Class 10:

  • Tri - 3 , Gono - sides , Metron - Measure
  • The study of relationship between sides and angles of a triangle .

Trigonometric ratios

  • sine = opp / hyp ,
  • cosecant = 1 / sin = hyp / opp ,
  • cosine = adj / hyp ,
  • secant = 1 / cos = hyp / adj ,
  • tangent = sin / cos = opp / adj ,
  • cotangent = 1 / tan = adj / opp.
  • The values of the trigonometric ratios of an angle do not vary with length of the sides of the triangles if the angles remains the same.
  • [!fact] sin or cos is always less than 1 because hyp is always the largest side in the triangle

Pythagoras theorem

  • opp2opp^2 + adj2adj^2 = hyp2hyp^2

Trigonometric ratios for specific angles

  • All angles which are integral multiples of π/2 are called quadrantal angles
    Pasted image 20240601171809

Trigonometric Quadratic identities

  • sin2sin^2 θ + cos2cos^2 θ = 1
    • sin2sin^2 θ = 1 − cos2cos^2 θ
    • cos2cos^2 θ = 1 − sin2sin^2 θ
  • tan2tan^2 θ - sec2sec^2 θ= 1
    • tan2tan^2 θ + 1 = sec2sec^2 θ
    • tan2tan^2 θ = sec2sec^2 θ − 1
  • cosec2cosec^2 θ - cot2cot^2 θ= 1
    • cot2cot^2 θ + 1 = cosec2cosec^2 θ
    • cot2cot^2 θ = cosec2cosec^2 θ − 1

Opposite Angle Identities

  • sin(−θ)=−sin(θ)sin(−θ) = −sin(θ)
  • cos(−θ)=cos(θ)cos(−θ) = cos(θ)
  • tan(−θ)=−tan(θ)tan(−θ) = −tan(θ)

Trigonometric Identities of Complementary Angles

  • Sin(90– θ)=Cos θSin (90 – θ) = Cos θ
  • Cos(90– θ)=Sin θCos (90 – θ) = Sin θ
  • Tan(90– θ)=Cot θTan (90 – θ) = Cot θ
  • Cot(90– θ)=Tan θCot ( 90 – θ) = Tan θ
  • Sec(90– θ)=Cosec θSec (90 – θ) = Cosec θ
  • Cosec(90– θ)=Sec θCosec (90 – θ) = Sec θ

Trigonometric Identities of Supplementary Angles

  • sin(180°−θ)=sinθsin (180°- θ) = sinθ
  • cos(180°−θ)=−cosθcos (180°- θ) = -cos θ
  • cosec(180°−θ)=cosecθcosec (180°- θ) = cosec θ
  • sec(180°−θ)=−secθsec (180°- θ)= -sec θ
  • tan(180°−θ)=−tanθtan (180°- θ) = -tan θ
  • cot(180°−θ)=−cotθcot (180°- θ) = -cot θ

Horizontal Axis Transforms (Function Unchanged)

AngleQuadrantsin(…)cos(…)tan(…)Explanation
−θ-\theta or 360∘−θ360^\circ - \thetaIV−sin⁡θ-\sin \thetacos⁡θ\cos \theta−tan⁡θ-\tan \thetaCos is + in Q4. Function stays same.
180∘−θ180^\circ - \thetaIIsin⁡θ\sin \theta−cos⁡θ-\cos \theta−tan⁡θ-\tan \thetaSin is + in Q2. Function stays same.
180∘+θ180^\circ + \thetaIII−sin⁡θ-\sin \theta−cos⁡θ-\cos \thetatan⁡θ\tan \thetaTan is + in Q3. Function stays same.

Vertical Axis Transforms (Function →\to Co-Function)

AngleQuadrantsin(…)cos(…)tan(…)Explanation
90∘−θ90^\circ - \thetaIcos⁡θ\cos \thetasin⁡θ\sin \thetacot⁡θ\cot \thetaAll + in Q1. Change to co-function.
90∘+θ90^\circ + \thetaIIcos⁡θ\cos \theta−sin⁡θ-\sin \theta−cot⁡θ-\cot \thetaSin is + in Q2. Change to co-function.
270∘−θ270^\circ - \thetaIII−cos⁡θ-\cos \theta−sin⁡θ-\sin \thetacot⁡θ\cot \thetaTan is + in Q3. Change to co-function.
270∘+θ270^\circ + \thetaIV−cos⁡θ-\cos \thetasin⁡θ\sin \theta−cot⁡θ-\cot \thetaCos is + in Q4. Change to co-function.

Double Angle Identities

  • sin⁡2x=2sin⁡xcos⁡x=2tan⁡x1+tan⁡2x\sin 2x = 2\sin x \cos x = \frac{2\tan x}{1 + \tan^2 x}
  • cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x
  • cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1
  • cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x
  • cos⁡2x=1−tan⁡2x1+tan⁡2x\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}
  • tan⁡2x=2tan⁡x1−tan⁡2x\tan 2x = \frac{2\tan x}{1 - \tan^2 x}

Triple Angles Identities

  • sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3 x
  • cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4\cos^3 x - 3\cos x
  • tan⁡3x=3tan⁡x−tan⁡3x1−3tan⁡2x\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}
  • cosec 3x=cosec3x3cosec2x−4\text{cosec } 3x = \frac{\text{cosec}^3 x}{3\text{cosec}^2 x - 4}
  • sec⁡3x=sec⁡3x4−3sec⁡2x\sec 3x = \frac{\sec^3 x}{4 - 3\sec^2 x}
  • cot⁡3x=cot⁡3x−3cot⁡x3cot⁡2x−1\cot 3x = \frac{\cot^3 x - 3\cot x}{3\cot^2 x - 1} Geek for Geeks - Proof

Angle Sum and Difference Identities

Note that ± means you can use plus or minus, and the ∓ means to use the opposite sign.

  • sin(A±B)=sin(A)cos(B)±cos(A)sin(B)sin(A ± B) = sin(A)cos(B) ± cos(A)sin(B)
  • cos(A±B)=cos(A)cos(B)∓sin(A)sin(B)cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B)
  • tan(A±B)= tan(A)±tan(B)/1∓tan(A)tan(B)tan(A ± B) = tan(A) ± tan(B) / 1 ∓ tan(A)tan(B)
  • cot(A±B)= cot(A)cot(B)∓1/cot(B)±cot(A)cot(A ± B) = cot(A)cot(B) ∓ 1 / cot(B) ± cot(A)

Product-Sum Trigonometric Identities - Four Factorisation Formulas

  • SinA+SinB=2Sin((A+B)/2).Cos((A−B)/2)Sin A + Sin B = 2 {Sin((A+B)/2) . Cos((A-B)/2)}
  • CosA+CosB=2Cos((A+B)/2).Cos((A−B)/2)Cos A + Cos B = 2 {Cos((A+B)/2) . Cos((A-B)/2)}
  • SinA–SinB=2Cos((A+B)/2).Sin((A−B)/2)Sin A – Sin B = 2 {Cos((A+B)/2) . Sin((A-B)/2)}
  • CosA–CosB=−2Sin((A+B)/2).Sin((A−B)/2)Cos A – Cos B = -2 {Sin((A+B)/2) . Sin((A-B)/2)}

Trigonometric Identities of Products - Four Defactorisation Formulas

  • SinA.SinB=Cos(A–B)–Cos(A+B)/2Sin A. Sin B = {Cos (A – B) – Cos (A + B)}/2
  • CosA.CosB=Cos(A+B)+Cos(A–B)/2Cos A. Cos B = {Cos (A + B) + Cos (A – B)}/2
  • SinA.CosB=Sin(A+B)+Sin(A–B)/2Sin A. Cos B = {Sin (A + B) + Sin (A – B)}/2
  • CosA.SinB=Sin(A+B)–Sin(A–B)/2Cos A. Sin B = {Sin (A + B) – Sin (A – B)} / 2

Trigonometric Ratios Derivation

https://www.onlinemath4all.com/trigonometric-ratio-table.html

Class 10

8.1find ratios from triangle
11 questionsshow that
8.2use the table and solve
4 questions
8.3use identities and other formulas
5 questions

RD Sharma new

Trigonometric ratios

  • find the values of other ratios - use the definition and pythagoras theorem
  • be sure where the theta is formed , which can be identified from the angle given
  • in case of prove that or show that questions, check whether any identity can be used, if so proceed ahead
  • if not then try to find the other side using the pythagoras theorem then substitute in the given LHS
  • if not then either divide by sin or cos, as depending upon the given, if tan is given divide by cos, if cot is give, divide by sin.
  • if needed, draw 2 right triangles, marking one angle at each diagram to avoid confusion
  • be sure of the table values, if the degree is given, then substitute and simplify
  • find the theta - get the value in the right side, then take inverse theta and get the angle

Trigonometric identities

  • always look the right hand side, before proceeding
  • rationalise if the identities cant be substituted
  • write the lhs in terms of rhs
  • remember the three square identities of sin, tan, cosec
  • if the rhs is one, try to substitute in its basic terms of sin and cos
  • if the value of the some function is asked, and one of the ratio is given, then plot it in the right triangle and find the other side. then from the definition, find the ratios, and subtitute in the function.

Class 12

Degrees, Minutes, Seconds

A degree is divided into 60 minutes, and a minute is divided into 60 seconds . One sixtieth of a degree is called a minute, written as 1′, and one sixtieth of a minute is called a second, written as 1″.

  • ! 1° = 60′, 1′ = 60″

Angle diagrams for different degrees

degrees class 12

Radian Measure

  • Another unit for measurement of an angle, called the radian measure.
  • radian measures and real numbers can be considered as one and the same.
  • Angle subtended at the centre by an arc of length 1 unit in a unit circle (circle of radius 1 unit) is said to have a measure of 1 radian.
  • The circumference of a circle of radius 1 unit is 2π. Thus, one complete revolution of the initial side subtends an angle of 2π radian.
    radian measure

If in a circle of radius r, an arc of length L subtends an angle θ radian at the centre, we have

  • Radian measure = {π / 180} ×Degree measure
  • Degree measure = {180 / π} ×Radian measure

Graphs

Sine function

Sine-Function-1

Cosecant function

cosec

Cosine function

cosine

Secant graph

sec graph

Tan function

tan

Cot function

cot

Periodicity of ratios

Trig Ratio (f(x))Domain (Valid x)Range (Output Values)PeriodGeneral Solution (if f(x)=f(y))Value at x=nπ (n∈Z)
sin⁡x\sin xR\mathbb{R}[−1,1][-1, 1]2π2\pix=nπ+(−1)nyx = n\pi + (-1)^n y00
cos⁡x\cos xR\mathbb{R}[−1,1][-1, 1]2π2\pix=2nπ±yx = 2n\pi \pm y(−1)n(-1)^n
tan⁡x\tan xR−{(2n+1)π2}\mathbb{R} - \{ (2n+1)\frac{\pi}{2} \}R\mathbb{R}π\pix=nπ+yx = n\pi + y00
cot⁡x\cot xR−{nπ}\mathbb{R} - \{ n\pi \}R\mathbb{R}π\pix=nπ+yx = n\pi + yNot Defined (∞\infty)
sec⁡x\sec xR−{(2n+1)π2}\mathbb{R} - \{ (2n+1)\frac{\pi}{2} \}R−(−1,1)\mathbb{R} - (-1, 1)2π2\pix=2nπ±yx = 2n\pi \pm y(−1)n(-1)^n
cosec x\text{cosec } xR−{nπ}\mathbb{R} - \{ n\pi \}R−(−1,1)\mathbb{R} - (-1, 1)2π2\pix=nπ+(−1)nyx = n\pi + (-1)^n yNot Defined (∞\infty)

Note:

For cos⁡(nπ)\cos(n\pi), the value oscillates:

  • it is 11 when nn is even (e.g., 0,2π0, 2\pi) and
  • −1-1 when nn is odd (e.g., π,3π\pi, 3\pi).
  • This is written as (−1)n(-1)^n

Domain excluded points

RatioFormulaDenominator becomes 0 when…Domain (Excluded Points)
sin⁡x\sin xOppHyp\frac{\text{Opp}}{\text{Hyp}}NeverR\mathbb{R}
cos⁡x\cos xAdjHyp\frac{\text{Adj}}{\text{Hyp}}NeverR\mathbb{R}
tan⁡x\tan xsin⁡xcos⁡x\frac{\sin x}{\mathbf{\cos x}}cos⁡x=0\cos x = 0 (at 90∘,270∘...90^{\circ}, 270^{\circ}...)R−{(2n+1)π2}\mathbb{R} - \{ (2n+1)\frac{\pi}{2} \}
sec⁡x\sec x1cos⁡x\frac{1}{\mathbf{\cos x}}cos⁡x=0\cos x = 0 (at 90∘,270∘...90^{\circ}, 270^{\circ}...)R−{(2n+1)π2}\mathbb{R} - \{ (2n+1)\frac{\pi}{2} \}
cot⁡x\cot xcos⁡xsin⁡x\frac{\cos x}{\mathbf{\sin x}}sin⁡x=0\sin x = 0 (at 0∘,180∘...0^{\circ}, 180^{\circ}...)R−{nπ}\mathbb{R} - \{ n\pi \}
cosec x\text{cosec } x1sin⁡x\frac{1}{\mathbf{\sin x}}sin⁡x=0\sin x = 0 (at 0∘,180∘...0^{\circ}, 180^{\circ}...)R−{nπ}\mathbb{R} - \{ n\pi \}

General solution

RatioQuadrants where value matchesHow we get thereGeneral Solution (if f(x)=f(y))Mathematical Trick
SinI & IIGo Forward from even π\pi, Backward from odd π\pix=nπ+(−1)nyx = n\pi + (-1)^n y(−1)n(-1)^n handles the sign flip.
CosI & IVGo Up (++) or Down (−-) from the start line (2nπ2n\pi)x=2nπ±yx = 2n\pi \pm y±\pm handles the direction.
TanI & IIIJust keep adding half-circles (π\pi)x=nπ+yx = n\pi + ySimple addition (nπn\pi).
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