Chapter 3 – Trigonometric Functions Questions and Answers: NCERT Solutions for Class 11 Maths

Exercise 3.1

1. Find the radian measures corresponding to the following degree measures:
(i) 25o25o

Ans: We know that 180o=π180o=π radian
Therefore 1∘=π1801∘=π180 radian
hence,
25o=π180×2525o=π180×25 radian
=5π36=5π36radian

(ii) -47o30′-47o30′

Ans: Here we have,
-47o30′=-4712o-47o30′=-4712o
=-952=-952 degree
Since we know that, 180o=π180o=π radian
Therefore 1o=π1801o=π180 radian
Hence,
-952-952 degree=π180×(-952)=π180×(-952) radian
=(-1936×2)π=(-1936×2)π radian
=-1972π=-1972πradian
Therefore,
-47o30′=-1972π-47o30′=-1972π radian

(iii) 240o240o

Ans: We know that,
180o=π180o=π radian
Therefore 1o=π1801o=π180 radian
Hence,
240o=π180×240240o=π180×240 radian
=43π=43πradian

(iv) 520o520o

Ans: We know that,
180o=π180o=π radian
Therefore 1o=π1801o=π180 radian
Hence,
520o=π180×520520o=π180×520 radian
=26π9=26π9radian

2. Find the degree measures corresponding to the following radian measures
(Useπ=227π=227 )
(i) 11161116

Ans: We know that,
ππ radian=180o=180o
Therefore 1 radian =180πo1 radian =180πo
Hence,
11161116 radian=180π×1116=180π×1116 degree
=45×11π×4=45×11π×4degree
=45×11×722×4=45×11×722×4
degree
=3158=3158degree
Further computing,
11161116 radian=3938=3938 degree
=39o+3×608=39o+3×608 minutes
Since 1o=60′1o=60′
11161116 radian =39o+22′+12=39o+22′+12minutes
Since 1′=60”1′=60”
11161116 radian=39o22′30”=39o22′30”

(ii) -4-4

Ans: We know that,
ππ radian=180o=180o
Therefore 1 radian =180πo1 radian =180πo
Hence,
-4-4 radian=180π×(-4)=180π×(-4) degree
=180×7(-4)22=180×7(-4)22degree
=-252011=-252011 degree
=-229111=-229111degree
Since 1o=60′1o=60′
We have,
-4-4 radian=-229o+1×6011=-229o+1×6011 minutes
=-229o+5′+511=-229o+5′+511 minutes
Since 1′=60′′1′=60′′
-4-4 radian=-229o5′27”=-229o5′27”

(iii) 5π35π3

Ans: We know that,
ππ radian=180o=180o
Therefore 1 radian =180πo1 radian =180πo
Hence,
5π35π3 radian=180π×5π3=180π×5π3 degree
=300o=300o

(iv)7π67π6

Ans: We know that,
ππ radian=180o=180o
Therefore 1 radian =180πo1 radian =180πo
Hence,
7π67π6 radian=180π×7π6=180π×7π6
=210o=210o

3. A wheel makes 360360 revolutions in one minute. Through how many radians does it turn in one second?

Ans: Number of revolutions the wheel makes in 11 minute=360=360
Number of revolutions the wheel make in 11 second=36060=36060
=6=6
In one complete revolution, the wheel turns an angle of
2π2π
radian.
Hence, it will turn an angle of 6×2π=12π6×2π=12π radian, in 66 complete revolutions.
Therefore, the wheel turns an angle of 12π12π radian in one second.

4. Find the degree measure of the angle subtended at the centre of a circle of radius 100100cm by an arc of length 2222 cm.
(Useπ=227π=227 )

Ans: We know that,
in a circle of radius rr unit, if an angle θθ radian at the centre is subtended by an arc of length ll unit then
θ=lrθ=lr ……(1)
Therefore,
Substituting r=100cmr=100cm ,
l=22cml=22cm
in the formula (1) , we have,
θ=22100θ=22100 radian
Since 1 radian=180π1 radian=180π
Therefore,
θ=180π×22100θ=180π×22100 degree
=180×7×2222×100=180×7×2222×100degree
=635=635 degree
=1235=1235degree
Since 1o=60′1o=60′ , we have,
θ=12o36′θ=12o36′
Hence , the required angle is 12o36′12o36′.

5. In a circle of diameter 4040 cm, the length of a chord is 2020 cm. Find the length of minor arc of the chord.

Ans: Given that, diameter of the circle=40=40 cm
Hence Radius (r)(r) of the circle=402cm=402cm
=20cm=20cm
Let ABAB be a chord of the circle whose length is 2020 cm.

In ΔOAB,ΔOAB,
OA=OBOA=OB
== Radius of the circle
=20cm=20cm
Now also, AB=20cmAB=20cm
Therefore, ΔOABΔOAB is an equilateral triangle.
∴θ=60o∴θ=60o
=π3=π3 radian
We know that,
in a circle of radius rr unit, if an angle θθ radian at the centre is subtended by an arc of length ll unit then
θ=lrθ=lr ……(1)
Substituting θ=π3θ=π3 in the formula (1),
π3=arc AB20π3=arc AB20
arc AB=20π3cmarc AB=20π3cm
Therefore, the length of the minor arc of the chord is 20π3cm20π3cm .

6. If in two circles, arcs of the same length subtend angles 60o60o and 75o75o at the centre, find the ratio of their radii.

Ans: Let the radii of the two circles be r1r1 and r2r2 . Let an arc of length l1l1 subtends an angle of 60o60o at the centre of the circle of radius r1r1 , whereas let an arc of length l2l2 subtends an angle of 75o75o at the centre of the circle of radius r2r2 .
Now, we have,
60o=π360o=π3 radian and
75∘=5π1275∘=5π12 radian
We know that,
in a circle of radius rr unit, if an angle θθ radian at the centre is subtended by an arc of length ll unit then
θ=lrθ=lr
l=rθl=rθ
Hence we obtain,
l=r1π3l=r1π3
and l=r25π12l=r25π12
according to the l1=l2l1=l2
thus we have,
r1π3π=r25π12r1π3π=r25π12
r1=r254r1=r254
r1r2=54r1r2=54
Hence , the ratio of the radii is 5:45:4 .

7. Find the angle in radian through which a pendulum swings if its length is 7575 cm and the tip describes an arc of length.
(i) 1010 cm

Ans: We know that,
in a circle of radius rr unit, if an angle θθ radian at the centre is subtended by an arc of length ll unit then
θ=lrθ=lr
Given that r=75cmr=75cm
And here, l=10cml=10cm
Hence substituting the values in the formula,
θ=1075θ=1075 radian
=215=215radian

(ii) 1515 cm

Ans: We know that,
in a circle of radius rr unit, if an angle θθ radian at the centre is subtended by an arc of length ll unit then
θ=lrθ=lr
Given that r=75cmr=75cm
And here, l=15cml=15cm
Hence substituting the values in the formula,
θ=1575θ=1575 radian
=15=15radian

(iii) 2121 cm

Ans: We know that,
in a circle of radius rr unit, if an angle θθ radian at the centre is subtended by an arc of length ll unit then
θ=lrθ=lr
And here, l=21cml=21cm
Hence substituting the values in the formula,
θ=2175θ=2175 radian
=725=725radian

Exercise 3.2

1. Find the values of the other five trigonometric functions if cos x=-12cos x=-12 , xx lies in the third quadrant.

Ans: Here given that, cos x=-12cos x=-12
Therefore we have,
sec x=1cos xsec x=1cos x
=1(-12)=1(-12)
=-2=-2
Now we know that,sin2x+cos2x=1sin2x+cos2x=1
Therefore we have, sin2x=1-cos2xsin2x=1-cos2x
Substituting cos x=-12cos x=-12 in the formula, we obtain,
sin2x=1-(-12)2sin2x=1-(-12)2
sin2x=1-14sin2x=1-14
=34=34
sin x=±3–√2sin x=±32
Since xx lies in the 3rd3rdquadrant, the value of sinxsin⁡x will be negative.
sin x=-3–√2sin x=-32
Therefore, cosec x=1sin xcosec x=1sin x
=1(-3–√2)=1(-32)
=-23–√=-23
Hence ,
tan x=sin xcos xtan x=sin xcos x
=(-3–√2)(-12)=(-32)(-12)
=3–√=3
And
cot x=1tan xcot x=1tan x
=13–√=13

2. Find the values of other five trigonometric functions if sin x=35sin x=35 , xx lies in second quadrant.

Ans:
Here given that, sin x=35sin x=35
Therefore we have,
cosec x=1sin xcosec x=1sin x
=1(35)=1(35)
=53=53
Now we know that , sin2x+cos2x=1sin2x+cos2x=1
Therefore we have, cos2x=1-sin2xcos2x=1-sin2x
Substituting sinx=35sin⁡x=35 in the formula, we obtain,
cos2x=1-(35)2cos2x=1-(35)2
cos2x=1-925cos2x=1-925
=1625=1625
cos x=±45cos x=±45
Since xx lies in the 2nd2ndquadrant, the value of cosxcos⁡x will be negative.
cos x=-45cos x=-45
Therefore, secx=1cosxsecx=1cos⁡x
=1(-45)=1(-45)
=-54=-54
Hence ,
tanx=sinxcosxtan⁡x=sin⁡xcos⁡x
=(35)(-45)=(35)(-45)
=-34=-34
And
cotx=1tanxcot⁡x=1tan⁡x
=-43=-43

3. Find the values of other five trigonometric functions if cot x=34cot x=34 , xx lies in third quadrant.

Ans: Here given that, cotx=34cot⁡x=34
Therefore we have,
tanx=1cotxtan⁡x=1cot⁡x
=1(34)=1(34)
=43=43
Now we know that ,
sec2x-tan2x=1sec2x-tan2x=1
Therefore we have, sec2x=1+tan2xsec2x=1+tan2x
Substituting tan x=43tan x=43 in the formula, we obtain,
sec2x=1+(43)2sec2x=1+(43)2
sec2x=1+169sec2x=1+169
=259=259
sec x=±53sec x=±53
Since xx lies in the 3rd3rdquadrant, the value of secxsec⁡x will be negative.
sec x=-53sec x=-53
Therefore, cosx=1secxcos⁡x=1sec⁡x
=1(-53)=1(-53)
=-35=-35
Now , tan x=sin xcos xtan x=sin xcos x
Therefore, sin x=tan xcos xsin x=tan xcos x
Hence we have, sin x=43×(-35)sin x=43×(-35)

=(-45)=(-45)

And
cosec x=1sin xcosec x=1sin x
=-54=-54

4. Find the values of other five trigonometric functions if sec x=135sec x=135 , xx lies in fourth quadrant.

Ans: Here given that, secx=135sec⁡x=135
Therefore we have,
cosx=1secxcos⁡x=1sec⁡x
=1(135)=1(135)
=513=513
Now we know that , sec2x-tan2x=1sec2x-tan2x=1
Therefore we have, tan2x=sec2x-1tan2x=sec2x-1
Substituting
sec x=135sec x=135
in the formula, we obtain,
tan2x=(135)2-1tan2x=(135)2-1
tan2x=16925-1tan2x=16925-1
=14425=14425

tanx=±125tanx=±125
Since xx lies in the 4th4th quadrant, the value of tanxtan⁡x will be negative.
tan x=-125tan x=-125
Therefore,
cot x=1tan xcot x=1tan x

=-512=-512
Now , tan x=sin xcos xtan x=sin xcos x
Therefore, sin x=tan xcos xsin x=tan xcos x
Hence we have, sin x=513×(-125)sin x=513×(-125)
=(-1213)=(-1213)
And
cosec x=1sin xcosec x=1sin x
=-1312=-1312

5. Find the values of other five trigonometric functions if tan x=-512tan x=-512 , xx lies in second quadrant.

Ans: Here given that, tan x=-512tan x=-512
Therefore we have,
cot x=1tan xcot x=1tan x
=1(-512)=1(-512)
=-125=-125
Now we know that , sec2x-tan2x=1sec2x-tan2x=1
Therefore we have, sec2x=1+tan2xsec2x=1+tan2x
Substituting tan x=-512tan x=-512 in the formula, we obtain,
sec2x=1+(-512)2sec2x=1+(-512)2
sec2x=1+25144sec2x=1+25144
=169144=169144
sec x=±1312sec x=±1312
Since xx lies in the 2nd2nd quadrant, the value of secxsec⁡x will be negative.
sec x=-1312sec x=-1312
Therefore, cos x=1sec xcos x=1sec x
=-1213=-1213
Now , tan x=sin xcos xtan x=sin xcos x
Therefore, sin x=tan xcos xsin x=tan xcos x
Hence we have, sin x=(-512)×(-1213)sin x=(-512)×(-1213)
=(513)=(513)
And
cosec x=1sin xcosec x=1sin x
=135=135

6. Find the value of the trigonometric function sin765osin765o .

Ans: We know that the values of sinxsin⁡x repeat after an interval of 2π2π or 360∘360∘ .
Therefore we can write,
sin765o=sin(2×360o+45o)sin765o=sin(2×360o+45o)
=sin45o=sin45o
=12–√.=12.

7. Find the value of the trigonometric function cosec(-1410o)cosec(-1410o)

Ans: We know that the values of cosec xcosec x repeat after an interval of 2π2π or 360∘360∘ .
Therefore we can write,
cosec(-1410o)=cosec(-1410o+4×360o)cosec(-1410o)=cosec(-1410o+4×360o)
=cosec(-1410o+1440o)=cosec(-1410o+1440o)
=cosec30o=cosec30o
=2=2

8. Find the value of the trigonometric function tan19π3tan19π3 .

Ans: We know that the values of tan xtan x repeat after an interval of ππ or
180∘180∘
.
Therefore we can write,
tan19π3=tan613πtan19π3=tan613π
=tan(6π+π3)=tan(6π+π3)

=tanπ3=tanπ3
=3–√=3

9. Find the value of the trigonometric function sin(-11π3)sin(-11π3)

Ans: We know that the values of sin xsin x repeat after an interval of 2π2π or 360∘360∘ .
Therefore we can write,
sin(-11π3)=sin(-11π3+2×2π)sin(-11π3)=sin(-11π3+2×2π)
=sin(π3)=sin(π3)
=3–√2=32

10. Find the value of the trigonometric function cot(-15π4)cot(-15π4)

Ans: We know that the values of cot xcot x repeat after an interval of ππ or
180∘180∘
.
Therefore we can write,
cot(-15π4)=cot(-15π4+4π)cot(-15π4)=cot(-15π4+4π)
=cotπ4=cotπ4
=1=1

Exercise 3.3

1. Prove that sin2π6+cos2π3-tan2π4=-12sin2π6+cos2π3-tan2π4=-12

Ans: Substituting the values of
sinπ6,cosπ6,tanπ4sinπ6,cosπ6,tanπ4
on left hand side,
sin2π6+cos2π3-tan2π4=(12)2+(12)2-(1)2sin2π6+cos2π3-tan2π4=(12)2+(12)2-(1)2

=14+14-1=14+14-1
=−12=−12
== R.H.S.
Hence proved.

2. Prove that 2sin2π6+cosec27π6cos2π3=322sin2π6+cosec27π6cos2π3=32

Ans: Substituting the values of sinπ6,cosec7π6,cosπ3sinπ6,cosec7π6,cosπ3 on left hand side,
L.H.S.=2sin2π6+cosec27π6cos2π3=2sin2π6+cosec27π6cos2π3
=2(12)2+cosec2(π+π6)(12)2=2(12)2+cosec2(π+π6)(12)2
=2×14+(-cosecπ6)2(14)=2×14+(-cosecπ6)2(14)
=12+(-2)2(14)=12+(-2)2(14)
Since cosec xcosec x repeat its value after an interval of 2π2π ,
we have, cosec7π6=-cosecπ6cosec7π6=-cosecπ6
L.H.S =12+44=12+44
=32=32
== R.H.S.
Hence proved.

3. Prove that cot2π6+cosec5π6+3tan2π6=6cot2π6+cosec5π6+3tan2π6=6

Ans: Substituting the values of cotπ6,cosec5π6,tanπ6cotπ6,cosec5π6,tanπ6 on left hand side,
L.H.S.=cot2π6+cosec5π6+3tan2π6=cot2π6+cosec5π6+3tan2π6
=(3–√)2+cosec(π-π6)+3(13–√)2=(3)2+cosec(π-π6)+3(13)2
=3+cosecπ6+3×13=3+cosecπ6+3×13
Since cosec xcosec x repeat its value after an interval of 2π2π ,
we have, cosec5π6=cosecπ6cosec5π6=cosecπ6
L.H.S =3+2+1=3+2+1
=1=1
== R.H.S.
Hence proved.

4. Prove that 2sin23π4+2cos2π4+2sec2π3=102sin23π4+2cos2π4+2sec2π3=10

Ans:
Substituting the values of sin3π4,cosπ4,secπ3sin3π4,cosπ4,secπ3 on left hand side,
L.H.S.=2sin23π4+2cos2π4+2sec2π3=2sin23π4+2cos2π4+2sec2π3
=2{sin(π-π4)}2+2(12–√)2+2(2)2=2{sin(π-π4)}2+2(12)2+2(2)2
=2{sinπ4}2+2×12+8=2{sinπ4}2+2×12+8
Since sin xsin x repeat its value after an interval of
2π2π
,
we have,
sin3π4=sinπ4sin3π4=sinπ4

L.H.S =1+1+8=1+1+8
=10=10
== R.H.S.
Hence proved.

5. Find the value of :
(i) sin75osin75o

Ans: We have,
sin75o=sin(45o+30o)sin75o=sin(45o+30o)
=sin45ocos30o+cos45osin30o=sin45ocos30o+cos45osin30o
Since we know that, sin(x+y)=sin x cos y+cos x sin ysin(x+y)=sin x cos y+cos x sin y
Therefore we have,
Sin75o=12–√×3–√2+12–√×12Sin75o=12×32+12×12
Sin75o=3–√+122–√Sin75o=3+122

(ii) tan15otan15o

Ans: We have,
tan15o=tan(45o-30o)tan15o=tan(45o-30o)
=tan45o-tan30o1+tan45otan30o=tan45o-tan30o1+tan45otan30o
Since we know, tan(x-y)=tan x-tan y1+tan x tan ytan(x-y)=tan x-tan y1+tan x tan y
Therefore we have,
tan15o=1-13–√1+1(13–√)tan15o=1-131+1(13)
=3–√-13–√3–√+13–√=3-133+13
=3–√-13–√+1=3-13+1
=(3–√-1)2(3–√+1)(3–√-1)=(3-1)2(3+1)(3-1)
Further computing we have,
tan15o=3+1-23–√(3–√)2-(1)2tan15o=3+1-23(3)2-(1)2

=4-23–√3-1=4-233-1

=2-3–√=2-3

6. Prove that cos(π4-x)cos(π4-y)-sin(π4-x)sin(π4-y)=sin(x+y)cos(π4-x)cos(π4-y)-sin(π4-x)sin(π4-y)=sin(x+y)

Ans: We know that, cos(x+y)=cos xcos y-sin xsin ycos(x+y)=cos xcos y-sin xsin y
cos(π4-x)cos(π4-y)-sin(π4-x)sin(π4-y)=cos[π4-x+π4-y]cos(π4-x)cos(π4-y)-sin(π4-x)sin(π4-y)=cos[π4-x+π4-y]
=cos[π2-(x+y)]=cos[π2-(x+y)]
=sin(x+y)=sin(x+y)
L.H.S == R.H.S.
Hence proved.

7. Prove that tan(π4+x)tan(π4-x)=(1+tanx1-tanx)2tan(π4+x)tan(π4-x)=(1+tanx1-tanx)2

Ans: We know that ,tan(A+B)=tan A+tan B1-tan Atan Btan(A+B)=tan A+tan B1-tan Atan B
and tan(A-B)=tan A-tan B1+tan Atan Btan(A-B)=tan A-tan B1+tan Atan B
L.H.S.=tan(π4+x)tan(π4-x)=tan(π4+x)tan(π4-x)
Using the above formula,
L.H.S=⎛⎝⎜tanπ4+tanx1-tanπ4tanx⎞⎠⎟tanπ4-tanx1+tanπ4tanxL.H.S=(tanπ4+tanx1-tanπ4tanx)tanπ4-tanx1+tanπ4tanx
=(1+tan x1-tan x)(1-tan x1+tan x)=(1+tan x1-tan x)(1-tan x1+tan x) substituting$tanπ4=1$substituting$tanπ4=1$
=(1+tan x1-tan x)2=(1+tan x1-tan x)2
== R.H.S.
Hence proved.

8. Prove that cos(π+x)cos(-x)sin(π-x)cos(π2+x)=cot2xcos(π+x)cos(-x)sin(π-x)cos(π2+x)=cot2x

Ans: Observe that cos xcos x repeats same value after an interval 2π2π
and sin xsin x repeats same value after an interval 2π2π.
L.H.S.=cos(π+x)cos(-x)sin(π-x)cos(π2+x)=cos(π+x)cos(-x)sin(π-x)cos(π2+x)
=[-cos x][cos x](sin x)(-sin x)=[-cos x][cos x](sin x)(-sin x)
=-cos2x-sin2x=-cos2x-sin2x
=cot2x=cot2x
Hence proved.

9. Prove that,
Cos(3π2+x)Cos(2π+x)[cot(3π2-x)+cot(2π+x)]=1Cos(3π2+x)Cos(2π+x)[cot(3π2-x)+cot(2π+x)]=1

Ans: We know that cot xcot x repeats same value after an interval 2π2π .
L.H.S.=Cos(3π2+x)Cos(2π+x)[cot(3π2−x)+cot(2π+x)]=Cos(3π2+x)Cos(2π+x)[cot(3π2−x)+cot(2π+x)]
=sin x cos x[tan x+cot x]=sin x cos x[tan x+cot x]
Substituting tan x=sin xcos xtan x=sin xcos x and
cot x=cos xsin xcot x=cos xsin x ,
L.H.S=sin xcos x(sin xcos x+cos xsin x)L.H.S=sin xcos x(sin xcos x+cos xsin x)
=(sin x cos x)[sin2x+cos2xsin x cos x]=(sin x cos x)[sin2x+cos2xsin x cos x]
=1=1
== R.H.S.
Hence proved.

10. Prove that sin(n+1)xsin(n+2)x+cos (n+1)x cos (n+2)x=cos xsin(n+1)xsin(n+2)x+cos (n+1)x cos (n+2)x=cos x

Ans: We know that , cos(x-y)=cosxcosy+sinxsinycos(x-y)=cosxcosy+sinxsiny
L.H.S.=sin(n+1)xsin(n+2)x+cos (n+1)x cos (n+2)x=sin(n+1)xsin(n+2)x+cos (n+1)x cos (n+2)x
=cos[(n+1)x-(n+2)x]=cos[(n+1)x-(n+2)x]
=cos(-x)=cos(-x)
=cosx=cosx
== R.H.S.

11. Prove that cos(3π4+x)-cos(3π4-x)=-2–√sinxcos(3π4+x)-cos(3π4-x)=-2sinx

Ans: We know that , cos A-cos B=-2sin(A+B2).sin(A-B2)cos A-cos B=-2sin(A+B2).sin(A-B2)
∴∴ L.H.S.=cos(3π4+x)-cos(3π4-x)=cos(3π4+x)-cos(3π4-x)
=-2sin⎧⎩⎨⎪⎪⎪⎪⎪⎪(3π4+x)+(3π4-x)2⎫⎭⎬⎪⎪⎪⎪⎪⎪.sin⎧⎩⎨⎪⎪⎪⎪⎪⎪(3π4+x)-(3π4-x)2⎫⎭⎬⎪⎪⎪⎪⎪⎪=-2sin{(3π4+x)+(3π4-x)2}.sin{(3π4+x)-(3π4-x)2}
=-2sin(3π4)sin x=-2sin(3π4)sin x
Since sin xsin x repeats the same value after an interval 2π2π ,
we have, sin(3π4)=sin(π-π4)sin(3π4)=sin(π-π4)
therefore,
L.H.S=-2sinπ4sin xL.H.S=-2sinπ4sin x
=-2×12–√×sinx=-2×12×sinx
=-2–√sin x=-2sin x
== R.H.S.
Hence proved.

12. Prove that sin26x-sin24x=sin 2x sin 10xsin26x-sin24x=sin 2x sin 10x

Ans: We know that,sinA+sinB=2sin(A+B2)cos(A-B2)sinA+sinB=2sin(A+B2)cos(A-B2)
And sin A-sin B=2cos(A+B2)sin(A-B2)sin A-sin B=2cos(A+B2)sin(A-B2)
∴∴ L.H.S.=sin26x-sin24xa=sin26x-sin24xa
=(sin 6x+sin 4x)(sin 6x-sin 4x)=(sin 6x+sin 4x)(sin 6x-sin 4x)
=[2sin(6x+4×2)cos(6x-4×2)][2cos(6x+4×2).sin(6x-4×2)]=[2sin(6x+4×2)cos(6x-4×2)][2cos(6x+4×2).sin(6x-4×2)]
=(2sin 5x cos x)(2cos 5x sin x)=(2sin 5x cos x)(2cos 5x sin x)
Now we know that, sin 2x=2sin x cos xsin 2x=2sin x cos x ,
Therefore we have,
L.H.S=(2sin 5x cos 5x)(2sin x cos x)L.H.S=(2sin 5x cos 5x)(2sin x cos x)
=sin 10x sin 2x=sin 10x sin 2x
== R.H.S.

13. Prove that cos22x-cos26x=sin 4x sin 8xcos22x-cos26x=sin 4x sin 8x

Ans: We know that,
cos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)
And cos A-cos B=-2sin(A+B2)sin(A-B2)cos A-cos B=-2sin(A+B2)sin(A-B2)
L.H.S.=cos22x-cos26x=cos22x-cos26x
=(cos 2x+cos 6x)(cos 2x-6x)=(cos 2x+cos 6x)(cos 2x-6x)
=[2cos(2x+6×2)cos(2x-6×2)][-2sin(2x+6×2)sin(2x-6×2)]=[2cos(2x+6×2)cos(2x-6×2)][-2sin(2x+6×2)sin(2x-6×2)]
Further computing, we have,
L.H.S=[2cos 4x cos(-2x)][-2sin 4xsin(-2x)]L.H.S=[2cos 4x cos(-2x)][-2sin 4xsin(-2x)]
=[2cos 4x cos 2x][-2sin 4x(-sin 2x)]=[2cos 4x cos 2x][-2sin 4x(-sin 2x)]
=(2sin 4x cos 4x)(2sin 2xcos 2x)=(2sin 4x cos 4x)(2sin 2xcos 2x)
Now we know that, sin 2x=2sin x cos xsin 2x=2sin x cos x
Therefore we have,
L.H.S=sin 8x sin 4xL.H.S=sin 8x sin 4x
== R.H.S.
.Hence proved.

14. Prove that sin 2x+2sin 4x+sin6=4cos2xsin 4xsin 2x+2sin 4x+sin6=4cos2xsin 4x

Ans: We know that,
sin A+sin B=2sin(A+B2)cos(A-B2)sin A+sin B=2sin(A+B2)cos(A-B2)
L.H.S.=sin 2x+2sin 4x+sin 6x=sin 2x+2sin 4x+sin 6x
. =[sin 2x+sin 6x]+2sin 4x=[sin 2x+sin 6x]+2sin 4x
=[2sin(2x+6×2)cos(2x-6×2)]+2sin4x=[2sin(2x+6×2)cos(2x-6×2)]+2sin4x
=2sin 4xcos(-2x)+2sin 4x=2sin 4xcos(-2x)+2sin 4x
Further computing,
We have, L.H.S=2sin 4x cos 2x+2sin 4xL.H.S=2sin 4x cos 2x+2sin 4x
=2sin 4x(cos 2x+1)=2sin 4x(cos 2x+1)
Now we know that, cos 2x+1=2cos2xcos 2x+1=2cos2x
Therefore we have,
L.H.S=2sin 4x(2cos2x)L.H.S=2sin 4x(2cos2x)
=4cos2xsin 4x=4cos2xsin 4x
= R.H.S.
Hence proved.

15. Prove that cot 4x(sin 5x+sin 3x)=cot x(sin 5x-sin 3x)cot 4x(sin 5x+sin 3x)=cot x(sin 5x-sin 3x)

Ans: We know that, sin A+sin B=2sin(A+B2)cos(A-B2)sin A+sin B=2sin(A+B2)cos(A-B2)
L.H.S.=cot 4x(sin 5x+sin 3x)=cot 4x(sin 5x+sin 3x)
=cot 4xsin 4x[2sin(5x+3×2)cos(5x+3×2)]=cot 4xsin 4x[2sin(5x+3×2)cos(5x+3×2)]
=(cos 4xsin 4x)[2sin 4x cos x]=(cos 4xsin 4x)[2sin 4x cos x]
=2cos 4x cos x=2cos 4x cos x
Now also ,we know that, sin A-sin B=2cos(A+B2)sin(A-B2)sin A-sin B=2cos(A+B2)sin(A-B2)
R.H.S.=cot x(sin 5x-sin 3x)=cot x(sin 5x-sin 3x)
=cos xsin x[2cos(5x+3×2)sin(5x-3×2)]=cos xsin x[2cos(5x+3×2)sin(5x-3×2)]

=cos xsin x[2cos 4x sin x]=cos xsin x[2cos 4x sin x]
=2cos 4x cos x=2cos 4x cos x
Therefore , we can conclude that,
L.H.S.=R.H.S.
Hence proved.

16. Prove that cos 9x-cos 5xsin 17x-sin 3x=-sin 2xcos 10xcos 9x-cos 5xsin 17x-sin 3x=-sin 2xcos 10x

Ans: We know that,
cos A-cos B=-2sin(A+B2)sin(A-B2)cos A-cos B=-2sin(A+B2)sin(A-B2)
And sin A-sin B=2cos(A+B2)sin(A-B2)sin A-sin B=2cos(A+B2)sin(A-B2)
L.H.S.=cos 9x-cos 5xsin 17x-sin 3x=cos 9x-cos 5xsin 17x-sin 3x

=-2sin(9x+5×2).sin(9x-5×2)2cos(17x+3×2).sin(17x-3×2)=-2sin(9x+5×2).sin(9x-5×2)2cos(17x+3×2).sin(17x-3×2)
(following the formula)

=-2sin 7x.sin 2x2cos 10x.sin 7x=-2sin 7x.sin 2x2cos 10x.sin 7x
=-sin 2xcos 10x=-sin 2xcos 10x
== R.H.S.
Hence proved.

17. Prove that:sin 5x+sin 3xcos 5x+cos 3x=tan 4xsin 5x+sin 3xcos 5x+cos 3x=tan 4x

Ans:
We know that
sin A+sin B=2sin(A+B2)cos(A-B2),sin A+sin B=2sin(A+B2)cos(A-B2),
cos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)
Now , L.H.S.=sin 5x+sin 3xcos 5x+cos 3x=sin 5x+sin 3xcos 5x+cos 3x
=2sin(5x+3×2)cos(5x-3×2)2cos(5x+3×2)cos(5x-3×2)=2sin(5x+3×2)cos(5x-3×2)2cos(5x+3×2)cos(5x-3×2) (using the formula)
=2sin(5x+3×2)cos(5x-3×2)2cos(5x+3×2)cos(5x-3×2)=2sin(5x+3×2)cos(5x-3×2)2cos(5x+3×2)cos(5x-3×2)
=2sin 4x cos x2cos 4x cos x=2sin 4x cos x2cos 4x cos x
Further computing we have,
L.H.S=tan 4xL.H.S=tan 4x
== R.H.S.

18. Prove that
sin x-sin ycos x+cos y=tanx-y2sin x-sin ycos x+cos y=tanx-y2

Ans: We know that,
sin A-sin B=2cos(A+B2)sin(A-B2),sin A-sin B=2cos(A+B2)sin(A-B2),
.cos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)
L.H.S.
=sin x-sin ycosx+cosy=sin x-sin ycosx+cosy
=2cos(x+y2).sin(x-y2)2cos(x+y2).cos(x-y2)=2cos(x+y2).sin(x-y2)2cos(x+y2).cos(x-y2)
=sin(x-y2)cos(x-y2)=sin(x-y2)cos(x-y2)
=tan(x-y2)=tan(x-y2)
Therefore L.H.S=R.H.SL.H.S=R.H.S
Hence proved.

19. Prove that sin x+sin 3xcos x+cos 3x=tan 2xsin x+sin 3xcos x+cos 3x=tan 2x

Ans: We know that
sin A+sin B=2sin(A+B2)cos(A+B2),sin A+sin B=2sin(A+B2)cos(A+B2),
.
cos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)
Now , L.H.S.=sinx+sin3xcos x+cos 3x=sinx+sin3xcos x+cos 3x
=2sin(x+3×2)cos(x-3×2)2cos(x+3×2)cos(x-3×2)=2sin(x+3×2)cos(x-3×2)2cos(x+3×2)cos(x-3×2) (using the formula)
=sin 2xcos 2x=sin 2xcos 2x
=tan 2x=tan 2x
Therefore L.H.S== R.H.S.
Hence proved.

20. Prove that sin x-sin 3xsin2x-cos2x=2sin xsin x-sin 3xsin2x-cos2x=2sin x

Ans: We know that,
sin A-sin B=2cos(A+B2)sin(A-B2)sin A-sin B=2cos(A+B2)sin(A-B2)
And
cos2A-sin2A=cos 2Acos2A-sin2A=cos 2A
L.H.S.=sin x-sin 3xsin2x-cos2x=sin x-sin 3xsin2x-cos2x

=2cos(x+3×2)sin(x-3×2)-cos2x=2cos(x+3×2)sin(x-3×2)-cos2x
=2cos2xsin(-x)-cos 2x=2cos2xsin(-x)-cos 2x
=-2×(-sinx)=-2×(-sinx)
Therefore , we have,
L.H.S=2sin xL.H.S=2sin x
== R.H.S.
Hence proved.

21. Prove that cos 4x+cos 3x+cos 2xsin 4x+sin 3x+sin 2x=cot 3xcos 4x+cos 3x+cos 2xsin 4x+sin 3x+sin 2x=cot 3x

Ans: We know that,
cos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)
And, sin A+sin B=2sin(A+B2)cos(A-B2)sin A+sin B=2sin(A+B2)cos(A-B2)
Now, L.H.S.=cos 4x+cos 3x+cos 2xsin 4x+sin 3x+sin 2x=cos 4x+cos 3x+cos 2xsin 4x+sin 3x+sin 2x

=(cos 4x+cos 2x)+cos 3x(sin4x+sin2x)+sin 3x=(cos 4x+cos 2x)+cos 3x(sin4x+sin2x)+sin 3x

=2cos(4x+2×2)cos(4x-2×2)+cos3x2sin(4x+2×2)cos(4x-2×2)+sin 3x=2cos(4x+2×2)cos(4x-2×2)+cos3x2sin(4x+2×2)cos(4x-2×2)+sin 3x
(using the formulas)

=2cos 3x cos x+cos 3x2sin 3x cos x+sin 3x=2cos 3x cos x+cos 3x2sin 3x cos x+sin 3x
Further computing, we obtain,
L.H.S=cos 3x(2cos x+1)sin 3x(2cos x+1)=cos 3x(2cos x+1)sin 3x(2cos x+1)

=cot 3x=cot 3x

== R.H.S.
Hence proved.

22. Prove that
cot x cot 2x-cot 2x cot 3x-cot 3x cot x=1cot x cot 2x-cot 2x cot 3x-cot 3x cot x=1

Ans:
We know that,
cot(A+B)=cotAcotB-1cot A+cot Bcot(A+B)=cotAcotB-1cot A+cot B
Now , L.H.S.=cot xcot 2x-cot 2x cot 3x-cot 3x cot x=cot xcot 2x-cot 2x cot 3x-cot 3x cot x

=cot x cot 2x-cot 3x(cot 2x+cot x)=cot x cot 2x-cot 3x(cot 2x+cot x)

=cot x cot 2x-cot(2x+x)(cot 2x+cot x)=cot x cot 2x-cot(2x+x)(cot 2x+cot x)

=cot x cot 2x-[cot 2x cot x-1cot x+cot 2x](cot 2x+cot x)=cot x cot 2x-[cot 2x cot x-1cot x+cot 2x](cot 2x+cot x)
Further computing we obtain,
L.H.S=cot x cot 2x-(cot 2x cot x-1)L.H.S=cot x cot 2x-(cot 2x cot x-1)

=1=1
== R.H.S.
Hence proved.

23. Prove that tan 4x=4tan x(1-tan2x)1-6tan2x+tan4xtan 4x=4tan x(1-tan2x)1-6tan2x+tan4x

Ans: We know that tan 2A=2tan A1-tan2Atan 2A=2tan A1-tan2A
L.H.S.=tan 4x=tan 4x
=tan2(2x)=tan2(2x)

=2tan 2×1-tan2(2x)=2tan 2×1-tan2(2x)
usingtheformulausingtheformula
=(4tan x1-tan2x)[1-4tan2x(1-tan2x)2]=(4tan x1-tan2x)[1-4tan2x(1-tan2x)2]
Further computing, we obtain,
L.H.S =(4tan x1-tan2x)[(1-tan2x)24tan2x(1-tan2x)2]=(4tan x1-tan2x)[(1-tan2x)24tan2x(1-tan2x)2]
=4tan x(1-tan2x)1+tan4x-2tan2x-4tan2x=4tan x(1-tan2x)1+tan4x-2tan2x-4tan2x
=4tan x(1-tan2x)1-6tan2x+tan4x=4tan x(1-tan2x)1-6tan2x+tan4x
== R.H.S.
Hence proved.

24. Prove that cos 4x=1-8sin2xcos2xcos 4x=1-8sin2xcos2x

Ans: We know that, cos 2x=1-2sin2xcos 2x=1-2sin2x
And sin 2x=2sin x cos xsin 2x=2sin x cos x
L.H.S.
=cos 4x=cos 4x
=cos 2(2x)=cos 2(2x)
=1-2sin22x=1-2sin22x
=1-2(2sin x cos x)2=1-2(2sin x cos x)2
Further computing we get,
L.H.S=1-8sin2xcos2x=1-8sin2xcos2x
==R.H.S.
Hence proved.

25. Prove that cos 6x=32xcos6x-48cos4x+18cos2x-1cos 6x=32xcos6x-48cos4x+18cos2x-1

Ans: We know that, cos 3A=4cos3A-3cosAcos 3A=4cos3A-3cosA
and cos 2x=1-2sin2xcos 2x=1-2sin2x
L.H.S.=cos 6x=cos 6x
=cos 3(2x)=cos 3(2x)

=4cos32x-3cos 2x=4cos32x-3cos 2x

=4[(2cos2x-1)3-3(2cos2x-1)]=4[(2cos2x-1)3-3(2cos2x-1)]
Further computing,
L.H.S=4[(2cos2x)3-(1)3-3(2cos2x)]-6cos2x+3=4[(2cos2x)3-(1)3-3(2cos2x)]-6cos2x+3
=4[(2cos2x)3-(1)3-3(2cos2x)2+3(2cos2x)]-6cos2x+3=4[(2cos2x)3-(1)3-3(2cos2x)2+3(2cos2x)]-6cos2x+3
=4[8cos6x-1-12cos4x+6cos2x]-6cos2x+3=4[8cos6x-1-12cos4x+6cos2x]-6cos2x+3
=32cos6x-48cos4x+18cos2x-1=32cos6x-48cos4x+18cos2x-1
Therefore we have,
L.H.S == R.H.S.
Hence proved.

Exercise 3.4

1. Find the principal and general solutions of the
tan x=3–√tan x=3 .

Ans: Here given that,
tan x=3–√tan x=3
We know that tanπ3=3–√tanπ3=3
and tan(4π3)=tan(π+π3)=tanπ3=3–√tan(4π3)=tan(π+π3)=tanπ3=3
Therefore, the principal solutions are
x=π3x=π3
and
4π34π3 .
Now,
tan x=tanπ3tan x=tanπ3
Which implies,
x=nπ+π3x=nπ+π3 , where
n∈Zn∈Z
Therefore, the general solution is
x=nπ+π3x=nπ+π3
, where n∈Zn∈Z .

2. Find the principal and general solutions of the equation secx=2secx=2

Ans: Here it is given that,
sec x=2sec x=2
Now we know that
secπ3=2secπ3=2 and
sec5π3=sec(2π-π3)sec5π3=sec(2π-π3)
=secπ3=secπ3
=2=2
Therefore, the principal solutions arex=π3x=π3 and
5π35π3
.
Now, sec x=secπ3sec x=secπ3
and we know ,
secx=1cosxsec⁡x=1cos⁡x
Therefore , we have,
cos x=cosπ3cos x=cosπ3
Which implies,
x=2nπ±π3x=2nπ±π3 , where n∈Zn∈Z .
Therefore, the general solution is x=2nπ±π3x=2nπ±π3 , where n∈Zn∈Z .

3. Find the principal and general solutions of the equation cot x=-3–√cot x=-3

Ans: Here it is given that,
cot x=-3–√cot x=-3
Now we know that cotπ6=3–√cotπ6=3
And
cot(π-π6)=-cotπ6cot(π-π6)=-cotπ6

=-3–√=-3
and cot(2π-π6)=-cotπ6cot(2π-π6)=-cotπ6

=-3–√=-3
Therefore we have,
cot5π6=-3–√cot5π6=-3
and cot11π6=-3–√cot11π6=-3
Therefore, the principal solutions are x=5π6x=5π6 and 11π611π6.
.Now, cot x=cot5π6cot x=cot5π6
And we know
cot x=1tan xcot x=1tan x
Therefore we have,
tan x=tan5π6tan x=tan5π6
Which implies,
x=nπ+5π6x=nπ+5π6 , where n∈Zn∈Z
Therefore, the general solution is x=nπ+5π6x=nπ+5π6 , where n∈Zn∈Z .

4. Find the general solution of cosec x=-2cosec x=-2

Ans: Here it is given that,
cosec x=-2cosec x=-2
Now we know that
cosecπ6=2cosecπ6=2
and
cosec(π+π6)=-cosecπ6acosec(π+π6)=-cosecπ6a
=-2=-2
and
cosec(2π-π6)=-cosecπ6cosec(2π-π6)=-cosecπ6

=-2=-2
therefore we have,
cosec7π6=-2cosec7π6=-2
and cosec11π6=-2cosec11π6=-2
Hence , the principal solutions arex=7π6x=7π6 and 11π611π6.
Now, cosec x=cosec7π6cosec x=cosec7π6
And we know,
cosec x=1sin xcosec x=1sin x
Therefore , we have,
sin x=sin7π6sin x=sin7π6
Which implies,
x=nπ+(-1)n7π6x=nπ+(-1)n7π6
,where n∈Zn∈Z.
Therefore, the general solution is x=nπ+(-1)n7π6x=nπ+(-1)n7π6 ,where n∈Zn∈Z.

5. Find the general solution of the equation cos 4x=cos 2xcos 4x=cos 2x

Ans: Here it is given that, cos 4x=cos 2xcos 4x=cos 2x
Which implies,
cos 4x-cos 2x=0cos 4x-cos 2x=0
Now we know that, cos A-cos B=-2sin(A+B2)sin(A-B2)cos A-cos B=-2sin(A+B2)sin(A-B2)
Therefore we have,
-2sin(4x+2×2)sin(4x-2×2)=0-2sin(4x+2×2)sin(4x-2×2)=0
sin 3x sin x=0sin 3x sin x=0
Hence we have, sin 3x=0sin 3x=0
Or, sin x=0sin x=0
Therefore, 3x=nπ3x=nπ
or x=nπx=nπ ,where n∈Zn∈Z
therefore,
x=nπ3x=nπ3

or x=nπx=nπ ,where n∈Zn∈Z.

6. Find the general solution of the equation cos 3x+cos x-cos 2x=0cos 3x+cos x-cos 2x=0.

Ans: Here given that,
cos 3x+cos x-cos 2x=0cos 3x+cos x-cos 2x=0
Now we know that, cos A+cos B=2cos(A+B2)cos(A-B2)cos A+cos B=2cos(A+B2)cos(A-B2)
Therefore cos 3x+cos x-cos 2x=0cos 3x+cos x-cos 2x=0 implies
2cos(3x+x2)cos(3x-x2)-cos 2x=02cos(3x+x2)cos(3x-x2)-cos 2x=0
2cos 2x cos x-cos 2x=02cos 2x cos x-cos 2x=0
cos 2x(2cos x-1)=0cos 2x(2cos x-1)=0
Hence we have,
Either cos 2x=0cos 2x=0
Or cos x=12cos x=12
Which in turn implies that,
Either 2x=(2n+1)π22x=(2n+1)π2
Or, cos x=cosπ3cos x=cosπ3 , where n∈Zn∈Z
Therefore,
Either x=(2n+1)π4x=(2n+1)π4
Or, x=2nπ±π3x=2nπ±π3 ,where
n∈Zn∈Z
.

7. Find the general solution of the equation
sin 2x+cos x=0sin 2x+cos x=0
.

Ans: Here it is given that,
sin 2x cos x=0sin 2x cos x=0
Now we know that, sin 2x=2sin x cos xsin 2x=2sin x cos x
Therefore we have,
2sin x cos x+cos x=02sin x cos x+cos x=0
Which implies,

cos x(2sin x+1)=0cos x(2sin x+1)=0
Therefore we have,
Either cos x=0cos x=0
Or, sin x=-12sin x=-12
Hence we have,

x=(2n+1)π2x=(2n+1)π2
, where n∈Zn∈Z .
Either
Or,
sin x=-12sin x=-12
=-sinπ6=-sinπ6
=sin(π-7π6)=sin(π-7π6)
=sin7π6=sin7π6
Which implies
x=nπ+(-1)n7π6x=nπ+(-1)n7π6 , where n∈Zn∈Z
Therefore, the general solution is
(2n+1)π2(2n+1)π2
or
nπ+(-1)n7π6,n∈Znπ+(-1)n7π6,n∈Z
.

8. Find the general solution of the equation
sec22x=1-tan 2xsec22x=1-tan 2x

Ans: Here given that ,
sec22x=1-tan 2xsec22x=1-tan 2x
Now we know that, sec2x-tan2x=1sec2x-tan2x=1
Therefore we have,

sec22x=1-tan 2xsec22x=1-tan 2x
implies

1+tan22x=1-tan 2×1+tan22x=1-tan 2x

tan22x+tan 2x=0tan22x+tan 2x=0

tan 2x(tan 2x+1)=0tan 2x(tan 2x+1)=0
Hence either tan 2x=0tan 2x=0
Or, tan 2x=-1tan 2x=-1
Which implies either
x=nπ2x=nπ2
, where n∈Zn∈Z ,
Or, tan 2x=-1tan 2x=-1
=-tanπ4=-tanπ4
=tan(π-π4)=tan(π-π4)
=tan3π4=tan3π4
Which in turn implies that,
2x=nπ+3π4,2x=nπ+3π4, where n∈Zn∈Z
i.e, x=nπ2+3π8,x=nπ2+3π8, where n∈Zn∈Z.
Therefore, the general solution is nπ2nπ2 or nπ2+3π8,n∈Znπ2+3π8,n∈Z.

9. Find the general solution of the equation
sin x+sin 3x+sin 5x=0sin x+sin 3x+sin 5x=0

Ans:
Here given that ,
sin x+sin 3x+sin 5x=0sin x+sin 3x+sin 5x=0
Now we know that, sin A+sin B=2sin(A+B2)cos(A-B2)sin A+sin B=2sin(A+B2)cos(A-B2)
Therefore ,

sin x+sin 3x+sin 5x=0sin x+sin 3x+sin 5x=0

(sin x+sin 3x)+sin 5x=0(sin x+sin 3x)+sin 5x=0

[2sin(x+5×2)cos(x-5×2)]+sin 3x=0[2sin(x+5×2)cos(x-5×2)]+sin 3x=0

2sin 3x cos (-2x)+sin 3x=02sin 3x cos (-2x)+sin 3x=0
Simplifying we get,
2sin 3xcos 2x+sin 3x=02sin 3xcos 2x+sin 3x=0

sin 3x(2cos 2x+1)=0sin 3x(2cos 2x+1)=0
Hence either sin 3x=0sin 3x=0
Or, cos 2x=-12cos 2x=-12
Which implies 3x=nπ3x=nπ , where n∈Zn∈Z
Or, cos 2x=-12cos 2x=-12
=-cosπ3=-cosπ3
=cos(π-π3)=cos(π-π3)
=cos2π3=cos2π3
i.e., either x=nπ3x=nπ3 , where n∈Zn∈Z
or, 2x=2nπ±2π32x=2nπ±2π3 ,where n∈Zn∈Z .
Therefore, the general solution is nπ3nπ3 or nπ±π3,n∈Znπ±π3,n∈Z.