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Trigonometric Ratios & Identities revision notes

Every idea in the Trigonometric Ratios & Identities chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 17 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Angles, signs and standard values

Radians and signs in the quadrants
3× asked, last 2026
π rad=180∘,1 rad≈57∘17′,s=rθI: all+,II: sin⁡,cosec⁡+,III: tan⁡,cot⁡+,IV: cos⁡,sec⁡+\begin{gathered}\pi\ \text{rad}=180^\circ,\qquad 1\ \text{rad}\approx 57^\circ17',\qquad s=r\theta\\ \text{I: all}+,\qquad \text{II: }\sin,\operatorname{cosec}+,\qquad \text{III: }\tan,\cot+,\qquad \text{IV: }\cos,\sec+\end{gathered}
s=rθs=r\theta needs θ\theta in radians. A reciprocal has the same sign. Fix the quadrant before taking any square root.
xxyy(−45,−35)(-\frac45,-\frac35)xxI: all ++II: sin⁡\sin, cosec⁡\operatorname{cosec} ++III: tan⁡\tan, cot⁡\cot ++IV: cos⁡\cos, sec⁡\sec ++cos⁡x\cos xsin⁡x\sin x
sin⁡x=−3/5\sin x=-3/5, π<x<3π/2\pi<x<3\pi/2: the point (cos⁡x,sin⁡x)(\cos x,\sin x) is in quadrant III, so cos⁡x=−4/5\cos x=-4/5 and tan⁡x=+3/4\tan x=+3/4
  • Not the positive root by habit. A square root gives only the size; the quadrant gives the sign. sin⁡x=−3/5\sin x=-3/5 with π<x<3π/2\pi<x<3\pi/2: cos⁡x=−4/5\cos x=-4/5 but tan⁡x=+3/4\tan x=+3/4.
Allied angles
6× asked, last 2025
f(n⋅90∘±θ)=±{f(θ),n evenco-f(θ),n oddf(n\cdot 90^\circ\pm\theta)=\pm\begin{cases} f(\theta), & n\ \text{even}\\ \text{co-}f(\theta), & n\ \text{odd}\end{cases}
Sign = sign of ff in the quadrant of n⋅90∘±θn\cdot90^\circ\pm\theta, taking θ\theta acute. Co-function: sin↔cos, tan↔cot, sec↔cosec.
xxyyθ\theta(cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta)90∘+θ90^\circ+\theta: (−sin⁡θ,cos⁡θ)(-\sin\theta,\cos\theta)180∘−θ180^\circ-\theta: (−cos⁡θ,sin⁡θ)(-\cos\theta,\sin\theta)
Turning by 90∘90^\circ swaps the coordinates with one sign change; 180∘−θ180^\circ-\theta mirrors in the yy-axis
  • Not cot⁡(90∘+θ)=tan⁡θ\cot(90^\circ+\theta)=\tan\theta. Odd nn swaps the function and quadrant II makes cot negative: −tan⁡θ-\tan\theta. So cot⁡105∘=−tan⁡15∘\cot105^\circ=-\tan15^\circ.
Values at 15°, 18°, 36°, 22.5°
8× asked, last 2026
sin⁡15∘=3−122,tan⁡15∘=2−3,tan⁡22.5∘=2−1sin⁡18∘=5−14,cos⁡36∘=5+14,sin⁡236∘=5−58\begin{gathered}\sin15^\circ=\frac{\sqrt3-1}{2\sqrt2},\qquad \tan15^\circ=2-\sqrt3,\qquad \tan22.5^\circ=\sqrt2-1\\ \sin18^\circ=\frac{\sqrt5-1}{4},\qquad \cos36^\circ=\frac{\sqrt5+1}{4},\qquad \sin^2 36^\circ=\frac{5-\sqrt5}{8}\end{gathered}
cos⁡72∘=sin⁡18∘\cos72^\circ=\sin18^\circ, sin⁡54∘=cos⁡36∘\sin54^\circ=\cos36^\circ, cos⁡15∘=(3+1)/22\cos15^\circ=(\sqrt3+1)/2\sqrt2, tan⁡75∘=2+3\tan75^\circ=2+\sqrt3.
  • Why 18°: with θ=18∘\theta=18^\circ, 2θ=90∘−3θ2\theta=90^\circ-3\theta, so sin⁡2θ=cos⁡3θ\sin2\theta=\cos3\theta gives 4s2+2s−1=04s^2+2s-1=0; keep the positive root. Hence cosec⁡18∘=5+1\operatorname{cosec}18^\circ=\sqrt5+1, a root of x2−2x−4=0x^2-2x-4=0.

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Practise Trigonometric Ratios & Identities

14 more concepts in 5 topics

Identities and elimination; Compound angles; Multiple and sub-multiple angles; Transformations, products and series; and Maximum and minimum values. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Identities and elimination

  • Pythagorean identities5×
  • a sin⁴θ + b cos⁴θ = ab/(a + b)2×

Compound angles

  • sin and cos of A ± B8×
  • tan(A ± B) and tan(A + B + C)6×

Multiple and sub-multiple angles

  • Double angle6×
  • Triple angle3×
  • Half angle and t = tan(θ/2)4×

Transformations, products and series

  • Sum ↔ product (C + D formulae)7×
  • cos A cos 2A cos 4A … product2×
  • θ, 60° − θ, 60° + θ products6×
  • Tangent sums and telescoping5×

Maximum and minimum values

  • a sin θ + b cos θ7×
  • Ranges via one variable6×
  • Bounded sides and AM–GM4×

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