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Trigonometry Class 11 Chapter 3

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TRIGONOMETRY : CLASS 11 : CHAPTER 3
Different systems of angles, radian measure, degree, AB and CD formulas,  transformations of angles in different quadrants and solution of trigonometric equations

Trigonometry:
It is the branch of mathematics in which we deal with the relation between the angle and sides of the right triangle.
Table 1
0o
30o
45o
60o
90o
180o
270o
360o
Sin
0
1/2
\[1/\sqrt{2}\]
\[\sqrt{3}/2\]
1
0
-1
0
Cos
1
\[\sqrt{3}/2\]
\[1/\sqrt{2}\]
1/2
0
-1
0
1
Tan
0
\[1/\sqrt{3}\]
1
\[\sqrt{3}\]
\[\infty\]
0
\[\infty\]
0
Cot
\[\infty\]
\[\sqrt{3}\]
1
\[1/\sqrt{3}\]
0
\[\infty\]
0
\[\infty\]
Sec
1
\[2/\sqrt{3}\]
\[\sqrt{2}\]
2
\[\infty\]
-1
\[\infty\]
1
cosec
\[\infty\]
2
\[\sqrt{2}\]
\[2/\sqrt{3}\]
1
\[\infty\]
-1
\[\infty\]


DIFFERENT SYSTEMS OF ANGLES
There are mainly three systems of measuring angles:
1) Centesimal System,  2) Sexagesimal System,  3) Circular System.

CENTESIMAL SYSTEM
\[1\; Right angle =100\; grades\; \; or\: \: 90^{o}=100^{g}\]\[1\; grade=100\; minutes\; \; or\: \: 1^{g}=100^{'}\]\[1\; minute=100\; seconds\; \; or\: \: 1^{'}=100^{''}\]
SEXAGESIMAL SYSTEM
\[1\; Right\; angle=90^{o}\]\[1\; degree=60\; minute\; \; or\: \: 1^{o}=60^{'}\]\[1\; minute=60\; second\; \; or\: \: 1^{'}=60^{''}\]
CIRCULAR SYSTEM
Radian Measure:- Angle made by an arc of unit length in a circle of unit radius is called one radian
\[One\; Radian =\frac{Unit\; length\; of\; arc}{Unit\; Radius}\]\[\theta \; Radian =\frac{length\; of\; arc}{Radius\; of\; Circle}\; \; or\: \: \theta =\frac{l}{r}\; \; or\; \; l=r\theta\]
RELATION BETWEEN DEGREE AND RADIAN
\[Angle\; at\; the\; centre\; of\; the\; circle =2\pi \; or\; 360^{o}\]\[2\pi \; Radian=360^{o}\Rightarrow \; \; \pi \; Radian=180^{o}\]\[1\: Radian=\frac{180}{\pi }\; \; \frac{180\times 7}{22}=57^{o}16^{'}\]\[180^{o}=\pi Radian\; \; \Rightarrow \; 1^{o}=\pi Radian=\frac{22}{7\times 180}Radian=0.01746\; Radian\]
Note : In order to convert degree measure into radian measure then we should multiply the angle in the degree measure by \[\frac{\pi }{180}\]
In order to convert a radian measure into a degree measure we should multiply the angle in the radian measure by \[\frac{180 }{\pi }\]
MOVEMENT OF ANGLE IN DIFFERENT  QUADRANT
https://dinesh51.blogspot.com

SIGN OF TRIGONOMETRIC RATIOS IN DIFFERENT QUADRANT
https://dinesh51.blogspot.com
TRIGONOMETRY-CBSE Mathematics
\[For\: odd\: multiples\: of\: \frac{\pi }{2}\: i.e.\; \; \; \frac{\pi }{2}\pm \theta ,\: or\: \frac{3\pi }{2}\pm \theta \\sin\theta \leftrightharpoons cos\theta ,\: \: tan\theta \leftrightharpoons cot\theta ,\: \: sec\theta \leftrightharpoons cosec\theta\]
COMPLETE  TRANSFORMATION OF ANGLES
Trigonometric Transformations at 90o (I and II Quadrant)
sin(90 - θ)
+cosθ

sin(90 + θ)
+cosθ
cos(90 – θ)
+Sinθ
cos(90 + θ)
-Sinθ
tan(90 - θ) 
+cotθ
tan(90 + θ) 
-cotθ
cot(90 - θ)
+tanθ
cot(90 + θ)
-tanθ
sec(90 - θ)
+cosecθ
sec(90 + θ)
-cosecθ
cosec(90 - θ)
+secθ
cosec(90 + θ)
+secθ

Trigonometric Transformations at 180o  (II and III Quadrant)
sin(180 - θ)
Sin θ

sin(180 + θ)
-Sin θ
cos(180 – θ)
-Cos θ
cos(180 + θ)
-Cos θ
tan(180 - θ) 
-tanθ
tan(180 + θ) 
+tanθ
cot(180 - θ)
-cotθ
cot(180 + θ)
+cotθ
sec(180 - θ)
-secθ
sec(180 + θ)
-secθ
cosec(180 - θ)
cosecθ
cosec(180 + θ)
-cosecθ

Trigonometric Transformations at 270o  (III and IV Quadrant)
sin(270 - θ)
-cosθ

sin(270 + θ)
-cosθ
cos(270 – θ)
-Sinθ
cos(270 + θ)
+Sinθ
tan(270 - θ) 
+cotθ
tan(270 + θ) 
-cotθ
cot(270 - θ)
+tanθ
cot(270 + θ)
-tanθ
sec(270 - θ)
-cosecθ
sec(270 + θ)
+cosecθ
cosec(270 - θ)
-secθ
cosec(270 + θ)
-secθ

Trigonometric Transformations at 360o  (IV and I Quadrant)
sin(360 - θ)
-Sin θ

sin(360 + θ)
+Sin θ
cos(360 – θ)
+Cos θ
cos(360 + θ)
+Cos θ
tan(360 - θ) 
-tanθ
tan(360 + θ) 
+tanθ
cot(360 - θ)
-cotθ
cot(360 + θ)
+cotθ
sec(360 - θ)
+secθ
sec(360 + θ)
+secθ
cosec(360 - θ)
-cosecθ
cosec(360 + θ)
+cosecθ

Trigonometric Transformations at 0o   (IV and I Quadrant)
sin(0 - θ)
-Sin θ

sin(0+θ)
+ Sin θ
cos(0 - θ)
+Cos θ
cos(0+θ)
+ Cos θ
tan(0 - θ) 
-tanθ
tan(0+θ) 
+ tanθ
cot( 0-θ)
-cotθ
cot(0+θ)
+ cotθ
sec( 0-θ)
+secθ
sec(0+θ)
+ secθ
cosec( 0-θ)
-cosecθ
cosec(0+θ)
+ cosecθ

STANDARD ANGLES IN DEGREE MEASURE AND RADIAN MEASURE
Table 2
Degree Measure
30o
45o
60o
90o
180o
270o
360o
Radian Measure
\[\frac{\pi }{6}\]
\[\frac{\pi }{4}\]
\[\frac{\pi }{3}\]
\[\frac{\pi }{2}\]
\[\pi\]
\[\frac{3\pi }{2}\]
\[2\pi\]


TRIGONOMETRIC FORMULAS WITH COMPOUND ANGLE
Sin(A+B)=SinACosB + CosASinB
Sin(A-B)=SinACosB - CosASinB
Cos(A+B)=CosACosB - SinA SinB
Cos(A-B)=CosACosB + SinA SinB
\[tan(A+B)=\frac{tanA+tanB}{1-tanA tanB}\]\[tan(A-B)=\frac{tanA-tanB}{1+tanA tanB}\]\[cot(A+B)=\frac{cotA cotB -1}{cotA + cotB}\]\[cot(A-B)=\frac{cotA cotB+1}{cotA - cotB}\]
TRIGONOMETRIC FORMULAS WITH MULTIPLE ANGLE
\[Sin2\theta =2Sin\theta\; Cos\theta \; \; \; OR\; \; \frac{2tan\theta }{1+tan^{2}\theta }\]\[Cos2\theta =Cos^{2}\theta -Sin^{2}\theta,\; \; or\: \: 2Cos^{2}\theta -1,\; \; or\; \; 1-2Sin^{2}\theta \; \; or\; \; \frac{1-tan^{2}\theta }{1+tan^{2}\theta }\]\[tan2\theta =\frac{2tan\theta }{1-tan^{2}\theta }\]\[Sin3\theta =3Sin\theta -4Sin^{3}\theta\]\[Cos3\theta =-\left ( 3Cos\theta -4Cos^{3}\theta \right )\: \: or\: \: \left ( 4Cos^{3}\theta -3Cos\theta \right )\]\[Tan3\theta =\frac{3tan\theta -tan^{3}\theta }{1-3tan^{2}\theta }\]\[Sin^{2}\theta =\frac{1-cos2\theta }{2},\: \: \: Sin^{2}4\theta =\frac{1-cos8\theta }{2}\]\[Cos^{2}\theta =\frac{1+cos\: 2\theta }{2},\: \: \: Cos^{2}\: 4\theta =\frac{1+cos\: 8\theta }{2}\]\[tan^{2}\; \theta =\frac{1-cos2\theta }{1+cos2\theta },\:\: \: \: \: tan^{2}\: 4\theta =\frac{1-cos\: 8\theta }{1+cos\: 8\theta }\]
TRIGONOMETRIC FORMULAS WITH SUB-MULTIPLE ANGLE
\[Sin\theta =2\: sin\: \frac{\theta }{2}\: cos\: \frac{\theta }{2},\:\: or\: \:\; \frac{2tan\frac{\theta }{2}}{1+tan^{2}\frac{\theta }{2}}\]\[Cos\theta =Cos^{2}\; \frac{\theta }{2} -Sin^{2}\;\frac{\theta }{2},\; \; or\; \; 2Cos^{2}\; \frac{\theta }{2} -1,\; \; or\; \; 1-2Sin^{2}\; \frac{\theta}{2} \; \; or\; \; \frac{1-tan^{2}\; \frac{\theta}{2} }{1+tan^{2}\; \frac{\theta}{2} }\]\[Tan\theta =\frac{2tan\frac{\theta }{2}}{1-tan^{2}\frac{\theta }{2}}\]
\[Sin^{2}\; \frac{\theta}{2} =\frac{1-cos\; \theta }{2},\: \:\; \; \: Cos^{2}\; \frac{\theta}{2} =\frac{1+cos\: \theta }{2}\]\[tan^{2}\; \frac{\theta}{2} =\frac{1-cos\theta }{1+cos\theta },\]
A, B FORMULAS:-
\[2SinA\: CosB=Sin(A+B)+Sin(A-B)\]\[2CosA\: SinB=Sin(A+B)-Sin(A-B)\]\[2CosA\: CosB=Cos(A+B)+Cos(A-B)\]\[2SinA\: SinB=-Cos(A+B)+Cos(A-B)\]
C,D FORMULAS:- 
\[SinC+SinD=2Sin\left (\frac{C+D}{2} \right )Cos\left ( \frac{C-D}{2} \right )\]\[SinC-SinD=2Cos\left (\frac{C+D}{2} \right )Sin\left ( \frac{C-D}{2} \right )\]\[CosC+CosD =2Cos\left (\frac{C+D}{2} \right )Cos\left ( \frac{C-D}{2} \right )\]\[CosC-CosD=-2Sin\left (\frac{C+D}{2} \right )Sin\left ( \frac{C-D}{2} \right )\]
SOME SPECIAL FORMULAS
\[Tan\left ( 45+\theta \right )=\frac{1+tan\theta }{1-tan\theta }\]\[Tan\left ( 45-\theta\right )=\frac{1-tan\theta }{1+tan\theta }\]\[Sin^{2}A-Sin^{2}B=Sin(A+B)Sin(A-B)\]\[Cos^{2}A-Cos^{2}B=Sin(A+B)Sin(A-B)\]
PERIODIC FUNCTIONS
All trigonometric ratios are periodic functions
\[Sin(x+2\pi )=Sin(x+4\pi )=Sin(x+6\pi )=Sin(x+8\pi )=..........\]\[Cos(x+2\pi )=Cos(x+4\pi )=Cos(x+6\pi )=Cos(x+8\pi )=..........\]\[Sec(x+2\pi )=Sec(x+4\pi )=Sec(x+6\pi )=Sec(x+8\pi )=..........\]\[Cosec(x+2\pi )=Cosec(x+4\pi )=Cosec(x+6\pi )=Cosec(x+8\pi )=..........\]\[On\; changing\; x\; to\; 2n\pi,\; sinx,\; cosx,\; secx,\; cosecx\; all\; remain\; unchanged.\]\[sinx,\; cosx,\; secx,\; cosecx\; all\; are\; periodic\; functions\; and\; their\; period\; is\; 2\pi\]\[tanx=tan(x\pm \pi )=tan(x\pm 2\pi )=tan(x\pm 3\pi )=........\]\[cotx=cot(x\pm \pi )=cot(x\pm 2\pi )=cot(x\pm 3\pi )=........\]\[tanx\; and\; cotx\; are\; periodic\; and\; their\; period\; is\; \pi\]
PRINCIPAL SOLUTION OF TRIGONOMETRIC FUNCTIONS
\[Solutions\; of\; trigonometric\; equations\; for\; which\;\; 0\leq x\leq 2\pi \; \\ are\; called\; principal\; solutions\]
PERIODIC VALUES:-
It is the least value which when added to the given function so that the value of that function remain unchanged.
TRIGONOMETRIC EQUATIONS:-
Equations involving trigonometric functions of a variable are called trigonometric equations.
GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS:-
\[If\; sinx=0,\; \; then\; x=n\pi\]\[If\; cosx=0,\; \; then\; x=(2n+1)\frac{\pi }{2}\]\[If\; tanx=0,\; \; then\; x= n\pi\]\[If\; sinx=siny\; \; then\; x=n\pi +(-1)^{n}y\]\[If\; cosx=cosy,\; \; then\; x=2n\pi \pm y\]\[If\; tanx=tany,\; \; then\; x=n\pi +y\]

Miscellaneous Exercise Chapter 3 Class 11
Q 8) If tanx = -4/3, and x lie in the II quadrant then find the value of sinx/2, cosx/2, tanx/2
Solution: Since angle lie in the II quadrant. Therefore 
  \[\frac{\pi }{2}<x<\pi \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}\]⇒ Angle  x/2 lie in the first quadrant and in first quadrant sinx/2,  cosx/2 and tanx/2 all are positive  \[tanx=\frac{-4}{3}=\frac{P}{B}\]\[\Rightarrow P= 4k,\: and\: B= -3k\]\[In\: \: \Delta OAB,\; OA=\sqrt{(4k)^{2}+(3k)^{2}}=5k\]\[Now\: \: cosx =\frac{B}{H}=\frac{-3k}{5k}=\frac{-3}{5}\]\[sin^{2}\; \frac{x}{2}=\frac{1-cosx}{2}=\frac{1-(\frac{-3}{5})}{2}\]\[sin^{2}\; \frac{x}{2}=\frac{1+\frac{3}{5}}{2}=\frac{8}{2\times 5}=\frac{4}{5}\]\[sin\; \frac{x}{2}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}\]\[cos^{2}\: \frac{x}{2}=\frac{1+cosx}{2}=\frac{1-\frac{3}{5}}{2}\]\[cos^{2}\: \frac{x}{2}=\frac{2}{2\times 5}=\frac{1}{5}\]\[cos\: \frac{x}{2}=\sqrt{\frac{1}{5}}=\frac{1}{\sqrt{5}}\]\[tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{2/\sqrt{5}}{1/\sqrt{5}}=2\]

Worksheet(1) for the students   (Pair the following)

Which do not have any pair write answer for them also

1) \[tan2x\]

a) \[sinx\]

2) \[\frac{tan2x}{1+tan^{2}x}\]

b) \[\frac{2tan\frac{x}{2}}{1-tan^{2}\frac{x}{2}}\]

3) \[2cos^{2}\frac{x}{2}-1\]

c) \[\frac{2tan\frac{x}{2}}{1+tan^{2}\frac{x}{2}}\]

4) \[sinx\]

d) \[\frac{2tanx}{1-tan^{2}x}\]

5) \[\sqrt{1-cos^{2}x}\]

e) \[|sinx|\]

6) \[\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}\]

f)\[sin2x\]

g)\[sinx\]

7) \[sec^{2}x-1\]

h)\[cosx\]

8) \[\frac{1-tan^{2}\frac{x}{2}}{1+tan^{2}\frac{x}{2}}\]

i) \[2sin\frac{x}{2}cos\frac{x}{2}\]

j) \[tan\left ( \frac{\pi }{4}-\frac{x}{2} \right )\]

9) \[cosAcosB+sinAsinB\]

k) \[cos(B-A)\]

10) \[tan\left (\frac{\pi }{4}-x \right )\]

l)\[cos(A+B)\]

m) \[\frac{1+tanx}{1-tanx}\]

Answer Key : (1, d), (2, f), (3, n), (4, c and i), (5, e), (6, j), (7, tan2x), (8, h), (9, k), (10, n)

Worksheet(2) for the students   Pair the following

1) \[\sqrt{1-cosx}\]

a\[\frac{tanA-tanB}{1-tanAtanB}\]

2) \[\sqrt{1+cosx}\]

b \[4cos^{3}x-3cosx\]

3) \[\sqrt{1-sin2x}\]

c \[\frac{3tan-tan^{3}x}{1-3tan^{2}x}\]

4) \[\sqrt{1+sin2x}\]

d \[sinAcosB-cosAsinB\]

5) \[sin3x\]

e \[3sinx-4sin^{3}x\]

6) \[cos3x\]

f \[|cosx-sinx|\]

7) \[tan3x\]

g \[\left |\sqrt{2}sin^{2}\frac{x}{2} \right |\]

8) \[sin(A-B)\]

h \[|cosx+sinx|\]

9) \[tan(A-B)\]

i \[\left |\sqrt{2}cos^{2}\frac{x}{2} \right |\]

Answer Key: (1, g), (2, i), (3, f), (4, h), 5, e), (6, b), (7, c), (8, d), (9, a)

THANKS FOR YOUR VISIT
PLEASE COMMENT BELOW
India Himachal Pradesh 174030, India

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