2=\dfrac{1-\cos A}{\sin A}=\dfrac{\sin A}{1+\cos A}

Common traps

sin⁡2A=2sin⁡Acos⁡A\sin 2A=2\sin A\cos A — not 2sin⁡A2\sin A, and (sin⁡A)2≠sin⁡2A(\sin A)^2\neq\sin 2A

Two slips: dropping the cos⁡A\cos A (writing sin⁡2A=2sin⁡A\sin 2A=2\sin A), and confusing the double angle with a square (sin⁡2A\sin 2A is not sin⁡2A\sin^2 A). The identity is sin⁡2A=2sin⁡Acos⁡A\sin 2A=2\sin A\cos A. Check with A=30°A=30°: sin⁡60°=32\sin 60°=\tfrac{\sqrt3}{2}, while 2sin⁡30°=12\sin 30°=1 and sin⁡230°=14\sin^2 30°=\tfrac14 — all three differ.

Watch the sign pattern: sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A=3\sin A-4\sin^3 A but cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A=4\cos^3 A-3\cos A

Students mix up the order and signs of the two triple-angle formulas. Sine starts with the linear term and subtracts the cube (3sin⁡A−4sin⁡3A3\sin A-4\sin^3 A); cosine starts with the cube and subtracts the linear term (4cos⁡3A−3cos⁡A4\cos^3 A-3\cos A). Writing sin⁡3A=4sin⁡3A−3sin⁡A\sin 3A=4\sin^3 A-3\sin A flips the whole sign — a guaranteed wrong answer.

The ±\pm on a half-angle is fixed BY the quadrant of A/2A/2, not free

From sin⁡A2=±1−cos⁡A2\sin\tfrac A2=\pm\sqrt{\tfrac{1-\cos A}{2}} students grab the positive root automatically. But the sign is decided by which quadrant A/2A/2 lies in. E.g. if A=300°A=300° then A/2=150°A/2=150° (quadrant II), so sin⁡A2>0\sin\tfrac A2>0 but cos⁡A2<0\cos\tfrac A2<0. Likewise 1±sin⁡A=∣sin⁡A2±cos⁡A2∣\sqrt{1\pm\sin A}=\big|\sin\tfrac A2\pm\cos\tfrac A2\big| — take the modulus, then resolve the sign from the quadrant.

Product-to-Sum & Sum-to-Product

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Product-to-sum formulas

The four product-to-sum identities

2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B),2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B),2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B),2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\cos B=\sin(A+B)+\sin(A-B),\qquad 2\cos A\cos B=\cos(A+B)+\cos(A-B),\qquad 2\cos A\sin B=\sin(A+B)-\sin(A-B),\qquad 2\sin A\sin B=\cos(A-B)-\cos(A+B)

Sum-to-product formulas

The four sum-to-product identities

sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2,cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2,sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2,cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2},\qquad \cos C-\cos D=-2\sin\tfrac{C+D}{2}\sin\tfrac{C-D}{2},\qquad \sin C-\sin D=2\cos\tfrac{C+D}{2}\sin\tfrac{C-D}{2},\qquad \cos C+\cos D=2\cos\tfrac{C+D}{2}\cos\tfrac{C-D}{2}

Conditional identities (A + B + C = 90° or 180°)

The two signature conditional identities

A+B+C=π: tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡CA+B+C=\pi:\ \tan A+\tan B+\tan C=\tan A\tan B\tan C

Common traps

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B=\cos(A-B)-\cos(A+B) — cosines, and the difference comes FIRST

The two-sine product is the most error-prone: it converts to cosines, not sines, and the order is cos⁡(A−B)−cos⁡(A+B)\cos(A-B)-\cos(A+B) (difference minus sum). Students write cos⁡(A+B)−cos⁡(A−B)\cos(A+B)-\cos(A-B) and get the whole sign backwards. Contrast 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B=\cos(A+B)+\cos(A-B), which is a plus. Tip: the 2sin⁡Asin⁡B2\sin A\sin B one is the only product-to-sum identity with a leading minus.

Half-SUM and half-DIFFERENCE go in fixed slots — don't swap them

In sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2}, the half-sum C+D2\tfrac{C+D}{2} sits inside the leading function and the half-difference C−D2\tfrac{C-D}{2} inside the trailing one. Students swap them, or use C−DC-D and C+DC+D without halving. Also remember cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C-\cos D=-2\sin\tfrac{C+D}{2}\sin\tfrac{C-D}{2} carries a leading minus (so if C>DC>D and both are acute, the difference is negative).

tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A+\tan B+\tan C=\tan A\tan B\tan C only when A+B+C=180°A+B+C=180°

These are conditional identities — they hold only under the stated angle-sum, not for arbitrary angles. The ∑tan⁡=∏tan⁡\sum\tan=\prod\tan relation needs A+B+C=180°A+B+C=180°; the ∑tan⁡Atan⁡B=1\sum\tan A\tan B=1 relation needs A+B+C=90°A+B+C=90°. Applying the wrong one (or applying either to angles that don't sum correctly) is the trap. Verify the angle-sum condition before invoking the identity.

Maximum & Minimum Values

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AM-GM for reciprocal-type minima

AM-GM minimum

u+v≥2uv  (u,v>0),equality at u=vu+v\ge 2\sqrt{uv}\ \ (u,v>0),\quad \text{equality at } u=v

Common traps

Max of asin⁡x+bcos⁡xa\sin x+b\cos x is a2+b2\sqrt{a^2+b^2}, NOT a+ba+b

Because sin⁡x\sin x and cos⁡x\cos x hit 11 at different values of xx, you cannot add their maxima — the peak is a2+b2\sqrt{a^2+b^2}, reached when the single sinusoid Rsin⁡(x+φ)R\sin(x+\varphi) equals 11. For 3sin⁡x+4cos⁡x3\sin x+4\cos x the maximum is 9+16=5\sqrt{9+16}=5, not 3+4=73+4=7. Whenever you see asin⁡x+bcos⁡xa\sin x+b\cos x, reach for the amplitude a2+b2\sqrt{a^2+b^2}, never the coefficient sum.

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