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JEE Mains Maths · Formula sheet

Trigonometric Identities formulas

12 formulas and 12 common traps for JEE Mains Maths Trigonometric Identities, grouped by subtopic.

Full notes with worked examples

Compound Angle Formulae

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Signs by quadrant

Pythagorean identities

sin⁡2x+cos⁡2x=1,1+tan⁡2x=sec⁡2x,1+cot⁡2x=csc⁡2x\sin^2x+\cos^2x=1,\quad1+\tan^2x=\sec^2x,\quad1+\cot^2x=\csc^2x

Sine and cosine of a sum

Compound angles

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

Tangent of a sum

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

The triangle gives the size, not the sign

sin⁡x=−35\sin x=-\frac35 in quadrant III gives cos⁡x=−45\cos x=-\frac45, not +45+\frac45. Fix every sign from the quadrant before substituting. For α+β\alpha+\beta, place the sum, not the parts.

The sign in cos(A ± B) flips

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A-B)=\cos A\cos B+\sin A\sin B, with a plus. When collapsing a pair, match its sign to the formula, or cos⁡(A−B)\cos(A-B) becomes cos⁡(A+B)\cos(A+B).

One tangent, two angles

tan⁡(A+B)=−1\tan(A+B)=-1 fits both 3π4\frac{3\pi}{4} and −π4-\frac{\pi}{4}. Use the ranges of AA and BB, and the signs of tan⁡A\tan A and tan⁡B\tan B, to choose the angle before taking its sine or cosine.

Standard Values and Multiple Angles

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Exact values at 15°, 18° and 36°

Values at 18° and 36°

sin⁡18∘=5−14,cos⁡36∘=5+14\sin18^\circ=\frac{\sqrt5-1}{4},\qquad\cos36^\circ=\frac{\sqrt5+1}{4}

Double and triple angles

cos⁡2θ=1−2sin⁡2θ,sin⁡3θ=3sin⁡θ−4sin⁡3θ,cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos2\theta=1-2\sin^2\theta,\quad\sin3\theta=3\sin\theta-4\sin^3\theta,\quad\cos3\theta=4\cos^3\theta-3\cos\theta

Common traps

sin 18° and cos 36° differ in one sign

sin⁡18∘=5−14≈0.31\sin18^\circ=\frac{\sqrt5-1}{4}\approx0.31 and cos⁡36∘=5+14≈0.81\cos36^\circ=\frac{\sqrt5+1}{4}\approx0.81. Check the size before substituting: a value above 12\frac12 cannot be sin⁡18∘\sin18^\circ.

sin 3θ and cos 3θ have opposite sign patterns

sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta=3\sin\theta-4\sin^3\theta, but cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta. Writing 3cos⁡θ−4cos⁡3θ3\cos\theta-4\cos^3\theta gives −cos⁡3θ-\cos3\theta.

Powers of Sine and Cosine

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Reducing through sin²θ cos²θ

Power sums

sin⁡4θ+cos⁡4θ=1−2p,sin⁡6θ+cos⁡6θ=1−3p,p=sin⁡2θcos⁡2θ\sin^4\theta+\cos^4\theta=1-2p,\quad\sin^6\theta+\cos^6\theta=1-3p,\quad p=\sin^2\theta\cos^2\theta

A condition that fixes sin²θ

The perfect-square condition

asin⁡4θ+bcos⁡4θ=aba+b ⇒ sin⁡2θ=ba+ba\sin^4\theta+b\cos^4\theta=\frac{ab}{a+b}\ \Rightarrow\ \sin^2\theta=\frac{b}{a+b}

Common traps

p runs from 0 to 1/4

p=sin⁡2θcos⁡2θ=14sin⁡22θp=\sin^2\theta\cos^2\theta=\frac14\sin^22\theta, so its largest value is 14\frac14, not 1. Using the range of sin⁡22θ\sin^22\theta in place of the range of pp gives a range four times too wide.

sin²θ takes the other coefficient

In asin⁡4θ+bcos⁡4θ=aba+ba\sin^4\theta+b\cos^4\theta=\frac{ab}{a+b}, sin⁡2θ=ba+b\sin^2\theta=\frac{b}{a+b}, with bb from the cosine term. Substitute back once to check before using it.

Sum-Product Formulas and Telescoping

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Cosine products with doubling angles

Doubling product

cos⁡θcos⁡2θcos⁡4θ⋯cos⁡2n−1θ=sin⁡2nθ2nsin⁡θ\cos\theta\cos2\theta\cos4\theta\cdots\cos2^{n-1}\theta=\frac{\sin2^n\theta}{2^n\sin\theta}

The 60° product identities

The 60° identity

sin⁡θsin⁡(60∘−θ)sin⁡(60∘+θ)=14sin⁡3θ\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\tfrac14\sin3\theta

Sum-to-product and telescoping

Sum to product

sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2,cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\sin C+\sin D=2\sin\frac{C+D}{2}\cos\frac{C-D}{2},\quad\cos C+\cos D=2\cos\frac{C+D}{2}\cos\frac{C-D}{2}

Common traps

Check where the last angle lands

The product is ±12n\pm\frac{1}{2^n} only when sin⁡2nθ=±sin⁡θ\sin2^n\theta=\pm\sin\theta, and the sign matters. At θ=π7\theta=\frac{\pi}{7}, 8θ=π+θ8\theta=\pi+\theta, so the three-factor product is −18-\frac18, not 18\frac18.

The angles must fit the pattern

sin⁡10∘sin⁡50∘sin⁡70∘\sin10^\circ\sin50^\circ\sin70^\circ fits with θ=10∘\theta=10^\circ. sin⁡10∘sin⁡20∘sin⁡40∘\sin10^\circ\sin20^\circ\sin40^\circ does not: no θ\theta gives those three angles, so the identity does not apply to it directly.

Multiply by the right sine

For cosines at angles spaced by dd, multiply by 2sin⁡d22\sin\frac d2, not 2sin⁡d2\sin d. With the wrong factor the new terms do not cancel in pairs.

Maximum, Minimum and Range

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The bound on a sin θ + b cos θ

The bound

−a2+b2≤asin⁡θ+bcos⁡θ≤a2+b2-\sqrt{a^2+b^2}\le a\sin\theta+b\cos\theta\le\sqrt{a^2+b^2}

Ranges through a bounded quantity

The range of p

sin⁡2θcos⁡2θ=14sin⁡22θ∈[0,14]\sin^2\theta\cos^2\theta=\tfrac14\sin^22\theta\in\left[0,\tfrac14\right]

Common traps

The bound needs one angle

a2+b2\sqrt{a^2+b^2} bounds asin⁡θ+bcos⁡θa\sin\theta+b\cos\theta only when both terms have the same angle. sin⁡θ+cos⁡2θ\sin\theta+\cos2\theta is not of this form; bounding each term on its own gives a range that is too wide.

Check that each end is reached

The range of 1−3p1-3p uses both ends of p∈[0,14]p\in\left[0,\frac14\right]. If the question removes some angles, for example where tan⁡2θ=1\tan^2\theta=1, an end can drop out and the interval becomes open there.

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