Trigonometric Functions Class 11
Master Trigonometric Functions Class 11 with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.
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Trigonometric Functions Class 11 – NCERT Solutions
Each question below opens its complete step-by-step Teachoo solution.
Ex 3.1
13 questionsEx 3.1, 1 (i)
Ex 3.1, 1 teachoo.com
Find the radian measures corresponding to the following degree
measures:
(i) 25°
Radian measure = ao x Degree measure
== ‘©
= a0 * 25
enx=
ENG
Sr .
= 36 radians
Ex 3.1, 1 (ii)
Find the radian measures corresponding to the following degree measures:
(ii) –47° 30’
Ex 3.1, 1 (iii)
Find the radian measures corresponding to the following degree measures:
(iii) 240°
Ex 3.1, 1 (iv)
Find the radian measures corresponding to the following degree measures:
(iv) 520°
Ex 3.1, 2 (i)
Ex 3.1, 2 teachoo.com
Find the degree measures corresponding to the following radian
22
measures (use T= =)
ait
(i) 3
We know that
. Te
Radian measure = 180 x Degree measure
11 Tw
— =-— x Degree measure
16 180
11 180
— x — = Degree measure
16 Tw
D 11 +180
egree measure =— x —
8 16 te
11 180
Degree measure =— x—— x7
16 22
Ex 3.1, 2 (ii)
Find the degree measures corresponding to the following radian measures ("use " π" = " 22/7)
(ii) –4
Ex 3.1, 2 (iii)
Find the degree measures corresponding to the following radian measures ("use " π" = " 22/7)
(iii) 5𝜋/3
Ex 3.1, 2 (iv)
Find the degree measures corresponding to the following radian measures ("use " π" = " 22/7)
(iv) 7𝜋/6
Ex 3.1, 3
Ex 3.1, 3 teachoo.com
A wheel makes 360 revolutions in one minute. Through how many
radians does it turn in one second?
Number of revolutions in 1 minute = 360
Number of revolutions in 60 second = 360
So,
er 360
Number of revolutions in 1 second = oO
=6
Now, finding angle made in 6 revolutions
Ex 3.1, 4
Ex 3.1, 4 teachoo.com
Find the degree measure of the angle subtended at the centre of a
. . 22
circle of radius 100 cm by an arc of length 22 cm. (use T= 2)
Given
l=22 cm Jo)
r=100 cm
22cm
6=?
We know that
o= +
Putting values
22
8= 100
Ex 3.1, 5
Ex3.1,5 teachoo.com
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find
the length of minor arc of the chord.
Given Diameter = 40 cm
Radius = r= 20cm
LN)
BN 20cm €
We need to find
length of minor arc of the chord. 8
Thus, we use formula
6= : to find length of arc
Ex 3.1, 6
teachoo.com
Ex 3.1, 6
If in two circles, arcs of the same length subtend angles 60° and 75° a
the centre, find the ratio of their radii.
$
length | length I
We know that
l=ré
Let the radius of the two circles ber, and r,
Ex 3.1, 7
Ex3.1,7 teachoo.com
Find the angle in radian though which a pendulum swings if its
length is 75 cm and the tip describes an arc of length
(i) 10cm
6?
Given, 75cm
r= Radius = Length of pendulum = 75 cm --"
10 cm
U= Length of arc = 10cm
We need to find 8
We know that
i
@=-
Tr
Ex 3.2
10 questionsEx 3.2, 1
Ex 3.2,1 teachoo.com
Find the values of other five trigonometric functions if
cosx = -5, x lies in third quadrant.
Rough
Since x is in 3" Quadrant
sin and cos will be negative $ A
But, tan will be positive
T c
Given
-1
cos x =—
2
We know that
sin? x + cos?x =1
Ex 3.2, 2
teachoo.com
Ex3.2, 2
Find the values of other five trigonometric functions if sin x = gx
lies in second quadrant.
Rough
Since x is in II"? Quadrant
a ~ S A
sin will be positive
But cos and tan will be negative T Cc
H inx=-
ere , sINnX = =
We know that
sin’x + cos’x=1
(5) + we
=) + cos’x =1
5
Ex 3.2, 3
Ex 3.2, 3 teachoo.com
3
Find the values of other five trigonometric functions if cot x = rare lie’
in third quadrant.
Rough
Since x lies in III"? Quadrant 9
sin and cos will be negative S A
But tan will be positive
T c
; 3
Given cot x =—
4
We know that
1+ cot? x = cosec” x
3\2
1+ (=) = cosec?x
4
Ex 3.2, 4
Ex 3.2, 4 teachoo.com
Find the values of other five trigonometric functions if sec x = ie Xx
lies in fourth quadrant.
Rough
Since x lies in the IV Quadrant
Where cos will be positive 5 A
But sin and tan will be negative T Cc
We know that
1+ tan’x = sec’x
13\2
1+tan’x = (=)
5
13\2
tan’x= (=) -1
5
Ex 3.2, 5
teachoo.com
Ex 3.2, 5
Find the values of other five trigonometric functions if tan x = D>
x lies in second quadrant.
Rough
Since x lies in II"4 Quadrant
: : at $s A
So, sin x will be positive
But tan x and cos x will be negative T Cc
We know that
1+tan’x = sec’x
2
-5
1+ (S) = sec?x
12
25
1+ — =sec’x
144
Ex 3.2, 6
Ex 3.2, 6 teackoo.com
Find the value of the trigonometric function sin 765°
sin 765°
= sin (765 x —_)
180 4
_ 17 V7
= sin (-) 16
1
. 1
= sin (4, 1)
. 1
= sin (47 + a Tt)
Values of sin x repeats after an interval of 21
Hence, ignoring 2 x (2n) i.e. 4n
. 7t
=sin G 1)
Ex 3.2, 7
Ex 3.2, 7 feachoo.com
Find the value of the trigonometric function cosec (-1410°)
ee
=—cosec (1410°) So, cosec(-x) = ~cosec x
Rough
we
=-cosec (1410 x =) 7
Ya7
42
=-cosec (27) 5
= —cosec |
475
Ere
5 5 1
=-cosec (7 —7 =7+~ org-i
6 6 6
We take 8 -: as it
1
=-cosec (an -5 nr) is even
Ex 3.2, 8
Ex 3.2, 8 teackoo.com
. . . . 1971
Find the value of the trigonometric function tan >
197”
tan —
3
1 6
=tan6 3 Tt J) 19
18
1 1
=tan (6n + 3 n)
1
= tan (3(2n) +3 n)
Values of tan x repeats after an interval of 2n
Hence, ignoring 3 x (2n)
=tan(; n)
=tan(> 1
Ex 3.2, 9
Ex 3.2,9 teachoo.com
. . . Lim
Find the value of the trigonometric function sin (—)
. (4 *)
sin (—_—
3 Rough
. 117 3
=-sin (=) (As sin(-x) = -sin x) 3 Ji
9
2 2
=-sin (3 3 nr)
41 2
1. So, - =3 3
=-sin(4n-—1
( 3 ) =3+2 or4-+
3 3
Values of sin x repeats after an interval of 2n.
Hence ignoring 47 i.e. 2 x (2n) We take 4 -; as it
1 is even
=-sin (> nr)
Ex 3.2, 10
Ex 3.2, 10 teachoo.com
Find the value of the trigonometric function cot (-=)
-157% Rough
cot (—*)
4 3
s 475
=-cot (=) (As cot(-x) = -cot x) 12
7 —
3
3 15 _33
=-cot (357) $0, 7537
=3+2 or 4—+
1 4 4
=-cot (4x -| n)
Values of cot x repeats after an interval of 2n. || We take 4 -; as it
Hence, ignoring 4ri.e. 2 x (2m) is even
--cot (Fn)
=-cot( [A
Ex 3.3
26 questionsEx 3.3, 1
Ex 3.3,1 teachoo.com
Prove that
in? = 27m pape Zit
sin’ — + cos*> — tan’ 7 5
Solving L.H.S
in2 = 2m_ 2m
sin’ + cos*> - tan? >
Putting m = 180°
— oa .9 180° 2 180° 4 180°
= sin’—— + cos*—— — tan*
= sin? 30° + cos? 60° - tan? 45°
; i, ot 7 o_t o
Putting sin 30° = 37 sin 60° = 3 tan 45°=1
Ex 3.3, 2
There is some mistake in this Video. Please check images above.
View solutionEx 3.3, 3
Ex 3.3, 3 teachoo.com
Prove that cot?t + cosec— +3 tan?= =6
Solving L.H.S.
TT 51 1
cot? — + cosec — + 3 tan? —
6 6 6
Putting m = 180°
180 5 x180 180
= cot?(“) + cosec(-~*) +3 tan’(~*)
6 6 6
= cot? 30° + cosec (150°) + 3tan2 30°
Ex 3.3, 4
Ex 3.3, 4 teachoo.com
93
Prove that 2sin? > + 2cos? . + asec? = =10
Solving L.H.S
2sin? = + 2cos? = + 2sec?=
4 4 3
Putting m = 180°
. 80 80 80
2 sin? (3 x = ) + 2cos? =) + 2sec? >)
= 2sin? (135°) + 2 cos? (45°) + 2sec?(60°} (Using sin (180 — 8) = sin 8
Here, sin 135°
= sin (180—45°)
cos 45° = 7
= sin 45°
oO 1
sec 60° = tos 60° =
va
1
= 7=2
2
Ex 3.3, 5 (i)
Ex 3.3, 5 teachoo.com
Find the value of:
(i) sin 75°
sin 75°
= sin (45° + 30°)
sin (x + y) = sinx cos y+cosxsiny
Putting x = 45° and y = 30°
= sin 45° cos 30° + cos 45° sin 30°
1,8 ,1,1
“V2 20 y2° 2
ai (3,2
2 ( 2 5)
_¥3+1
22
Ex 3.3, 5 (ii)
Find the value of:
(ii) tan 15°
Ex 3.3, 6
Ex 3.3, 6 teachoo.com
Prove that:
TT Tw : Tw : Tw .
cos G - x) cos G- y) -sin G- x) sin G- y) =sin(x + y)
Solving L.H.S
We know that
cos (A+B) =cosA cosB-sinA sinB
The equation given in Question is of this form
we we
Where A= G —x) B= G -y)
Hence
(F—x) 0s (7 y) -sin Fx) sin(F—y)
cos{7 —X)}cos(7—y)-sin{; —xJsin(o —y
Ex 3.3, 7
Ex 3.3, 7 teachoo.com
tan(= + x 1+t 2
Prove that: tan(f + x) = a)
tan(> _ x) 1-tan x.
Solving L.H.S.
tan G + x)
tan(+ -x)
Ex 3.3, 8
Ex 3.3, 8 teachoo.com
cos (w+x)cos(—x) _ 2
Prove that Sin Grow cos ("+ x) (+x) = cot? x
Solving L.H.S.
cos(# + x) cos (—x)
sin(at — x) cos G + x)
Putting m= 180°
___¢0s(180° + x) cos (-x)
~ sin(180° — x) cos (90° + x)
Using
cos (180° + x) =—cos x
cos (-x) = cos x
sin (180° —x) = sinx
& cos(90° + x) =-sin x
Ex 3.3, 9
teachoo.com
Ex 3.3,9
Prove cos F + x) cos (21 + x)[cot (= - x) + cot (2+ x)] =1
Solving L.H.S.
Now,
3n +
cos (F+ x) =sinx
cos (27 + x) = cos x
cot (2m + x} = cot x
cot G- x) =tanx
Ex 3.3, 10
Ex 3.3, 10 teachoo.com
Prove that
sin(n + 1)xsin(n + 2)x+cos(n + 1)xcos(n + 2)x =cosx
Solving L.H.S.
We know that
cos (A-B)=cosA cos B+sinAsinB
Here, A= (n+1)x,B=(n+2)x
Hence
sin(n + 1)xsin(n + 2)x + cos(n + 1)xcos(n + 2)x
=cos [(n+1)x — (n+2)x]
Ex 3.3, 11
Ex 3.3, 11 teachoo.com
310 310 +
Prove that cos (= + x) —cos (= _ x) =-¥2 sinx
Solving L.H.S.
310 310
cos (=F + x) — cos (= _ x)
Using cos x — cos y =—2 sin — sin <>
Putting x = +x & y= =x
4 4
31 31 31 31
= +x)+(-x =+x)-{—-x
wasn (CED CE=9) ,, (CE2I=CE=9)
2 2
31 31 31 31
—-+—)+@-x) —+x-S4+x
=-2 sin (Greens 4__4 ) ) sin (soe 5 4 )
Ex 3.3, 12
Ex 3.3, 12 teachoo.com
Prove that sin? 6x - sin? 4x = sin 2x sin 10x
Solving L.H.S.
sin? 6x — sin? 4x
= (sin 6x + sin 4x) (sin 6x — sin 4x}
Lets calculate (sin 6x + sin 4x) and (sin 6x — sin 4x) separately
sin 6x + sin 4x sin 6x — sin 4x
Using sin x + sin y =2 sin ai cos aa Using sin x — sin y =2 cos ay sin aa
Putting x = 6x & y = 4x Putting x = 6x & y = 4x
= dein (Okt 6x—4x 6xt+4x\ . (6x-4x
=2sin ( 2 ) cos ( 2 ) =2 cos (=) sin(*)
= i 10%" 2x) 10x . 2x
= 2sin ( 2 \ cos (=) =2 cos (*) sin ()
= 2sin 5x cos x = 2 cos 5xsinx
Ex 3.3, 13
Ex 3.3, 13 teachoo.com
Prove that cos? 2x - cos? 6x = sin 4x sin8x
Solving L.H.S.
cos? 2x — cos? 6x
= (cos 2x + cos 6x) (cos 2x — cos 6x)
Lets calculate (cos 2x + cos 6x) and (cos 2x — cos 6x) separately
cos 2x + cos 6x cos 2x — cos 6X
Using cos x + cos y = 2 cos = cos oa Using cos x — cos y = —2 sin sin a
Putting x = 2x & y = 6x Putting x = 2x & y = 6x
2x+6) 2x-6)
=2 cos GS *) cos A *) =—2sin (=) sin(75*)
2 2 2 2
8. -4
=2 cos () cos (=) =—2sin () sin (=)
2 2 2 2
= 2 cos 4x cos (-2x) =-2 sin 4x sin (—2x)
Ex 3.3, 14
teachoo.com
Ex 3.3, 14
Prove that sin 2x + 2sin 4x + sin 6x = 4cos? x sin 4x
Solving L.H.S.
sin 2x + 2sin 4x + sin 6x
= (sin 6x + sin 2x) + 2sin 4x
Using sin x + sin y= 2 sin ae cos >
Putting x = 6x & y = 2x
=2sin es) cos on) + 2sin 4x
=2sin &) cos os) + 2sin 4x
=2 sin 4x cos 2x + 2sin 4x
Ex 3.3, 15
Ex 3.3, 15 teachoo.com
Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x — sin 3x)
Solving L.H.S.
cot 4x (sin 5x + sin 3x}
Using sin x + sin y = 2 sin — cos <>
Putting x = 5x & y = 3x
= cot 4x x[2 sin ee) cos ms) ]
= cot 4x x [ 2sin &) cos eo)
= 2 cot 4x sin 4x cos x
= 2 8 x gin ax x cos X
sin 4x
=2 cos 4x cos x
Ex 3.3, 16
Ex 3.3, 16 teachoo.com
cos 9x — cos 5x sin 2x
Prove that ———————_ = ———_
sin 17x — sin 3x cos 10x
Solving L.H.S
cos 9x — cos 5x
sin 17x — sin 3x
We solve cos 9x — cos5x & sin 17x — sin 3x seperately
cos 9x — cos 5x sin 17x — sin 3x
- an ed . AY em
Using cos x — cos y = —2 sin=+2 sin =—2| Using sin x— sin y = 2 cos ZH
2 2
Putting x = 9x & y = 5x Putting x = 17x & y = 3x
_ 17x43. . {17X-3
=—2sin (=) sin(= =) =2 0s ( x *) sin( Xx *)
2 2 2 2
20. . {14
=—2sin (=) sin (=) =2 cos F) sin (=)
2 2 2 2
=-2 sin 7x sin (2x) =2 cos 10x sin 7x
Ex 3.3, 17
Ex 3.3, 17 teachoo.com
sin 5x + sin 3x
Prove that ——————_ = tan 4x
cos 5x + cos 3x
Solving L.H.S.
sin 5x + sin 3x
cos 5x + cos 3x
We solve sin 5x + sin 3x & cos 5x + cos 3x seperately
sin 5x + sin 3x cos 5x + cos 3x
Using sin x + sin y =2 sin =~ cos aa Using cos x + cosy =2 cos= 5% cos=>*
Putting x = 5x & y = 3x Putting x = 2x & y = 6x
. 5x+3x 5x—3x 5x+3 5x—3.
=2sin S) cos CS) =2 cos C=) cos CS)
2 2 2 2
+ 8x 2x 8x 2x
=2sin F) cos (=) =2 cos () cos (=)
2 2 2 2
=2 sin 4x cos x = 2 cos 4x cos (x)
Ex 3.3, 18
Ex 3.3, 18 teachoo.com
Prove that ——— = tan =
cosx+cosy 2
Solving L.H.S
sin x — siny
cosx+cosy
We Know that
sinx—sin y =2cos =) sin G)
& cos X + cos y = 2 cos G2) cos CG)
x+y _ ({xX-y¥
- 2 cos(*3*) sin(*>*)
x+y x-y
2 cos(*=*) cos(*>*)
Ex 3.3, 19
teachoo.co
Ex 3.3, 19 ”
sin xX + sin 3x
Prove that ————— = tan 2x
cosx + cos 3x
Solving L.H.S.
sin xX + sin 3x
cos X + cos 3x
We solve sin x + sin 3x & cosx + cos 3x seperately
sinx + sin 3x cos X + cos 3x
Using sin x + sin y = 2 sin = cos oa Using cos x + cos y = 2 cos = cos oa
Putting x = x & y = 3x Putting x = x & y = 3x
_ + X+3x X—3X x+3x 5x—3x
=2sin ( 2 ) cos ( 2 ) =2 cos (*) cos (=)
_ + Ax —2x Ax 2x
=2sin (=) cos (=) =2 cos () cos (=)
=2 sin 2x cos (-x) = 2 cos 2x cos (—x)
Ex 3.3, 20
Ex 3.3, 20 teachoo.com
sin x — sin 3x .
Prove that —=————_ = 2 sin x
sin” X— cos’ xX
Solving L.H.S.
sin x — sin 3x
sin’ x — cos’? x
We solve sin x — sin 3x & sin? x — cos? x separately
sin x — sin 3x sin? x — cos? x
ton ct : +y xe
Using sin x — sin y = 2 cos = sin 3 We know that
; cos 2x = cos? x — sin? x
Putting x = x & y = 3x
cos 2x = — ( sin? x — cos? x)
=2 X + 3x . {kK — 3x
=2Zcos ~~, sin ~~ Thus,
-( sin? x — cos? x) = cos 2x
AX + —2x
=2cos (2) sin (=) 9 2
2 2 — (sin? x-—cos* x) =— cos 2x
= 2 cos 2x sin (-x) =- cos 2x
Ex 3.3, 21
Ex 3.3, 21 teachoo.com
cos 4x + cos 3x + cos 2x
Prove that ————_—_——_———— = cot 3x
sin 4x + sin 3x + sin 2x
Solving L.H.S
Solving Numerator and Denominator separately
We know that
x+ y x-y
cos X + COS y = 2cos =) cos =)
Replacing x by 4x and y by 2x
4x + 2x 4x — 2x
cos 4x + cos 2x = 2cos (=). cos (=)
6x 2x
= 2c0s (=). cos (=)
2 2
Ex 3.3, 22
Ex 3.3, 22 teachoo.com
Prove that cot x cot 2x — cot 2x cot 3x— cot 3x cotx=1
Solving L.H.S.
cot x cot 2x — cot 2x cot 3x — cot 3x cot x
= cot x cot 2x — cot 3x (cot 2x + cot x)
= cot x cot 2x — cot (2x + x) (cot 2x + cot x)
cot A cotB-1
; 4B) = coe cot B- 1
(Using cot (A 8) cotA+cotB )
cot 2x cotx—-—1
= cot x cot 2x — (ae) (cot 2x + cot x)
cot x + cot 2x
= cot x cot 2x — (cot 2x cot x -— 1)
= cot x cot 2x — cot 2x cotx +1
Ex 3.3, 23
MB teachoo.com
Prove that tan 4x = A tan x (i-tan’x)
1-6 tan* x+tan* x
Solving L.H.S.
tan 4x
We know that
2tanx
tan 2x = ————
1-tan’x
Replacing x with 2x
2tan 2x
tan (2 x 2x) = ———~—
1 — tan? 2x
2tan 2x
tan 4x =, —_
1 — tan? 2x
Ex 3.3, 24
Ex 3.3, 24 teachoo.com
Prove that cos 4x = 1- 8sin? x cos?x
Solving L.H.S.
cos 4x
We know that
cos 2x = 2 cos? x-1
Replacing by 2x
cos 2(2x) = 2 cos? (2x) -1
cos 4x = 2 cos? 2x -1
= 2(cos 2x -1
Using cos 2x = 2 cos? x-1
=2(2cos?x-1)?-1
Ex 3.3, 25
Ex 3.3, 25 teachoo.com
Prove that: cos 6x = 32 cos®° x — 48 cos*x + 18 cos*x-1
Solving L.H.S.
cos 6x
We know that
cos 2x = 2 cos? x-1
Replacing by 3x
cos 2(3x) = 2 cos? (3x) -1
cos 6x = 2 cos? 3x -1
= 2(cos 3x)?-1
Putting cos 3x = 4 cos? x — 3 cos x
=2(4cos?x-3cosx)/?-1
Examples
29 questionsExample 1
Example 1 teachoo.com
Convert 40° 20’ into radian measure.
40° 20° 1 degree = 60 min
4 20°
= 40° + <0
oP
=40°+ 3
_ 120°+1
"3
= ze
~ 3
We know that
Radian measure = ao x Degree measure
Example 2
teachoo.co
Example 2 tia!
Convert 6 radians into degree measure.
We know that
. Te
Radian measure = Te0 * Degree measure
6 =— x Degree measure
~ 180 8
6 x180
— = Degree measure
SxtOX7 = Degree measure
22 anes 343
1173780
540 x7
——— __ = Degree measure 33
1 48
44
540 x7
D aoe 40
egree measure Ta 33
7
3780
Degree measure = TT
Example 3
Example 3 teachoo.com
Find the radius of the circle in which a central angle of 60°
intercepts an arc of length 37.4 cm (use m= 2 )
Given 1 = 37.4cm
Cao
and 8 = 60° ce>
Converting angle into radians 37.4cm
Radian measure = ino x Degree measure
=— x 60
180
no.
=3 radian
Example 4
Example 4 teachoo.com
The minute hand of a watch is 1.5 cm long. How far does its tip move
in 40 minutes? (Use nm = 3.14).
We know that i
l=ré 9
Here, r = minute hand of clock = 1.5 cm
i
Finding Angle @ a
Now we know that
Minute hand make 1 revolution in 1 hour
Hence we can say that,
Minute hand completes 360° in 1 hour
Example 5
teachoo.com
Example 5
If the arcs of the same lengths in two circles subtend angles 65° and
110° at the center, find the ratio of their radii.
A
length | length I
We know that
l=ré
Let the radius of the two circles ber, and r,
Example 6
Example 6 teachoo.com
If cos x = -:, x lies in third quadrant, find the values of other five
trigonometric functions.
Since x is in III"? Quadrant Rough
sin and cos will be negative $ (II) A(l)
But tan will be positive
TUM) | cc)
| -3
Given, cos x = —
5
We know that
sin? x + cos?x=1
2
-3
sin?x + (=) =1
5
Example 7
Example 7 teachoo.com
If cotx = -= , X lies in second quadrant, find the values of other
five trigonometric functions.
Rough
Since x lies in II"? Quadrant
S(II) | A (I)
Where cos x and tan x will be negative
But sin x will be Positive T (IM) | C(IV)
We know that
1+ cot?x = cosec? x
-5\2
1+ (5) = cosec?x
12
25
1+— =cosec’*x
144
Example 8
teachoo.co
Example 8 eae om
Find the value of sin a
in Boe
sin
. 1
= sin (10n + 3 )
: 1
=sin(5x (2n)+57)
Values of sin x repeats after an interval
of 27, hence ignoring 5 x (2n)
. fl
= sin G n)
: 1 °
= sin G x 180 )
Example 9
Example 9 teachoo.com
Find the value of cos (-1710°).
cos(—1710°)
= cos(1710°)
= Cos (1710 x i)
180
=cos (>)
7 2
1
= Cos (9 2 )
=Cos (107 - 4)
2
Value of cos x repeats after an interval of 2m,
Hence, ignoring 5 x (2n) i.e. 10n
Example 10
teachoo.com
Example 10
Prove that
asin = sec — asin cote a1
sin secs sin—— cot =
Solving LHS
asin = sec —a sine cot
sin = sec > sin—— cot?
1 . ca
=3x5x2-4sin (a —2) x1
. TE
=3-4sin—
1
=3-4x—
=1
=R.H.S.
Example 11
Example 11 teachoo.com
Find the value of sin 15°.
sin 15° = sin (45° — 30°}
sin (x —y) = sin x cos y—cos x siny
Putting x = 45° y = 30°
= sin 45° cos 30° — cos 45° sin 30°
1 v3 #121
=x -=x-
v2 2 V2 2
4 (4%)
= 35 5
_v3-1
=
Example 12
teachoo.com
Example 12
. 13
Find the value of tan =
t Wie t (n + =)
an— 5 = tan D
t Te
=tan—
12
t G =)
=tan(———
4 6
tanx — tan
Using tan (x - y) = ——eeeus
1+ tanxtany
TT TT
_ tan " - tan=
~ TT TT
1+ tan 7 tan ra
Example 13
teachoo.co
Example 13 mn
sin(x+y) _ tanx+tany
Prove that ———— = —————
sin(x+y) tanx—tany
Taking L.H.S.
sinxt+y) _ sin x cosy + cosxsiny
sin(x+y) "sin x cos y-—cosxsiny
Dividing the numerator and denominator by cos x cos y,
sinxcos y + cosxsiny
_ cos x cos y
~ sin x cos y — cos x siny
cos x cos y
sinx cosy + cosxsiny
— cosxcos y cos x cos y
~ sinx cos y _ cosxsiny
cos x cos y cos xcos y
Example 14
Example 14 teachoo.com
Show that
tan 3x tan 2x tan x = tan 3x —tan 2x —-tanx
We know that 3x = 2x+ x
Therefore,
tan3x = tan(2x + x)
tan 2x + tanx
tan 3x = ————_—_
1—-tan 2x tan x
tan 3x -tan3x tan 2x tanx = tan2x + tanx
tan 3x -tan 2x -tanx = tan3x tan 2x tanx
tan 3x tan 2xtanx = tan3x- tan2x- tanx
Example 15
Example 15 teachoo.com
Prove that
cos (E+ x)+cos (F- x)= V2 cosx
Solving L.H.S.
cos (G+ x) + cos G - x)
cosx+ cos y = 2 cos *—* cos*—
Putting x = 7x & y = 5x
Stxtinx E+x-G-x)
=2co0s ( 4 cos AA
2 2
=2 cost cosx
1
=2 x B cos x
=V2 cosx
=RHS.
Example 16
Example 16 teachoo.com
cos 7x+ cos 5x
Prove that ———————— = cot x
sin 7x — sin 5x
Solving L.H.S.
We solve cos 7x + cos5x & sin 7x —sin 5x separately
cos 7x + cos 5x sin 7x — sin 5x
+ - . XO
cos x + cos y = 2 cos“ cos=—* sin x -sin y = 2 cos== sin
Putting x = 7x & y = 5x Putting x = 7x & y = 5x
=2 (e + =) @ - **) =2 = + =) . ‘ - **)
=2 cos | — } cos |, — = 2 cos (> } sin{
= 2.08 (5*) cos (=) = 20s (**) sin (5)
= 2 cos({——} cos (> =2cos{——} sin\>
= 2 cos 6x Cos X =2 cos 6x sin x
Example 17
teachoo.co
Example 17 mn
sin 5x — 2sin 3x + sinx
Prove that —————_—————— = tan x
cos 5X — cosx
Solving L.H.S.
sin 5x + sin x — 2sin 3x
cos 5X — cosx
_ (sin 5x + sin x) — 2 sin 3x
~ cos 5x — cosx
Solving numerator and denominator separately
Example 18
Example 18 teachoo.com
; 3 12 —_
If sinx = = cosy= re where x and y both lie in second
quadrant, find the value of sin (x + y).
We know that
sin (x + y}=sinx cos y+cosx siny
We know that value of sin x and cos y
but we do not know of cos x and sin y
Let us first find cos x
We know that
sin’x + cos’x=1
Example 19
Example 19 teachoo.com
Prove that cos 2x cos > — cos 3x cos = = sin 5x sin =
Solving L.H.S
We know that
2 cos x cos y = cos (x + y) + cos (x— y)
cos x cosy = ; (cos (x + y) + cos (x -y))
. x 9x
Solving cos 2x cos 3 and cos 3x cos z separately
Example 20
teachoo.com
Example 20
find the value of tan >
t Te
ane
Putting m = 180°
180°
= tan ——
8
=t 45°
=tan—
. 45°
We find tan => using tan 2x formula
2tanx
tan 2x = ——___
1-tan*x
Example 21
Example 21 teachoo.com
iftanx=2,n<x< , find the value of sin=, cos ~ and tan~
4 4 2 2 2
Given that
Rough
exe 3n
m<xX<—
2 s A
180° <x< 3 180°
—x
x9 T c
180° <x < 270°
Dividing by 2 all sides
180° x 270°
eee Se
2 2° 2
x
90° <-< 135°
2
x
So, 3 lies in 2"4 quadrant
Example 22
Example 22 teachoo.com
tT 1a 3
Prove that cos” x + cos? (x + *) + cos? (x - *) =5
Lets first calculate all 3 terms separately
We know that
cos 2x = 2 cos? x- 1
cos 2x + 1 = 2cos? x
cos2x+1
——— = cos? x
2
2 cos2x+1
cos? x = ————_
2
. . Te
Replacing x with (x + 4)
Question 1
teachoo.com
Example 18
Find the principal solutions of the equation sin x = 7
: v3
sin x =—
2
We know that sin 60° = 8
Here sin is positive,
Ss A
180-6 8
T c
180+6 360-8
sin is positive in I and UI"? quadrant
Value in Ist quadrant = 60°
Value in II"? Quadrant = 180 - 60
= 120°
Question 2
Example 19 teachoo.com
Find the principal solutions of the equation tan x = “Fe
tanx=—
an x= B
We know that
tan 30° = 4
an =F
Ss A
180-8 8
Since tan x is negative T Cc
180+98 360-9
So, x will lie in [IP? and Iv" Quadrant
Question 3
teachoo.com
Example 20
Find the solution of sin x =— a
Rough
We know that
Let sinx = sin y (4) sino? =23
2
. -v3 RE
also sin x = 8 (2) But we need - “=
S A
From (1) and (2) 180-9 9
siny = 22
ye T c
: 4 _
siny=sin +n 180+6] 360-6
>y= 4 Tt So, angle is in 3" and 4th
3 quadrant
8 = 60°
180 + 6 = 180 +60
= 240°
= 240 x=
180
4
=o1
3
Question 4
teackoo.com
Example 21
1
Solve cos x = —
2
Let cos x = cosy .(1)
; 1
Given cos x => (2)
Rough
From (1) and (2) We know that
ai 60° =~
cos y= 5 cos =5
Tw
cos y = cos —
3 TT
So, 60° = 60 x ——
- 180
Y=3 a=
"3
Question 5
Example 22 teachoo.com
Tw
Solve tan 2x =— cot (x + *)
cia
tan 2x =—cot (x + =)
We need to make both in terms of tan
Rough
tan (90° + 8) =—cot 6
—cot 8 = tan (90° + 8}
—cot 6 = tan G + )
. T
Replacing 8 by x + 3
Tw TC ‘T
—cot (x + *) =tan (G+ X+ =)
3 2 3
Question 6
teachoo.com
Example 23
Solve sin 2x — sin 4x + sin 6x = 0
sin 2x—sin 4x + sin 6x = 0
(sin 6x + sin 2x) — sin 4x =0
We know that
+ + _ 2 {tty x-y
sin (x) + sin y = 2sin Ce ) cos CS )
Replacing x by 6x & y by 2x
2 sin F) cos es) —sin 4x=0
2 sin & cos ) —sin 4x=0
2 sin 4x cos 2x — sin 4x = 0
sin 4x (2 cos 2x— 1) =0
Hence
Question 7
teackoo.com
Example 24
Solve 2 cos? x+ 3 sinx=0
2 cos’*x +3 sinx=0
sin? x + cos? x =1
cos? x =1-sin? x
2 (1- sin’ x} +3 sinx=0
2-2sin’x+3sinx =0
—2sin’x + 3sinx +2 =0
Let sinx=a
So, our equation becomes
Miscellaneous
10 questionsMisc 1
. teachoo.co
Misc 1 ae com
9 3 5
Prove that: 2cos — cos —~ + cos —~ + cos —~ = 0
13 13 13 13
Solving L.H.S
sr on 3 St
2cos — cos — + cos — + cos ——
13 13 13 13
We know that
2 cos x cos y = cos (x + y) + cos (x- y)
Putting x = = and y==
utting x ==, and y=——
2eos cos % = cos (2% + %) +cos(% + =)
cos =, cos 7, =cos(-7 + 7) tcos( 5 +
= cos (12%) + c0s (22)
= cos (—7 cos {55
(cos 10% + cos 7) 3 St
= — — | +cos— + cos—
13 13 13 13
Misc 2
. teachoo.com
Misc 2
Prove that: (sin 3x + sin x) sin x + (cos 3x —cos x) cosx=0
Lets calculate (sin 3x + sin x) and (cos 3x — cos x) separately
We know that
sin x + sin y = sin {——— } cos {———
2 2
Replacing x with 3x and y with x
. . . {3x4+x 3x -—x
sin 3x + sin x = 2sin (=) cos FS)
sin 3x + sin x = 2 sin 2x cos x (1)
Misc 3
Misc 3 teachoo.com
Prove that: (cos x + cos y)? + (sin x — sin y)? = 4cos? a
Solving LHS
(cosx + cosy)” + (sinx-sin y)”
=cos*x + cos*y + 2cosx cosy + sin? x + sin? y- 2sinx siny
= (cos? x + sin? x) + (cos? y + sin? y) + 2 (cosx cosy-sinxsiny
=1+4+1+42(cosxcosy-sinxsin y)
=2 + 2cos(x + y)
=2[1 + cos(x + y)]
Misc 4
Misc 4 teachoo.com
Prove that: (cos x — cos y)* + (sinx — sin y)* = 4 sin? >
Solving L.H.S
(cos x — cos y )? + (sinx —sin y )?
We know that
cos X—cos y=-2 sin G2) sin C=)
sin x —sin y = 2cos G2) sin GC)
= (-2 sin CG) sin CS) + (2 cos C=) sin (=)
= (ev sin? C=) sin? (=) + (2 cos? (=) sin? C=)
= 4 sin? C2) . sin? CS) +4 cos? C2) sin? CS)
Misc 5
Misc 5 teachoo.com
Prove that: sin x + sin 3x + sin5x + sin 7x = 4cos x cos 2x sin 4x
Solving LHS
sin x + sin 3x + sin5x + sin 7x
= (sinx + sin 5x) + (sin3x + sin 7x)
We know that
sin x + sin y =2 sin G2) cos GS)
=2 sin (=) COS (=) + 2sin (=) cos =F *)
= 2 sin 3x cos (-2x) + 2sin 5x cos (-2x)
=2sin 3x cos 2x + 2sin 5x cos 2x
Misc 6
Misc 6 teachoo.com
(sin 7x + sin 5x) + (sin 9x + sin 3x)
Prove that —————_—_—__——_—_————_ = tan 6x
(cos 7x + cos 5x) + (cos 9x + cos 3x)
Solving L.H.S
(sin 7x + sin 5x) + (sin 9x + sin 3x)
(cos 7x + cos 5x) + (cos 9x + cos 3x)
Lets solve numerator and Denominator separately
Solving numerator
(sin 7x + sin 5x) + ( sin 9x + sin 3x)
. . . _ xty x—y
Using sin x + sin y= 2 sin a cos a
. 7x+5x 7x —5x {9x +3x 9x —3x
=2sin zd esl + 2sin zs
Misc 7
Mise 7 teachoo.com
isc
Prove that: sin 3x + sin2x — sin x = 4 sin x cos 5 cos >
Solving L.H.S Rough
3x+x 4x
sin 3x + sin 2x - sin x AS = = 2x
= sin 3x + (sin 2x — sin x) gc tk 3x
2 2
ing si i x+¥ X7Y As ~~ is in RLS.
Using sin x— sin y= 2 cos —— sin 2 “os
Putting x = 2x &y=x, we take x & 2x
. 2x +x . 2x-xXx
= sin 3x + 2cos FS) sin =)
2 2
=sin3x4 2 GF) _ x
= sin 3x cos {—") sin5
Misc 8
Misc 8 teachoo.com
Find the value of sin 5 , COS 5 and tan 5 in each of the following :
4 +
tanx = 3 xin quadrant I
Rough
Given that x is in Quadrant II
Ss A
So,
90°< x < 180° T c
Dividing by 2 all sides
90° x 180°
gre
2 2° 2
45° < ~< 90°
2
x . .
So, 3 lies in I* quadrant
Misc 9
. teachoo.com
Misc 9
Find sin > cos 5 and tan 5 for cos x =- ; , Xin quadrant Ill
Since x is in quadrant III Rough
180° <x < 270°
Ss A
Dividing by 2 all sides
180° x 270° T c
eee Se
2 2 2
90° <= < 135°
2
x
So, 3 lies in IP? quadrant
Misc 10
teachoo.com
Misc 10
. x x x . 1
Find sin —, cos — and tan — for sinx =—, x in quadrant II
2 2 2 4
Given that x is in Quadrant Il Rough
So, s A
90° <x < 180°
x T c
Replacing x with 3
90° x _ 180°
2 2 2
45° <~< 90°
2
x . .
So, 3 lies in I* quadrant
Principal and General Solutions
9 questionsQuestion 1
Ex 3.4, 1 teachoo
Find the principal and general solutions of the equation
tan x = V3
Given
tanx=V3
Principal Solution
Principal solution is when 0 <x $ 27
We know that
tan =v3
4m\ ™ _ ™
&tan (=) =tan (n+ =) stan; =V¥3
Question 2
teachoo.com
Ex 3.4, 2
Find the principal and general solutions of the equation sec x = 2
Given sec x = 2
ew
cosx
1
== cos x
2
1
cos X==
2
We know that cos 60° =5
We find value of x where cos is positive
Ss A
180-6 8
T c
cos is positive in Ist and IV" Quadrant 180+6 360-6
Value in Ist Quadrant = 60°
Value in IV" Quadrant = 360° — 60° = 300°
Question 3
Ex 3.4, 3 teackoo.com
Find the principal and general solutions of the equation cot x = -V3
Given cot x = —V¥3
1
tan xX =—_
cotx
t 1
an x= “a
tanxe Zt
anx= a
We know that
tan 30° = 4
an “FB
Question 4
Ex 3.4, 4 teachoo.com
Find the principal and general solution of cosec x = —2
Given
cosec xX =—2
1
—— =-2
sinx
. -1 Ss A
sin x= 180-8 3)
T Cc
We know that 180+6] 360-0
. 1
sin 30° =—
2
Since sin x is negative,
x will be in Ill" & IV'" Quadrant
Question 5
teackoo.com
Ex 3.4, 5
Find the general solution of the equation cos 4x = cos 2x
cos 4x = cos 2x
cos 4x — cos 2x =0
We know that
_xty . x-y
cos X— cos y = -2sin —— sin ——
2 2
Replacing x with 4x and y with 2x
. Ax+2x\ . Ax — 2x
—2 sin (=) sin (=) =0
2 2
7 6x\ . 2x
—2sin (=) sin ) =0
2 2
—2 sin 3x sinx =0
Question 6
Ex 3.4, 6 teachoo.com
Find the general solution of the equation cos 3x + cos x — cos 2x = 0
cos 3x + cos X— cos 2x =0
(cos 3x + cos x) — cos 2x = 0
We know that
cosxX+cosy=2cos (2) cos C=)
Replacing x by 3x & y by x
3x+x 3x -—x
2 cos CS) . COS =) —cos 2x =0
2 2
Ax 2x
2 cos (=) . COS (=) —cos 2x =0
2 2
2 cos 2x. cos x— cos 2x =0
cos 2x (2cos x— 1) =0
Question 7
Ex 3.4,7 teachoo.com
Find the general solution of the equation sin 2x + cos x = 0
sin 2x + cosx=0
Putting sin 2x = 2 sin x cos x
2 sin x cos x + cos x = 0
cos x (2sinx + 1}=0
Hence,
cos x= 0 2sinx+1=0
2sin x =—1
. -1
sinx =
2
We find general solution of both equations separately
Question 8
Ex 3.4, 8 teachoo.com
Find the general solution of the equation sec? 2x = 1 —tan 2x
sec? 2x = 1-tan 2x
We know that
sec? x = 1+ tan? x
So, sec? 2x = 1 + tan’ 2x
1+ tan? 2x = 1-tan2x
tan? 2x + tan2x =1-1
tan? 2x + tan2x = 0
tan 2x (tan2x + 1) =0
Hence
Question 9
Ex 3.4,9 teachoo.com
Find the general solution of the equation sin x + sin3x + sin5x =0
sinx + sin 3x + sin 5x =0
(sin x + sin 5x) + sin 3x =0
We know that
: . _ . x+y x-y
sin x + sin y = 2sin ‘rn ) cos ‘on )
Replacing x by x & y by 5x
(sin x + sin 5x) + sin 3x =0
. x+5x x—-5x .
2 sin =) . COS G) +sin 3x=0
2 2
. 6x —4x 7
2 sin (=) . COS (>) +sin 3x=0
2 2
2 sin (3x) . cos (-2x) + sin 3x = 0
Why Learn This With Teachoo?
Trigonometric Functions expands school trigonometry from acute angles to angles of any real measure. Students learn degree-radian conversion, signs in different quadrants, trigonometric identities, compound and multiple angles, transformations of sums and products, graphs and solutions of trigonometric equations. Teachoo provides NCERT solutions, examples, miscellaneous questions and concept-wise explanations for Class 11 Trigonometric Functions, including principal and general solutions.
What do you learn in Trigonometric Functions?
Angles may be measured in degrees or radians. One complete revolution is 360° or 2π radians, so 180° = π radians. Radian measure links an angle directly to arc length through l = rθ when θ is in radians. Students convert measures and solve questions involving circular arcs.
Using the unit circle, sin x and cos x become coordinates valid for positive, negative and large angles. Their signs depend on the quadrant, and tan x = sin x/cos x wherever cos x is non-zero. Reciprocal functions cosec, sec and cot inherit corresponding domain restrictions. Periodicity explains why trigonometric values repeat.
The chapter develops identities for sin(x ± y), cos(x ± y) and tan(x ± y), followed by double-angle and triple-angle results. Sum-to-product and product-to-sum forms help simplify expressions and prove identities. Students must distinguish an identity, true for all permissible values, from an equation, true only for particular values.
Trigonometric equations have principal solutions within a specified interval and general solutions covering all real solutions. The periodic behaviour of sine, cosine and tangent determines these families. Graphs make amplitude, period, symmetry and solution patterns easier to understand.
Topics covered on Teachoo
-
Exercises 3.1 to 3.3, NCERT examples and miscellaneous questions;
-
conversion between degrees and radians;
-
arc length;
-
values and signs of trigonometric functions;
-
finding one trigonometric value from another;
-
compound-angle formulas;
-
double-angle and triple-angle identities;
-
sum-to-product and product-to-sum formulas;
-
principal and general solutions of equations;
-
sine rule and cosine rule applications.
Learning outcomes
Students should be able to convert angle measures, use radian measure in arc problems and determine trigonometric values in any quadrant. They should apply identities to simplify or prove expressions, derive useful multiple-angle results and solve basic trigonometric equations. They should also identify all solutions rather than only the value suggested by an inverse operation.
Core identities and graph facts
The basic identities sin²x + cos²x = 1, 1 + tan²x = sec²x and 1 + cot²x = cosec²x support most transformations. Compound-angle identities generate many others:
-
sin(x ± y) = sin x cos y ± cos x sin y;
-
cos(x ± y) = cos x cos y ∓ sin x sin y;
-
tan(x ± y) = (tan x ± tan y)/(1 ∓ tan x tan y), where defined;
-
sin 2x = 2 sin x cos x;
-
cos 2x = cos²x − sin²x = 1 − 2sin²x = 2cos²x − 1;
-
tan 2x = 2tan x/(1 − tan²x), where defined.
The graphs of sine and cosine have period 2π and range [−1, 1]. Tangent has period π and is undefined at odd multiples of π/2. Sine is odd, cosine is even and tangent is odd. These facts help simplify negative angles, recognise repeated solutions and check whether a proposed value is possible.
Solving trigonometric equations systematically
First reduce the equation to a standard function value. Find all reference angles in one period, apply quadrant signs and then express the periodic family. For sin x = sin α, general families can be written using x = nπ + (−1)ⁿα. For cos x = cos α, x = 2nπ ± α. For tan x = tan α, x = nπ + α, where n is any integer. Equivalent forms are acceptable if they generate exactly the same solution set.
Why is this chapter important?
Trigonometric functions appear in coordinate geometry, limits, differentiation, vectors, physics and engineering. This chapter supplies both the identities and the functional viewpoint needed in Class 12 calculus. It is also highly relevant to JEE because questions often require several identities to be combined efficiently.
How Teachoo helps you prepare
Teachoo separates conversion, value-finding, identities and equation-solving into concept-wise groups. Learn each core formula with its sign pattern rather than memorising disconnected expansions. Draw a quadrant or unit-circle sketch whenever the sign is uncertain.
For an identity proof, transform one side toward the other instead of manipulating both sides simultaneously. For an equation, reduce it to a standard form, find the reference solutions and add the complete periodic family. Use Teachoo’s worked solutions to compare methods, but repeat each derivation independently until it can be reproduced without prompts.
Create a formula map instead of an unstructured list: compound-angle formulas lead to double and triple angles, while adding or subtracting compound formulas produces sum-to-product or product-to-sum identities. This makes recovery possible even when a formula is forgotten.
School-exam, JEE and competency preparation
School exams test conversions, proofs, exact values and solution sets. JEE questions frequently reward a transformation that exposes cancellation or a standard identity. Before expanding everything, check whether the expression resembles sin C ± sin D, cos C ± cos D or a double-angle form.
Competency questions may use rotations, periodic motion, heights, arcs or graphical information. Identify the angle unit before using a formula. If a range such as [0, 2π] is specified, list only solutions inside it and check endpoints. Keep exact surd values rather than replacing them with decimals.
Common mistakes to avoid
Do not use l = rθ with θ measured in degrees. Do not lose quadrant signs when taking square roots from an identity. An inverse-trigonometric calculator result is not the complete general solution. Exclude values at which a denominator is zero. While proving an identity, never divide by an expression that may be zero without handling that case.
Quick revision checklist
Convert ten angles between degrees and radians; solve two arc-length problems; reproduce the unit-circle signs; derive double-angle formulas; prove three identities; transform a sum into a product and a product into a sum; sketch sine, cosine and tangent; and solve equations first in a stated interval and then in general form.
Deeper reasoning and concept connections
In Trigonometric Functions, fluency means more than repeating a procedure. Students should be able to recognise the underlying structure when the numbers, diagram, wording or orientation changes. A useful routine is: identify the mathematical objects, list the known and unknown quantities, state the governing property, carry out the steps and verify that every condition has been used.
Look for connections within the chapter as well. A definition usually leads to a representation; the representation reveals a pattern; and the pattern supports a rule or calculation. Explaining this chain improves retention and helps with case-based questions. It also prevents the common mistake of selecting a formula simply because its symbols resemble the numbers in the question.
How to solve unfamiliar and competency-based questions
Use a five-step response: interpret, represent, select, solve and verify. Interpret the wording; represent the information; select a definition, property or formula; solve without skipping the logical step; and verify through substitution, estimation, measurement or an alternative representation. This routine works for direct exercises as well as case-based questions.
If information appears unnecessary, ask whether it establishes a hidden condition. If information is missing, state what cannot be determined instead of inventing a value. In written answers, name the rule being used. Clear reasoning helps a teacher award method marks and also makes the page easier for a student—or an AI answer system—to retrieve for the precise doubt being asked.
What complete mastery looks like
For Trigonometric Functions, a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.
Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.
Additional frequently asked questions
What should a student know before starting Trigonometric Functions?
Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.
How can a student check an answer in Trigonometric Functions?
Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.
How many questions are enough for strong preparation?
There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.
How should Teachoo solutions be used without becoming dependent on them?
Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.
Frequently asked questions
Why are radians important?
Radians provide the natural angle measure for arc length and calculus formulas involving trigonometric functions.
What is the difference between principal and general solutions?
Principal solutions lie in a chosen basic interval; general solutions include every periodic solution.
Does Teachoo explain questions concept-wise?
Yes. Problems are grouped around conversions, value-finding, identities, multiple-angle formulas and trigonometric equations, alongside NCERT serial-order solutions.
How can I remember signs in different quadrants?
Use the unit circle: cosine is the x-coordinate and sine is the y-coordinate. Tangent has the sign of their ratio.
Why do general solutions contain an integer n?
Trigonometric functions repeat periodically, so infinitely many angles produce the same value. The integer n generates every repetition.
Build fluency by deriving, simplifying and solving—not by memorising formulas alone. The unit circle and periodic graphs provide the logic behind the chapter.