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Class–XI–CBSE-Mathematics Trigonometric Functions
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CBSE NCERT Solutions for Class 11 Mathematics Chapter 03
Back of Chapter Questions
1. Find the radian measures corresponding to the following degree measures: (i)25o
(ii)−47o30′
(iii)240o
(iv)520o
Solution:
(i) Step1: Given degree is 25o
It is known that 180o = πradian
Therefore, 25o =𝜋
180× 25radian
=5𝜋
36 radian
Overall Hint: Convert the degree measures to radians by using the formula Angle in
degree = 𝜋
180× 𝑀𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝐴𝑛𝑔𝑙𝑒 𝑖𝑛 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 radian
(ii) Step1: Given degree is −47o30’
−47o30′ = −471
2degree [∵ 1o = 60′]
= − 95
2
o
As180o = 𝜋radian
⇒ − 95
2
o=
𝜋
180× (
−95
2)radian= (
−19
72) 𝜋radian
Therefore, −47o30’ = (−19
72) 𝜋radian
Overall Hint: Convert the degree measures to radians by using the formula Angle in
degree = 𝜋
180× 𝑀𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝐴𝑛𝑔𝑙𝑒 𝑖𝑛 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 radian
(iii) Step1: Given degree is240o
It is known that 180o = 𝜋radian
⸫240o =𝜋
180× 240radian
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=4𝜋
3 radian
Overall Hint: Convert the degree measures to radians by using the formula Angle in
degree = 𝜋
180× 𝑀𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝐴𝑛𝑔𝑙𝑒 𝑖𝑛 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 radian
a. Step1: Given degree is520o
As 180o = 𝜋radian
Therefore, 520o =𝜋
180× 520radian=
26𝜋
9radian
Overall Hint: Convert the degree measures to radians by using the formula Angle in
degree = 𝜋
180× 𝑀𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝐴𝑛𝑔𝑙𝑒 𝑖𝑛 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 radian
2. Find the degree measures corresponding to the following radian measures(Useπ =22
7)
(i)11
16
(ii)−4
(iii)5π
3
(iv)7π
6
Solution:
(i) Step1:
Given radian is11
16
As 𝜋radian= 180o
⇒ 11
16radian=
180
π×
11
16degree=
45×11
𝜋×4degree
=45×11×7
22×4degree
=315
8degree
= 393
8degree
Step2:
= 39o +3×60
8Minutes [1o = 60′]
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= 39o + 22’ +1
2minutes
=39o22′30” [1’ = 60”]
Overall Hint: Convert the given radian angle into degrees by using the formula Rad × 180/π = Deg
(ii) Step1: Given radian is −4
As 𝜋radian= 180𝑜
−4radian=180
π× (−4)radian=
180×7×(−4)
22degree
=−2520
11degree= −229
1
11degree
= −229o +1×60
11minutes [1o = 60′]
Step2:
= −229o + 5’ +5
11minutes
= −229o + 5’ +5×60
11 minutes
= −229o5′27" [1’ = 60”]
Overall Hint: Convert the given radian angle into degrees by using the formula Rad ×
180/π = Deg
(iii) Step1:
Given radian is 5𝜋
3
As πradian= 180o
5𝜋
3radian=
180
π×
5𝜋
3radian
= 300o Overall Hint: Convert the given radian angle into degrees by using the formula Rad × 180/π = Deg
(iv) Step1:
Given radian is7𝜋
6
As 𝜋radian= 180o
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7𝜋
6radian=
180
π×
7𝜋
6radian= 210o
Overall Hint: Convert the given radian angle into degrees by using the formula Rad × 180/π = Deg
3. A wheel makes 360revolutions in one minute. Through how many radians does it turn in one second?
Solution:
Step1:
Given,
Numberofrevolutionsmadebythewheelin1minute= 360
Therefore, Numberofrevolutionmadebythewheelin1second=360
60= 6
For one completerevolution,thewheel willturnanangleof2πradian.
Therefore,in6completerevolutions,itwillturnanangleof= 6 × 2π = 12π radian
Hence,inonesecond,thewheelturnsanangleof12πradian.
Overall Hint: Convert the revolutions per minute to radians per second by using the formula 1 rpm = 2(pi)/60 = pi/30 radians per second
4. Find the degree measure of the angle subtended at the Centre of a circle of
radius100 cm by an arc of length22 cm (Useπ =22
7)
Solution:
Step1:
Given,
radiusofcircle= 100 cm,
lengthofarc= 22 cm.
Consider theanglesubtendedbythearcattheCentreofthecircleasθ.
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Weknow, θ = arc
radius
Step2:
θ =22
100radian=
180
π×
22
100degree=
180×7×22
22×100degree
=126
10degree
= 123
5degree
= 123×60
5 degree [1o = 60′]
= 12o36′
Therefore,therequiredangleis12o36’. Overall DL: M
Overall Hint: Find the angle subtended at the center of the circle by using the formula
θ = arc
radius then find the angle measure in degree by using θ =
180×7×22
22×100degree
5. In a circle of diameter40 cm,the length of a chord is20 cm.Find the length of minor arc of the chord.
Solution:
Step1:
Step1:
Step2:
Given, Diameter of the circle= 40 cm
So, Radius of the circle=40
2cm = 20 cm
Let AB be a chord of the circle.
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InΔOAB, OA = OB =radiusofcircle= 20 cm
Also, AB = 20 cm
Therefore ΔOAB is equilateral triangle.
⸫θ = 60o = 𝜋
3radian
Step3:
We know that θ = arc
radius
Arc= θ ×Radius
= 𝜋
3× 20 cm
=20𝜋
3cm
Therefore, the length of the minor arc of the chord is 20𝜋
3cm.
Overall Hint: Draw a rough figure making use of the given data then find the measure
of the angle in radians. Also find the arc length using the formula Arc= θ ×Radius
6. If in two circles,arcs of the same length subtend angles60oand75oat the Centre,find the ratio of their radii.
Solution:
Step1:
Let the angle formed by the arc of first circle θ1 = 60o = 60 × (π
180)radian=
60𝜋
180
Let the angle formed by the arc of second circle θ2 = 75o = 75 × (π
180) =
75π
180
Let the radius of first circle be 𝑟1 and radius of second circle be 𝑟2
It is known that, θ = arc
radius
Step2:
Radius=arc
θ
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𝑟1
𝑟2 =
𝑎𝑟𝑐
𝜃1𝑎𝑟𝑐
𝜃2
= 𝜃2
𝜃1
r1
r2=
75π
18060π
180
=75
60=
5
4
Therefore, the ratio of their radii is 5: 4 Overall Hint:First find the angles formed by the arcs of the two circles in radians then
make use of the formula 𝑟1
𝑟2 =
𝑎𝑟𝑐
𝜃1𝑎𝑟𝑐
𝜃2
= 𝜃2
𝜃1 to find ratio of their radii.
7. Find the angle in radian through which a pendulum swings if its length is75 cm and the tip describes an arc of length.
(i)10 cm
(ii)15 cm
(iii)21 cm
Solution:
We know that in a circle of radius 𝑟, if an arc of length 𝑙 unit subtends an angle θ radian
at the Centre, then 𝜃 = 𝑙
𝑟
Given that 𝑟 = 75 cm (i) Step1:
Here,𝑙 = 10 cm
𝜃 = 𝑙
𝑟
𝜃 = 10
75radian=
2
15radian
Overall Hint: Find the angle using formula
𝜃 = 𝑙
𝑟 radian
(ii) Step1:
Here,𝑙 = 15 cm
𝜃 = 𝑙
𝑟
⇒ θ = 15
75radian=
1
5radian
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Overall Hint: Find the angle using formula
𝜃 = 𝑙
𝑟 radian
(iii) Step1:
Here,𝑙 = 21 cm
𝜃 = 𝑙
𝑟
θ = 21
75radian
=7
25radian
Overall Hint: Find the angle using formula
𝜃 = 𝑙
𝑟 radian
Exercise3.2
Find the values of other five trigonometric functions in Exercise 1 to 5.
1. cos 𝑥 = −1
2, 𝑥 lies in third quadrant.
Solution:
Step1:
Given, cos 𝑥 =1
2
Therefore, sec 𝑥 = 1
cos 𝑥=
1
−1
2
= −2
We know that sin2𝑥 + cos2𝑥 = 1
Step2:
⇒ sin2𝑥 = 1 − cos2𝑥
⇒ sin2𝑥 = 1 − (−1
2)
2
⇒ sin2𝑥 = 1 −1
4
⇒ sin2𝑥 =3
4
Therefore, sin 𝑥 = ±√3
2
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The value of sin𝑥 is negative as 𝑥 lies in third quadrant. So, sin 𝑥 = −√3
2
Step3:
We know that cosec 𝑥 =1
sin𝑥
⇒ cosec 𝑥 = −2
√3
We know that tan 𝑥 =sin𝑥
cos𝑥=
−√3
2
−1
2
Therefore, tan 𝑥 = √3
We know that cot 𝑥 =1
tan𝑥
Therefore, cot𝑥 =1
√3
Overall Hint: As cos x is given we can find sec x as its reciprocal then we will find sin x using 1 – cos^2(x) and then cosec x as its reciprocal the cot x and tan x can be found using the ratio of cos x and sin x
2. sin 𝑥 =3
5, 𝑥Liesinsecondquadrant.
Solution:
Step1:
Given, sin 𝑥=3
5
cosec 𝑥 =1
sin𝑥
⇒ cosec 𝑥 =5
3
We know that, sin2𝑥 + cos2𝑥 = 1
Step2:
⇒ cos2𝑥 = 1 − sin2𝑥
⇒ 𝑐𝑜𝑠2𝑥 = 1 − (3
5)
2
⇒ cos2𝑥 = 1 −9
25
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⇒ cos2𝑥 =16
25
⇒ cos 𝑥 = ±4
5
The value of cos is negative as 𝑥 lies in second quadrant. So, cos 𝑥 = −4
5
Step3:
We know that, sec 𝑥 =1
cos𝑥
⇒ sec 𝑥 = −5
4
We know that, tan 𝑥 =sin𝑥
cos𝑥
=
3
5
−4
5
= −3
4
We know that, cot 𝑥 =1
tan𝑥= −
4
3
Overall Hint: As sin x is given we can find cosec x as its reciprocal then we will find cos x using 1 – sin^2(x) and then sec x as its reciprocal the cot x and tan x can be found using the ratio of cos x and sin x
3. cot 𝑥 =3
4, 𝑥 lies in third quadrant.
Solution:
Step1:
Given, cot 𝑥 =3
4
We know that, tan 𝑥 =1
cot𝑥
⇒ tan 𝑥 =4
3
We know that cosec2𝑥 = 1 + cot2𝑥
⇒ cosec2𝑥 = 1 + (3
4)
2
⇒ cosec2𝑥 = 1 +9
16=
25
16
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⇒ cosec 𝑥 = ±√25
16
Therefore, cosec 𝑥 = ±5
4
Step2:
But x lies in third quadrant so cosec𝑥 will be negative.
⇒ cosec 𝑥 = −5
4
We know that sin 𝑥 =1
cosec𝑥
sin 𝑥 = −4
5
Also,sin2𝑥 + cos2𝑥 = 1
⇒ cos2𝑥 = 1 − sin2𝑥
⇒ cos2𝑥 = 1 − (−4
5)
2
⇒ cos2𝑥 = 1 −16
25
⇒ cos2𝑥 =9
25
⇒ cos 𝑥 = ±3
5
Step3:
But 𝑥 lies in third quadrant,so value of cos𝑥 is negative.
Therefore, cos 𝑥 = −3
5
sec 𝑥 = 1
cos 𝑥= −
5
3
Overall Hint: As cot x is given we can find tan x as its reciprocal then we will find cosec x using 1 + cot^2(x) and then sin x as its reciprocal the cos x and sec x can be found using the 1- sin^(2)x and the reciprocal of cos x will be sec x
4. sec 𝑥 =13
5, 𝑥 lies in fourth quadrant.
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Solution:
Step1:
Given, sec 𝑥 = 13
5
⇒ cos 𝑥 =1
sec𝑥=
5
13
We know that sin2𝑥 + cos2𝑥 = 1
⇒ sin2𝑥 = 1 − cos2𝑥
⇒ sin2𝑥 = 1 − (5
13)
2
⇒ sin2𝑥 =144
169
⇒ sin 𝑥 = ±12
13
Step2:
But 𝑥 lie in a fourth quadrant, so value of sin 𝑥 is negative.
Therefore, sin 𝑥 = −12
13
Also,cosec 𝑥 =1
sin𝑥= −
13
12
tan 𝑥 =sin 𝑥
cos 𝑥=
−12
135
13
= −12
5
cot 𝑥 =1
tan𝑥
Therefore, cot 𝑥 = −5
12
Overall Hint: As sec x is given we can find cos x as its reciprocal then we will find sin x using 1 – cos^2(x) and then cosec x as its reciprocal the cot x and tan x can be found using the ratio of cos x and sin x and vice versa
5. tan x = −5
12, x in second quadrant.
Solution:
Step1:
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Given, tan 𝑥 = −5
12
cot 𝑥 =1
tan𝑥
⇒ cot 𝑥 = −12
5
We know that cosec2𝑥 = 1 + cot2𝑥
⇒ cosec2𝑥 = 1 + (−12
5)
2
⇒ cosec2𝑥 = 1 +144
25=
169
25
⇒ cosec𝑥 = ±√169
25
⇒ cosec𝑥 = ±13
5
Step2:
But 𝑥 lies in second quadrant, so cosec𝑥 is positive.
⸫cosec𝑥 =13
5
We know that sin 𝑥 =1
cosec 𝑥
sin 𝑥 =5
13
Also,sin2𝑥 + cos2𝑥 = 1
⇒ cos2𝑥 = 1 − sin2𝑥
⇒ cos2𝑥 = 1 − (5
13)
2
⇒ cos2𝑥 = 1 −25
169
⇒ cos2𝑥 =144
169
⇒ cos 𝑥 = ±12
13
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Step3:
But 𝑥 lies in these con quadrant, so value of cos𝑥 is negative.
Therefore, cos 𝑥 = −12
13
We know that, sec 𝑥 = 1
cos 𝑥
Therefore, sec 𝑥 = − 13
12
Overall Hint: As tan x is given we can find cot x as its reciprocal then we will find cosec x using 1 + cot^2(x) and then sin x as its reciprocal the cos x can be found using 1 – sin^2(x) and sec x as its reciprocal
Find the values of the trigonometric functions in Exercises 6 to 10
6. sin 765o
Solution:
Step1:
sin 765o can be written as sin(2 × 360o + 45o)
= sin 45o[∵ sin is positive in first quadrant]
=1
√2
Therefore, sin 765o = 1
√2
Overall Hint: Write sin 765o as sin (2*360 + 45) using Sin(2𝜋+ x) = sin x we get sin
765 as 1
√2
7. cosec (−1410o)
Solution:
Step1:
Given,cosec(−1410o)
= −cosec(1410o)
= − cosec(4 × 360o − 30o)
= −cosec(−30o)
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= cosec(30o)
= 2
Overall Hint: Write cosec (−1410o) as cosec (4 × 360o − 30o) using cosec
(2n𝜋+x) = Cosec we get cosec (−1410o) as 2
8. tan19π
3
Solution:
Step1:
Given, tan19𝜋
3
tan19𝜋
3= tan (6𝜋 +
𝜋
3)
= tan𝜋
3 [We know that value of tan 𝑥 repeats after an interval of 𝜋 or 180o]
= √3
Therefore, tan19π
3= √3
Overall Hint: Write tan19π
3 as tan(6𝜋 +
𝜋
3) and using tan(2n𝜋 + 𝑥) = tan x we get
tan19π
3= √3
9. sin (−11π
3)
Solution:
Step1:
Given, sin (−11𝜋
3)
sin (−11𝜋
3) = − sin (
11𝜋
3) [∵sin is negative in fourth quadrant]
= − sin (4𝜋 −𝜋
3)
= − [− sin (𝜋
3)] [∵sin is negative in fourth quadrant]
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=sin (𝜋
3)
=√3
2
Overall Hint: Write sin (−11π
3) as - sin (
11𝜋
3) = - sin (4𝜋 −
𝜋
3) then using sin (2n𝜋 - x)
= -sin x we get sin (−11π
3) as
√3
2
10. cot (−15π
4)
Solution:
Step1:
Given, cot (−15𝜋
4)
cot (−15𝜋
4) = − cot (
15𝜋
4) [∵cot is negative in fourth quadrant]
= − cot (4𝜋 −𝜋
4)
= − [− cot (𝜋
4)] [∵cot is negative in fourth quadrant]
= cot (𝜋
4)
= 1
Overall Hint: Write cot (−15π
4) as - cot (
15𝜋
4) then as -- cot (4𝜋 −
𝜋
4) using cot (2n𝜋-
x) = cot x we get cot (−15π
4) as 1
Exercise:3.3
1. sin2 π
6+ cos2 π
3− tan2 π
4= −
1
2
Solution:
L.H.Ssin2 𝜋
6+ cos2 𝜋
3− tan2 𝜋
4
= (1
2)
2
+ (1
2)
2
− (1)2
= 1
4+
1
4−1
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=1
2− 1
= −1
2
=R.H.S
Hence LHS = RHS.
Overall Hint: Substitute the values of sinπ
6cos
π
3, tan
π
4 in the LHS . This will be equal
to what’s given in the RHS hence proved
2. Prove that : 2 sin2 π
6+ cosec2 7π
6cos2 π
3=
3
2
Solution:
L.H.S= 2 sin2 π
6+ cosec2 7π
6cos2 π
3
= 2 (1
2)
2
+ cosec2 (𝜋 +𝜋
6) (
1
2)
2
= 2 × 1
4+ (−cosec
𝜋
6)
2
(1
4)
=1
2+ (−2)2 (
1
4)
=1
2+ 1
=3
2
=R.H.S
Hence LHS = RHS.
Overall Hint: Substitute the values of sinπ
6, cosec
7π
6, cos
π
3 in the LHS . This will be
equal to what’s given in the RHS
3. Prove that cot2 π
6+ cosec5 π
6+ 3 tan2 π
6= 6
Solution:
L.H.S= cot2 𝜋
6+ cosec
5𝜋
6+ 3tan2 𝜋
6
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= (√3)2
+ cosec (𝜋 −𝜋
6) + 3 (
1
√3)
2
= 3 + cosec (𝜋
6) + 3 ×
1
3
= 3 + 2 + 1
= 6
=R.H.S
Hence, LHS = RHS.
Overall Hint: Substitute the values of cotπ
6, cosec
π
6, tan
π
6 in the LHS . This will be
equal to what’s given in the RHS
4. Prove that 2 sin2 3π
4+ 2 cos2 π
4+ 2 sec2 π
3= 10
Solution:
GIVEN, L.H.S= 2sin2 3𝜋
4+ 2cos2 𝜋
4+ 2sec2 𝜋
3
2sin2 (𝜋 −𝜋
4) + 2 (
1
√2)
2
+ 2(2)2
= 2sin2 (𝜋
4) + 2 ×
1
2+ 2 × 4
= 2 (1
√2)
2
+ 1 + 8
= 1 + 1 + 8
= 10
=R.H.S
Hence LHS = RHS.
Overall Hint: Substitute the values of sin3π
4, cos
π
4, sec
π
3 in the LHS . This will be
equal to what’s given in the RHS
5. Find the value of: (i)sin 75o
(ii)tan 15o
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Solution:
(i) Step1: sin75𝑜 = sin(45𝑜 + 30𝑜)
= sin 45ocos 30o + cos 45osin 30o [∵ sin(𝑋 + 𝑌) = sin 𝑋 cos𝑌 +cos𝑋 sin𝑌]
=1
√2×
√3
2+
1
√2×
1
2
=√3
2√2+
1
2√2
=√3 + 1
2√2
Overall Hint: The value of sin 75 can be found using sin (60+15) with the help of
formula sin(𝑋 + 𝑌) = sin 𝑋 cos𝑌 + cos𝑋 sin𝑌 by taking 60 and 15 as X and Y respectively
(ii) Step1:
tan15𝑜 = tan(45𝑜 − 30𝑜)
=tan45o−tan30𝑜
1+tan45otan30o [∵ tan(𝑥 − 𝑦) =tan 𝑥−tan 𝑦
1+tan 𝑥 tan 𝑦]
=1 −
1
√3
1 + 1 ×1
√3
=√3 − 1
√3 + 1
= (√3 − 1)
2
(√3 − 1) (√3 + 1)=
3 + 1 − 2√3
(√3)2
− 12=
4 − 2√3
3 − 1
= 2 − √3 Overall Hint: The value of tan 15 can be found using tan(45-30) with the help of
formula tan(𝑥 − 𝑦) =tan 𝑥−tan 𝑦
1+tan 𝑥 tan 𝑦 by taking 45 and 30 as x and y respectively
6. Prove the following :cos (π
4− x) cos (
π
4− y) − sin (
π
4− x) sin (
π
4− y) = sin (x + y)
Solution:
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Step1:
L.H.S= cos (𝜋
4− 𝑥) cos (
𝜋
4− 𝑦) − sin (
𝜋
4− 𝑥) sin (
𝜋
4− 𝑦)
=1
2[2cos (
𝜋
4− 𝑥) cos (
𝜋
4− 𝑦)] +
1
2[−2sin (
𝜋
4− 𝑥) sin (
𝜋
4− 𝑦)]
=1
2[cos {(
𝜋
4− 𝑥) + (
𝜋
4− 𝑦)}] +
1
2[cos {(
𝜋
4− 𝑥) − (
𝜋
4− 𝑦)}]
+1
2[cos {(
𝜋
4− 𝑥) + (
𝜋
4− 𝑦)}] −
1
2[cos {(
𝜋
4− 𝑥) − (
𝜋
4− 𝑦)}]
[∵ 2cos𝐴cos𝐵 = cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵) and −2sin𝐴 sin𝐵 = cos(𝐴 + 𝐵) −cos(𝐴 − 𝐵)]
Step2:
=1
2× 2cos (
𝜋
2− (𝑥 + 𝑦))
= cos [𝜋
2− (𝑥 + 𝑦)]
= sin (𝑥 + 𝑦)
=R.H.S
Hence RHS = LHS Overall Hint: By multiplying the LHS throughout with 2 and dividing it with ½ and
using the identity 2cos𝐴cos𝐵 = cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵) and −2sin𝐴 sin𝐵 = cos(𝐴 + 𝐵) − cos(𝐴 − 𝐵) We can prove the LHS and RHS as equal
7. Prove that:tan(
π
4+x)
tan(π
4−x)
= (1+tanx
1−tanx)
2
Solution:
Step1:
We know that,
tan(𝑥 + 𝑦) =tan 𝑥+tan 𝑦
1−tan 𝑥 .tan 𝑦 andtan(𝑥 − 𝑦) =
tan 𝑥−tan 𝑦
1+tan 𝑥.tan 𝑦
Given, L.H.S=tan(
𝜋
4+𝑥)
tan(𝜋
4−𝑥)
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=
tan𝜋
4+tan 𝑥
1−tan𝜋
4tan 𝑥
tan𝜋
4−tan 𝑥
1+tan𝜋
4tan 𝑥
=
1+tan 𝑥
1−tan 𝑥1−tan 𝑥
1+tan 𝑥
= (1 + tan 𝑥
1 − tan 𝑥)
2
=R.H.S
Hence, LHS = RHS
Overall Hint:Use the identities tan(𝑥 + 𝑦) =tan 𝑥+tan 𝑦
1−tan 𝑥 .tan 𝑦 andtan(𝑥 − 𝑦) =
tan 𝑥−tan 𝑦
1+tan 𝑥.tan 𝑦
In the numerator and denominator of The LHS to prove the RHS
8. Prove that cos(π+𝑥)cos(−𝑥)
sin(π−𝑥)cos(π
2+𝑥)
= cot2𝑥
Solution:
Step1:
Given, L.H.S=cos(𝜋+𝑥)cos (−𝑥)
sin(𝜋−𝑥)cos(𝜋
2+𝑥)
=(−cos 𝑥)( cos𝑥)
(sin 𝑥) (−sin𝑥)
=cos2𝑥
sin2𝑥
= cot2𝑥
=R.H.S
Hence LHS = RHS
Overall Hint: Cos(𝜋+x) = - cos x and cos (-x) = cos x and sin(𝜋 − 𝑥) = sinx, cos(𝜋
2+
𝑥) – sin x in the numerator and denominator of the LHS to prove the RHS
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9. cos (3π
2+ 𝑥) cos(2π + 𝑥) [cot (
3π
2− 𝑥) + cot(2π + 𝑥)] = 1
Solution:
Step1:
Given, L.H.S.= cos (3𝜋
2+ 𝑥) cos (2𝜋 + 𝑥) [cot (
3𝜋
2− 𝑥) + cot (2𝜋 + 𝑥)]
= sin 𝑥 cos 𝑥 [tan 𝑥 + cot 𝑥]
= sin 𝑥 cos 𝑥 [𝑠𝑖𝑛 𝑥
cos 𝑥+
cos 𝑥
sin 𝑥]
= sin 𝑥 cos 𝑥 [sin2 𝑥 + cos2 𝑥
cos 𝑥 sin 𝑥]
= sin2𝑥 + cos2𝑥
= 1
=R.H.S.
Hence LHS = RHS.
Overall Hint: Use Cos(2𝑛+1𝜋
2+ 𝑥) = sin x , cos(2𝑛𝜋 + 𝑥 ) = cos x and [cot (
3𝜋
2− 𝑥) +
cot (2𝜋 + 𝑥)] = tan 𝑥 + cot 𝑥 in the LHS to prove the RHS
10. Prove that sin(n + 1) 𝑥 sin(n + 2) 𝑥 + cos(n + 1) 𝑥 cos(n + 2) 𝑥 = cos 𝑥
Solution:
Step1:
Given, L.H.S.= sin(𝑛 + 1) 𝑥 sin(𝑛 + 2)𝑥 + cos(𝑛 + 1) 𝑥 cos(𝑛 + 2)𝑥
=1
2(2 sin(𝑛 + 1) 𝑥 sin(𝑛 + 2)𝑥 + 2 cos(𝑛 + 1) 𝑥 cos(𝑛 + 2)𝑥)
=1
2[cos{(𝑛 + 1)𝑥 − (𝑛 + 2)𝑥} − cos{(𝑛 + 1)𝑥 + (𝑛 + 2)𝑥}]
+1
2[cos{(𝑛 + 1)𝑥 + (𝑛 + 2)𝑥} + cos{(𝑛 + 1)𝑥 − (𝑛 + 2)𝑥}]
=1
2× 2cos {(𝑛 + 1)𝑥 − (𝑛 + 2)𝑥}
= cos(−𝑥)
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= cos 𝑥
=R.H.S
Hence LHS = RHS Overall Hint: Multiply the LHS of the equation throughout by 2 and divide with ½ then
use the formula cos(A − B) = cos A cos B + sin A sin B to prove the RHS
11. Prove that cos (3π
4+ 𝑥) − cos (
3π
4− 𝑥) = −√2 sin 𝑥
Solution:
Step1:
L.H.S= cos (3𝜋
4+ 𝑥) − cos (
3𝜋
4− 𝑥)
We know that,cos𝐴 − cos𝐵 = −2sin (𝐴+𝐵
2) sin (
𝐴−𝐵
2)
So, L.H.S.= −2sin {(
3𝜋
4+𝑥)+(
3𝜋
4−𝑥)
2} sin {
(3𝜋
4+𝑥)−(
3𝜋
4−𝑥)
2}
= −2sin (3𝜋
4) sin𝑥
= −2sin (𝜋 −𝜋
4) sin𝑥
= −2sin (𝜋
4) sin𝑥
= −2 ×1
√2sin𝑥
= −√2sin𝑥
=R.H.S
Hence Proved
Overall Hint: Use the formula cos𝐴 − cos𝐵 = −2sin (𝐴+𝐵
2) sin (
𝐴−𝐵
2) in the LHS to
prove the RHS. Also use sin (𝜋 −𝜋
4) = sin (
𝜋
4)
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12. Prove that sin2 6𝑥 − sin2 4𝑥 = sin 2𝑥 sin 10𝑥
Solution:
Step1:
Given, L.H.S.= sin26𝑥 − sin24𝑥
= (sin6𝑥 − sin4𝑥)(sin6𝑥 + sin4𝑥)
sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2)and sin𝐴 − sin𝐵 = 2cos (
𝐴+𝐵
2) sin (
𝐴−𝐵
2)
L.H.S.= [2cos (6𝑥+4𝑥
2) sin (
6𝑥−4𝑥
2)] [2sin (
6𝑥+4𝑥
2) cos (
6𝑥−4𝑥
2)]
= [2 cos 5𝑥 sin 𝑥][2 sin 5𝑥 cos 𝑥]
= [2 sin 5𝑥 cos 5𝑥][2 sin 𝑥 cos 𝑥]
= sin 10𝑥 sin 2𝑥
=R.H.S.
Hence RHS = LHS
Overall Hint: Write the LHS using a^2 – b^2 = (a+b)*(a-b) then make use of the
formulae sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2)and sin𝐴 − sin𝐵 =
2cos (𝐴+𝐵
2) sin (
𝐴−𝐵
2)
to prove the RHS
13. Prove that cos2 2𝑥 − cos2 6𝑥 = sin 4𝑥 sin 8𝑥
Solution:
Step1:
L.H.S.= cos22𝑥 − cos26𝑥
= (cos 2𝑥 + cos 6𝑥)(cos 2𝑥 − cos 6𝑥)
We know that,
cos𝐴 − cos𝐵 = −2sin (𝐴+𝐵
2) sin (
𝐴−𝐵
2)andcos𝐴 + cos𝐵 = 2cos (
𝐴+𝐵
2) cos (
𝐴−𝐵
2)
So,L.H.S.= [−2sin (2𝑥+6𝑥
2) sin (
2𝑥−6𝑥
2)] [2cos (
2𝑥+6𝑥
2) cos (
2𝑥−6𝑥
2)]
= [−2sin(4𝑥)sin(−2𝑥)][2cos(4𝑥)cos(−2𝑥)]
= [2 sin 4𝑥 sin 2𝑥][2 cos 4𝑥 cos 2𝑥]
= [2 sin 4𝑥 cos 4𝑥][2 sin 2𝑥 cos 2𝑥]
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= sin 8𝑥 sin 4𝑥
=R.H.S.
Hence LHS = RHS.
Overall Hint: Write the LHS in the form of a^2 – b^2 = (a+b)*(a-b) and make use of
the formulae cos𝐴 − cos𝐵 = −2sin (𝐴+𝐵
2) sin (
𝐴−𝐵
2)andcos𝐴 + cos𝐵 =
2cos (𝐴+𝐵
2) cos (
𝐴−𝐵
2)
To prove the RHS
14. Prove that sin 2𝑥 + 2 sin 4𝑥 + sin 6𝑥 = 4 cos2 𝑥 sin 4𝑥
Solution:
Step1:
Given, L.H.S.= sin 2𝑥 + 2 sin 4𝑥 + sin 6𝑥
= (sin 2𝑥 + sin 6𝑥) + 2 sin 4𝑥
We know that sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2)
So,L.H.S.= 2sin (2𝑥+6𝑥
2) cos (
2𝑥−6𝑥
2) + 2 sin 4𝑥
Step2:
= 2 sin 4𝑥 cos(−2𝑥) + 2 sin 4𝑥
= 2 sin 4𝑥 [cos 2𝑥 + 1]
= 2 sin 4𝑥 [2cos2𝑥 − 1 + 1]
= 2 sin 4𝑥(2 cos2𝑥)
= 4cos2𝑥. sin 4𝑥
=R.H.S
Hence LHS = RHS.
Overall Hint: Write sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2) to prove the RHS
15. Prove that cot 4𝑥 (sin 5𝑥 + sin 3𝑥) = cot 𝑥 (sin 5𝑥 − sin 3𝑥)
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Solution:
Step1:
Given, L.H.S.= cot 4𝑥 (sin 5𝑥 + sin 3𝑥)
We know that sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2)
So,L.H.S.=cos 4𝑥
sin 4𝑥[2sin (
5𝑥+3𝑥
2) cos (
5𝑥−3𝑥
2)]
=cos 4𝑥
sin 4𝑥(2 sin 4𝑥 cos 𝑥)
= 2 cos 4𝑥 cos 𝑥
R.H.S.= cot𝑥(sin5𝑥 − sin3𝑥)
Step2:
We know that sin𝐴 − sin𝐵 = 2cos (𝐴+𝐵
2) sin (
𝐴−𝐵
2)
So,R.H.S.=𝑐𝑜𝑠𝑥
sin𝑥[2cos (
5𝑥+3𝑥
2) sin (
5𝑥−3𝑥
2)]
=cos𝑥
sin𝑥[2cos4𝑥 sin𝑥]
= 2cos4𝑥 cos𝑥
L.H.S.=R.H.S.
Hence Proved
Overall Hint: Write cot 4𝑥 (sin 5𝑥 + sin 3𝑥) as cos 4𝑥
sin 4𝑥[2sin (
5𝑥+3𝑥
2) cos (
5𝑥−3𝑥
2)] using
formula sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2) in the LHS and cot 𝑥 (sin 5𝑥 −
sin 3𝑥) as 𝑐𝑜𝑠𝑥
sin𝑥[2cos (
5𝑥+3𝑥
2) sin (
5𝑥−3𝑥
2)] using the formula sin𝐴 − sin𝐵 =
2cos (𝐴+𝐵
2) sin (
𝐴−𝐵
2)
In the RHS
16. Prove thatcos 9𝑥−cos 5𝑥
sin 17𝑥−sin 3𝑥= −
sin 2𝑥
cos 10𝑥
Solution:
Step1:
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Given, L.H.S=cos 9𝑥−cos 5𝑥
sin 17𝑥−sin 3𝑥
We know that,
sin𝐴 − sin𝐵 = 2cos (𝐴+𝐵
2) sin (
𝐴−𝐵
2)andcos𝐴 − cos𝐵 = −2sin (
𝐴+𝐵
2) sin (
𝐴−𝐵
2)
L.H.S =− 2sin(
9𝑥+5𝑥
2)sin(
9𝑥−5𝑥
2)
2cos(17𝑥+3𝑥
2)sin(
17𝑥−3𝑥
2)
= −2sin 7𝑥. sin 2𝑥
2cos10𝑥. sin 7𝑥
= −sin 2𝑥
cos10𝑥
=R.H.S.
Hence LHS = RHS.
Overall Hint: Change the numerator and denominator in the LHS as cos 9𝑥−cos 5𝑥
sin 17𝑥−sin 3𝑥 =
− 2sin(9𝑥+5𝑥
2)sin(
9𝑥−5𝑥
2)
2cos(17𝑥+3𝑥
2)sin(
17𝑥−3𝑥
2) with the help of formula sin𝐴 − sin𝐵 =
2cos (𝐴+𝐵
2) sin (
𝐴−𝐵
2)andcos𝐴 − cos𝐵 = −2sin (
𝐴+𝐵
2) sin (
𝐴−𝐵
2)
For the numerator and denominator respectively then simplify it to show that its equal to RHS
17. Prove that :sin5𝑥+sin3𝑥
cos5𝑥+cos3𝑥= tan4𝑥
Solution:
Step1:
L.H.S=sin5𝑥+sin3𝑥
cos5𝑥+cos3𝑥
We know that,
sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2)andcos𝐴 + cos𝐵 = 2cos (
𝐴+𝐵
2) cos (
𝐴−𝐵
2)
Given, L.H.S = 2sin(
5𝑥+3𝑥
2)cos(
5𝑥−3𝑥
2)
2cos(5𝑥+3𝑥
2)cos(
5𝑥−3𝑥
2)
= 2sin(4𝑥)cos(𝑥)
2cos(4𝑥)cos(𝑥)
=sin4𝑥
cos4𝑥
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= tan4𝑥
=R.H.S
Hence LHS = RHS
Overall Hint: Change the numerator and denominator of the LHS (sin5𝑥+sin3𝑥
cos5𝑥+cos3𝑥) as
2sin(5𝑥+3𝑥
2)cos(
5𝑥−3𝑥
2)
2cos(5𝑥+3𝑥
2)cos(
5𝑥−3𝑥
2) with the help of the formulae sin𝐴 + sin𝐵 =
2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2)andcos𝐴 + cos𝐵 = 2cos (
𝐴+𝐵
2) cos (
𝐴−𝐵
2) then simplify it to
show that its equal to RHS
18. Prove that sin 𝑥−sin 𝑦
cos 𝑥+cos 𝑦= tan
x−y
2
Solution:
Step1:
Given, L.H.S=sin 𝑥−sin 𝑦
cos 𝑥+cos 𝑦
We know that sin𝐴 − sin𝐵 = 2cos (𝐴+𝐵
2) sin (
𝐴−𝐵
2)andcos𝐴 + cos𝐵 =
2cos (𝐴+𝐵
2) cos (
𝐴−𝐵
2)
L.H.S = 2cos(
𝑥+𝑦
2)sin(
𝑥−𝑦
2)
2cos(𝑥+𝑦
2)cos(
𝑥−𝑦
2)
=sin (
𝑥−𝑦
2)
cos (𝑥−𝑦
2)
= tan (𝑥 − 𝑦
2)
=R.H.S.
Hence LHS = RHS
Overall Hint: Change the numerator and denominator of LHS (sin 𝑥−sin 𝑦
cos 𝑥+cos 𝑦) to
2cos(𝑥+𝑦
2)sin(
𝑥−𝑦
2)
2cos(𝑥+𝑦
2)cos(
𝑥−𝑦
2) using the formulae sin𝐴 − sin𝐵 = 2cos (
𝐴+𝐵
2) sin (
𝐴−𝐵
2)andcos𝐴 +
cos𝐵 = 2cos (𝐴+𝐵
2) cos (
𝐴−𝐵
2) and then simplify it to show that its equal to RHS
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19. Prove thatsin𝑥+sin3𝑥
cos𝑥+cos3𝑥= tan 2𝑥
Solution:
Step 1:
Given, L.H.S.=sin𝑥+sin3𝑥
cos𝑥+cos3𝑥
=2sin(
𝑥+3𝑥
2)cos(
𝑥−3𝑥
2)
2cos(𝑥+3𝑥
2)cos(
𝑥−3𝑥
2) [∵ sin𝐴 + sin𝐵 = 2sin (
𝐴+𝐵
2) cos (
𝐴−𝐵
2)]
=sin 2𝑥
cos 2𝑥 [∵ cos𝐴 + cos𝐵 = 2cos (
𝐴+𝐵
2) cos (
𝐴−𝐵
2)]
= tan 2𝑥
=R.H.S.
Hence LHS = RHS.
Overall Hint: Change the numerator and denominator of the LHS ( sin𝑥+sin3𝑥
cos𝑥+cos3𝑥) to
2sin(𝑥+3𝑥
2)cos(
𝑥−3𝑥
2)
2cos(𝑥+3𝑥
2)cos(
𝑥−3𝑥
2) using the formulae [sin𝐴 + sin𝐵 = 2sin (
𝐴+𝐵
2) cos (
𝐴−𝐵
2)] and
[cos𝐴 + cos𝐵 = 2cos (𝐴+𝐵
2) cos (
𝐴−𝐵
2)] to prove the LHS equal to RHS
20. Prove that sin x−sin 3x
sin2 x−cos2 x= 2 sin x
Solution:
Step1:
Given, L.H.S.=sin𝑥−sin3𝑥
sin2𝑥−cos2𝑥
=2cos(
𝑥+3𝑥
2)sin(
𝑥−3𝑥
2)
− cos 2𝑥 [∵ sin𝐴 − sin𝐵 = 2cos (
𝐴+𝐵
2) sin (
𝐴−𝐵
2)]
=2 cos 2𝑥 sin(−𝑥)
− cos 2𝑥[∵ cos2𝑥 − sin2𝑥 = cos 2𝑥]
=−2 cos 2𝑥 sin 𝑥
− cos 2𝑥
= 2 sin 𝑥
=R.H.S.
Hence LHS = RHS.
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Overall Hint: Change the numerator and denominator of LHS (sin x−sin 3x
sin2 x−cos2 x) to
2cos(𝑥+3𝑥
2)sin(
𝑥−3𝑥
2)
− cos 2𝑥 using the formulae [sin𝐴 − sin𝐵 = 2cos (
𝐴+𝐵
2) sin (
𝐴−𝐵
2)] and
[cos2𝑥 − sin2𝑥 = cos 2𝑥] 𝑎𝑛𝑑 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦 𝑖𝑡 to prove the LHS equal to RHS
21. Prove thatcos4𝑥+cos3𝑥+cos2𝑥
sin4𝑥+sin3𝑥+sin2𝑥= cot 3𝑥
Solution:
Step1:
Given, L.H.S.=cos4𝑥+cos3𝑥+cos2𝑥
sin4𝑥+sin3𝑥+sin2𝑥
=(cos4𝑥 + cos2𝑥) + cos3𝑥
(sin4𝑥 + sin2𝑥) + sin3𝑥
We know that,
sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2)andcos𝐴 + cos𝐵 = 2cos (
𝐴+𝐵
2) cos (
𝐴−𝐵
2)
=2cos (
4𝑥+2𝑥
2) cos (
4𝑥−2𝑥
2) + cos3𝑥
2sin (4𝑥+2𝑥
2) cos (
4𝑥−2𝑥
2) + sin3𝑥
Step2:
=2 cos 3𝑥 cos 𝑥 + cos 3𝑥
2 sin 3𝑥 cos 𝑥 + sin 3𝑥
=cos 3𝑥 (cos 𝑥 + 1)
sin 3𝑥 (cos 𝑥 + 1)
=cos 3𝑥
sin 3𝑥
= cot 3𝑥
=R.H.S.
Hence LHS = RHS
Overall Hint: Change the numerator and denominator of LHS( cos4𝑥+cos3𝑥+cos2𝑥
sin4𝑥+sin3𝑥+sin2𝑥) to
2cos(4𝑥+2𝑥
2)cos(
4𝑥−2𝑥
2)+cos3𝑥
2sin(4𝑥+2𝑥
2)cos(
4𝑥−2𝑥
2)+sin3𝑥
using the formulae sin𝐴 + sin𝐵 =
2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2)andcos𝐴 + cos𝐵 = 2cos (
𝐴+𝐵
2) cos (
𝐴−𝐵
2) and simplify it to prove
the RHS
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22. Prove that cot 𝑥 cot 2𝑥 − cot 2𝑥 cot 3𝑥 − cot 3𝑥 cot 𝑥 = 1
Solution:
Step1:
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)(cot2𝑥 + cot𝑥)
= cot 𝑥 cot 2𝑥 − [cot 2𝑥 cot 𝑥−1
cot 2𝑥+cot 𝑥] (cot 2𝑥 + cot 𝑥) [∵cot(𝐴 + 𝐵) =
[cot𝐴cot𝐵−1
cot𝐴+cot𝐵]
= cot 𝑥 cot 2𝑥 − cot 2𝑥 cot 𝑥 + 1
= 1
=R.H.S.
Hence LHS = RHS. Overall Hint: take cot3x in the 2nd and 3rd term and Use the formula cot(𝐴 + 𝐵) =
[cot𝐴cot𝐵−1
cot𝐴+cot𝐵] to show that LHS is equal to RHS
23. Prove that tan 4𝑥 =4 tan 𝑥(1−tan2 𝑥)
1−6 tan2 𝑥+tan4 𝑥
Solution:
Step1:
We know that,tan 2𝐴 =2 tan 𝐴
1−tan2𝐴
L.H.S.= tan4𝑥 = tan2(2𝑥)
=2 tan 2𝑥
1 − tan2(2𝑥)
=2 (
2 tan 𝑥
1−tan2𝑥)
1 − (2 tan 𝑥
1−tan2𝑥)
2
Step2:
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=
4 tan 𝑥
1−tan2𝑥(1−tan2𝑥)2−4tan2𝑥
(1−tan2𝑥)2
=4 tan 𝑥(1−tan2𝑥)
(1−tan2𝑥)2−4tan2𝑥
=4 tan 𝑥(1−tan2𝑥)
1+tan4𝑥−2tan2𝑥−4tan2𝑥
=4 tan 𝑥(1−tan2𝑥)
1−6 tan2𝑥+tan4𝑥
=R.H.S.
Hence LHS = RHS
Overall Hint: Change the LHS tan4x as 2(
2 tan 𝑥
1−tan2𝑥)
1−(2 tan 𝑥
1−tan2𝑥)
2 using the formula tan 2A =
2 tan 𝐴
1−tan2𝐴 and then expand it to prove its equal to RHS
24. Prove thatcos 4𝑥 = 1 − 8 sin2 𝑥 cos2 𝑥
Solution:
Step1:
Given, L.H.S= cos 4𝑥 = cos2(2𝑥)
= 1 − 2sin2(2𝑥) [∵ cos2𝐴 = 1 − 2sin2𝐴]
= 1 − 2(2sin𝑥 cos𝑥)2] [∵ sin2𝐴 = 2sin𝐴cos𝐴
= 1 − 8sin2𝑥cos2𝑥
= R.H.S.
Hence LHS = RHS Overall Hint: Write cos(4x)= cos2(2x) and then apply formulae cos2𝐴 = 1 − 2sin2𝐴
sin2𝐴 = 2sin𝐴cos𝐴 to show it equal to RHS
25. Prove that cos 6𝑥 = 32 cos6 𝑥 − 48 cos4 𝑥 + 18 cos2 𝑥 − 1
Solution:
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SteP1:
Given, L.H.S. = cos 6𝑥
= cos3(2𝑥)
= 4𝑐𝑜𝑠32𝑥 − 3 cos 2𝑥
[∵ cos 3𝐴 = 4cos3𝐴 − 3 cos 𝐴]
= 4[(2cos2𝑥 − 1)3 − 3(2cos2𝑥 − 1)]
[∵ 𝑐𝑜𝑠 2𝑥 = 2 cos2 𝑥 − 1]
Step2:
= 4[(2cos2𝑥)3 − (1)3 − 3(2cos2𝑥)2 + 3(2cos2𝑥)] − 6cos2𝑥 + 3
= 4[8cos6𝑥 − 1 − 12cos4𝑥 + 6cos2𝑥] − 6cos2𝑥 + 3
= 32 cos6𝑥 − 48 cos4𝑥 + 18 cos2𝑥 − 1
=R.H.S.
Hence LHS = RHS. Overall Hint: Change the terms in LHS by using formulae cos 3𝐴 = 4cos3𝐴 − 3 cos 𝐴
𝑐𝑜𝑠 2𝑥 = 2 cos2 𝑥 − 1 and simplify it to prove that its equal to RHS
Exercise: 3.4
Find the principal and general solutions of the following equations:
1. tan 𝑥 = √3
Solution:
Step1:
We know that tan𝜋
3= √3and tan
4𝜋
3= tan (𝜋 +
𝜋
3) = tan
𝜋
3= √3
Thus ,the principal solutions are 𝑥 = 𝜋
3and𝑥 =
4𝜋
3
And,tan𝑥 = tan𝜋
3
⇒ 𝑥 = 𝑛𝜋 +𝜋
3,where 𝑛 ∈ 𝑍
Thus, the general solution is= 𝑛𝜋 +𝜋
3, where 𝑛 ∈ 𝑍
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Overall Hint : The principal solution is 𝜋 and 4𝜋
3 and the general solution is 𝑛𝜋 +
𝜋
3 , 𝑛 ∈ 𝑍
2. Find the principal and general solutions of the following equations:
sec 𝑥 = 2
Solution:
Step1:
Given, sec 𝑥 = 2
We know that sec𝜋
3= 2 andsec
5𝜋
3= sec (2𝜋 −
𝜋
3) = sec
𝜋
3= 2
Thus, the principal solutions are 𝑥 =𝜋
3and𝑥 =
5𝜋
3
Step2:
Now,sec 𝑥 = sec𝜋
3
⇒ cos 𝑥 = cos𝜋
3 [∵ sec 𝑥 =
1
cos 𝑥]
⇒ 𝑥 = 2𝑛𝜋 ±𝜋
3,where𝑛 ∈ 𝑍
Thus, the general solution is = 2𝑛𝜋 ±𝜋
3,where𝑛 ∈ 𝑍
Overall Hint: The principal solutions are 𝑥 =𝜋
3and𝑥 =
5𝜋
3 and the general solutions
are 2𝑛𝜋 ±𝜋
3, where 𝑛 ∈ 𝑍
3. Find the principal and general solutions of the following equations:
cot 𝑥 = −√3
Solution:
Step1:
We know that cot𝜋
6= √3
So,cot (𝜋 −𝜋
6) = −cot
𝜋
6= −√3andcot (2𝜋 −
𝜋
6) = −cot
𝜋
6= −√3
i.e.,cot (5𝜋
6) = −√3andcot (
11𝜋
6) = −√3
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Thus, the principal solutions are 𝑥 =5𝜋
6and𝑥 =
11𝜋
6
Step2:
Now,cot 𝑥 = cot (5𝜋
6)
⇒ tan 𝑥 = tan5𝜋
6[tan 𝑥 =
1
cot 𝑥]
⇒ 𝑥 = 𝑛𝜋 +5𝜋
6,where 𝑛 ∈ 𝑍
Thus, the general solution is= 𝑛𝜋 +5𝜋
6,where 𝑛 ∈ 𝑍
Overall Hint: The principal solution is 𝑥 =5𝜋
6and𝑥 =
11𝜋
6 and the general solution is
𝑛𝜋 +5𝜋
6,where 𝑛 ∈ 𝑍
4. Find the principal and general solution
cosec 𝑥 = −2
Solution:
Step1:
cosec 𝑥 = −2
We know that cosec𝜋
6= 2
Also, cosec (𝜋 +𝜋
6) = −cosec
𝜋
6= −2andcosec (2𝜋 −
𝜋
6) = −cosec
𝜋
6= −2
i.e., cosec (7𝜋
6) = −2andcosec (
11𝜋
6) = −2
Thus, the principal solutions are 𝑥 =7𝜋
6and𝑥 =
11𝜋
6
Step2:
Now,cosec 𝑥 = cosec (7𝜋
6)
⇒ sin𝑥 = sin7𝜋
6[sin 𝑥 =
1
cosec 𝑥]
⇒ 𝑥 = 𝑛𝜋 + (−1)𝑛 7𝜋
6,where𝑛 ∈ 𝑍
Thus, the general solution is= 𝑛𝜋 + (−1)𝑛 7𝜋
6 , where 𝑛 ∈ 𝑍
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Overall Hint: The principal solution is 𝑥 =7𝜋
6and𝑥 =
11𝜋
6 and the general solution is
𝑛𝜋 + (−1)𝑛 7𝜋
6,where𝑛 ∈ 𝑍
5. Find the general solution for each of the following equations:
cos 4𝑥 = cos 2𝑥
Solution:
Step1:
Given, cos4𝑥 = cos2𝑥
⇒ cos4𝑥 − cos2𝑥 = 0
cos4𝑥 − cos2𝑥 = −2sin (4𝑥+2𝑥
2) sin (
4𝑥−2𝑥
2)
[∵ cos𝐴 − cos𝐵 = −2sin (𝐴 + 𝐵
2) sin (
𝐴 − 𝐵
2)]
Step2:
⇒ −2 sin 3𝑥 sin 𝑥 = 0
⇒ sin3𝑥 = 0 orsin𝑥 = 0
⇒ 3𝑥 = 𝑛𝜋 or 𝑥 = 𝑛𝜋, where n ∈ Z
Thus, 𝑥 =𝑛𝜋
3or 𝑥 = 𝑛𝜋, where n ∈ Z
Overall Hint: The general solution is 𝑥 =𝑛𝜋
3or 𝑥 = 𝑛𝜋, where n ∈ Z
Use cos𝐴 − cos𝐵 = −2sin (𝐴+𝐵
2) sin (
𝐴−𝐵
2)
6. Find the general solution for each of the following equations:
cos 3𝑥 + cos 𝑥 − cos 2𝑥 = 0
Solution:
Step1:
Given, cos 3𝑥 + cos 𝑥 − cos 2𝑥 = 0
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⇒ 2cos (3𝑥 + 𝑥
2) cos (
3𝑥 − 𝑥
2) − cos 2𝑥 = 0
[∵ cos𝐴 + cos𝐵 = 2cos (𝐴 + 𝐵
2) cos (
𝐴 − 𝐵
2)]
Step2:
⇒ 2 cos 2𝑥 cos 𝑥 − cos 2𝑥 = 0
⇒ cos 2𝑥 (2 cos 𝑥 − 1) = 0
⇒ cos 2𝑥 = 0 or 2 cos 𝑥 − 1 = 0
⇒ cos 2𝑥 = 0 or cos 𝑥 =1
2
Therefore, 2𝑥 = (2𝑛 + 1)𝜋
2orcos 𝑥 = cos
𝜋
3, where n ∈ Z
Thus,𝑥 = (2𝑛 + 1)𝜋
4or 𝑥 = 2𝑛𝜋 ±
𝜋
3,where n ∈ Z
Overall Hint: The general solution is 𝑥 = (2𝑛 + 1)𝜋
4or 𝑥 = 2𝑛𝜋 ±
𝜋
3,where n ∈ Z
Use cos𝐴 + cos𝐵 = 2cos (𝐴+𝐵
2) cos (
𝐴−𝐵
2)
7. Find the general solution for each of the following equations:
sin 2𝑥 + cos 𝑥 = 0
Solution:
Step1:
Given, sin 2𝑥 + cos 𝑥 = 0
⇒ 2 sin 𝑥 cos 𝑥 + cos 𝑥 = 0
[∵ sin𝐴𝑥 = 2sin𝐴 cos𝐴]
⇒ cos 𝑥 (2 sin 𝑥 + 1) = 0
⇒ cos 𝑥 = 0 or 2 sin 𝑥 + 1 = 0
Step2:
⇒ cos 𝑥 = 0 ⇒ 𝑥 = (2𝑛 + 1)𝜋
2,where n ∈ Z
Also,2sin𝑥 + 1 = 0
⇒ sin𝑥 = −1
2
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⇒ sin𝑥 = −sin𝜋
6 ⇒ sin 𝑥 = sin (𝜋 +
𝜋
6) = sin
7𝜋
6
⇒ 𝑥 = 𝑛𝜋 + (−1)𝑛 7𝜋
6,where n ∈ Z
Thus, the general solution is 𝑥 = (2𝑛 + 1)𝜋
2or𝑥 = 𝑛𝜋 + (−1)𝑛 7𝜋
6,where n ∈ Z
Overall Hint: The general solution is 𝑥 = (2𝑛 + 1)𝜋
2or𝑥 = 𝑛𝜋 + (−1)𝑛 7𝜋
6,where n ∈
Z ,Use sin𝐴𝑥 = 2sin𝐴 cos𝐴
8. Find the general solution for each of the following equations
sec2 2𝑥 = 1 − tan 2𝑥
Solution:
Step1:
Given, sec22𝑥 = 1 − tan 2𝑥
1 + tan22𝑥 = 1 − tan 2𝑥
⇒ tan22𝑥 = − tan 2𝑥
⇒ tan22𝑥 + tan 2𝑥 = 0
⇒ tan 2𝑥 (tan 2𝑥 + 1) = 0
⇒ tan 2𝑥 = 0 Or tan 2𝑥 + 1 = 0
Now, tan 2𝑥 = tan 0
⇒ 2𝑥 = 𝑛𝜋 + 0, Where n ∈ Z
Step2:
⇒𝑥 =𝑛𝜋
2, where n ∈ Z
tan 2𝑥 + 1 = 0
⇒ tan 2𝑥 = −1
⇒ tan 2𝑥 = −tan𝜋
4 ⇒ tan 2𝑥 = tan (𝜋 −
𝜋
4) = tan
3𝜋
4
So,2𝑥 = 𝑛𝜋 +3𝜋
4, where n ∈ Z
⇒ 𝑥 =𝑛𝜋
2+
3𝜋
8, where n ∈ Z
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Thus, the general solution is 𝑥 =𝑛𝜋
2or 𝑥 =
𝑛𝜋
2+
3𝜋
8, where n ∈ Z
Overall Hint: The general solution is 𝑥 =𝑛𝜋
2or 𝑥 =
𝑛𝜋
2+
3𝜋
8, where n ∈ Z , Use sec22𝑥
=
1 + tan22𝑥. Take tan2x common.
9. Find the general solution for each of the following equation
sin 𝑥 + sin 3𝑥 + sin 5𝑥 = 0
Solution:
Step1:
Given, sin 𝑥 + sin 3𝑥 + sin 5𝑥 = 0
(sin𝑥 + sin5𝑥) + sin3𝑥 = 0
⇒ 2sin (𝑥+5𝑥
2) cos (
𝑥−5𝑥
2) + sin3𝑥 = 0
[∵ sin𝐴 + sin𝐵 = 2sin (𝐴 + 𝐵
2) cos (
𝐴 − 𝐵
2)]
⇒ 2 sin 3𝑥 cos(−2𝑥) + sin 3𝑥 = 0
⇒ sin 3𝑥 (2 cos 2𝑥 + 1) = 0
⇒ sin 3𝑥 = 0 or2 cos 2𝑥 + 1 = 0
⇒ sin3𝑥 = 0 or cos 2𝑥 = −1
2
sin 3𝑥 = 0
Step2:
3𝑥 = 𝑛𝜋, where n ∈ Z
⇒ 𝑥 =𝑛𝜋
3, where n ∈ Z
cos 2𝑥 = −1
2
cos 2𝑥 = −cos𝜋
3⇒ cos 2𝑥 = cos (𝜋 −
𝜋
3) = cos
2𝜋
3
So,2𝑥 = 2𝑛𝜋 ±2𝜋
3, where n ∈ Z
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⇒ 𝑥 = 𝑛𝜋 ±𝜋
3, where n ∈ Z
Thus, the general solution is𝑥 =𝑛𝜋
3or 𝑥 = 𝑛𝜋 ±
𝜋
3, where n ∈ Z
Overall Hint: The general solution is𝑥 =𝑛𝜋
3or 𝑥 = 𝑛𝜋 ±
𝜋
3, where n ∈ Z , Use sin𝐴 +
sin𝐵 = 2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2)
Miscellaneous Exercise
1. Prove that:2 cosπ
13cos
9π
13+ cos
3π
13+ cos
5π
13= 0
Solution:
Step1:
Given, L.H.S.= 2cos𝜋
13cos
9𝜋
13+ cos
3𝜋
13+ cos
5𝜋
13
2cos𝜋
13cos
9𝜋
13+ 2cos (
3𝜋
13+
5𝜋
13
2) cos (
3𝜋
13−
5𝜋
13
2)
[cos 𝑥 + cos 𝑦 = 2 cos (𝑥 + 𝑦
2) cos (
𝑥 − 𝑦
2)]
⇒ 2cos𝜋
13cos
9𝜋
13+ 2cos (
4𝜋
13) cos (−
𝜋
13)
Step2:
⇒ 2cos𝜋
13cos
9𝜋
13+ 2cos (
4𝜋
13) cos (
𝜋
13)
⇒ 2cos𝜋
13(cos
9𝜋
13+ cos (
4𝜋
13))
= 2cos𝜋
13(2cos (
9𝜋
13+
4𝜋
13
2) cos (
9𝜋
13−
4𝜋
13
2))
= 4cos𝜋
13(cos (
𝜋
2) cos (
5𝜋
26))
= 4cos𝜋
13(0 × cos (
5𝜋
26))
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= 0
=R.H.S.
Hence LHS = RHS.
Overall Hint: Use cos 𝑥 + cos 𝑦 = 2 cos (𝑥+𝑦
2) cos (
𝑥−𝑦
2) in the :LHS to prove it is
equal to RHS
2. Prove that:(sin 3𝑥 + sin 𝑥) sin 𝑥 + (cos 3𝑥 – cos 𝑥) cos 𝑥 = 0
Solution:
Step1:
Given, L.H.S.= (sin 3𝑥 + sin 𝑥) sin 𝑥 + (cos 3𝑥 − cos 𝑥) cos 𝑥
= sin 3𝑥 sin 𝑥 + sin2𝑥 + cos 3𝑥 cos 𝑥 − cos2 𝑥
= (sin 3𝑥 sin 𝑥 + cos 3𝑥 cos 𝑥)+(sin2𝑥 − cos2𝑥)
= cos(3𝑥 − 𝑥) − cos 2𝑥) [∵ cos(𝐴 − 𝐵) = cos𝐴 cos𝐵 + sin𝐴 sin𝐵 &cos2𝐴 =cos2𝐴 − sin2𝐴 ]
= cos2𝑥 − cos2𝑥
= 0
=R.H.S.
Hence LHS = RHS..
Overall Hint: Use cos(𝐴 − 𝐵) = cos𝐴 cos𝐵 + sin𝐴 sin𝐵 &cos2𝐴 = cos2𝐴 − sin2𝐴 to prove LHS is equal to RHS
3. Prove that:(cos 𝑥 + cos 𝑦)2 + (sin 𝑥 − sin 𝑦)2 = 4 cos2 x+𝑦
2
Solution:
Step1:
Given, L.H.S.= (cos𝑥 + cos𝑦)2 + (sin𝑥 − sin𝑦)2
= cos2𝑥 + cos2𝑦 + 2 cos 𝑥 cos 𝑦 + sin2𝑥 + sin2𝑦 − 2 sin 𝑥 sin 𝑦
= (sin2𝑥 + cos2𝑥)+(sin2𝑦 + cos2𝑦) + 2(cos 𝑥 cos 𝑦 − sin 𝑥 sin 𝑦)
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= (1) + (1) + 2cos (𝑥 + 𝑦) [∵ cos(𝐴 + 𝐵) = cos𝐴 cos𝐵 − sin𝐴 sin𝐵]
Step2:
= 2[1 + cos(𝑥 + 𝑦)]
= 2 [2cos2 (𝑥+𝑦
2)] [ ∵ 1 + cos 2𝐴 = 2cos2𝐴 ]
= 4cos2 (𝑥 + 𝑦
2)
=R.H.S.
Hence LHS = RHS. Overall Hint: Use cos(𝐴 + 𝐵) = cos𝐴 cos𝐵 − sin𝐴 sin𝐵 , 1 + cos 2𝐴 = 2cos2𝐴 , (a+b)^2 = (a^2+b^2+2ab) , (a-b)^2 = (a^2+b^2-2ab) to prove LHS is equal to RHS
4. Prove that:(cos 𝑥 − cos 𝑦)2 + (sin 𝑥 − sin 𝑦)2 = 4 sin2 𝑥−𝑦
2
Solution:
Step1:
Given, L.H.S.= (cos 𝑥 − cos 𝑦)2 + (sin 𝑥 − sin 𝑦)2
= cos2𝑥 + cos2𝑦 − 2 cos 𝑥 cos 𝑦 + sin2𝑥 + sin2𝑦 − 2 sin 𝑥 sin 𝑦
= (sin2𝑥 + cos2𝑥)+(sin2𝑦 + cos2𝑦) − 2(cos𝑥 cos𝑦 + sin𝑥 sin𝑦)
= (1) + (1) − 2 cos(𝑥 − 𝑦)
Step2:
[∵ cos(𝐴 − 𝐵) = cos𝐴 cos𝐵 + sin𝐴 sin𝐵]
= 2[1 − cos(𝑥 − 𝑦)]
= 2 [1 − (1 − 2sin2 (𝑥−𝑦
2))] [∵ 1 − cos2𝐴 = 2sin2𝐴]
= 4sin2 (𝑥 − 𝑦
2)
=R.H.S.
Hence LHS = RHS. Overall Hint: Use (a-b)^2 = (a^2+b^2-2ab) , 1 − cos2𝐴 = 2sin2𝐴 ,
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cos(𝐴 − 𝐵) = cos𝐴 cos𝐵 + sin𝐴 sin𝐵 to prove LHS equal to RHS
5. Prove that :sin 𝑥 + sin 3𝑥 + sin 5𝑥 + sin 7𝑥 = 4 cos 𝑥 cos 2𝑥 sin 4𝑥
Solution:
Step1:
Given, L.H.S.= sin 𝑥 + sin 3𝑥 + sin 5𝑥 + sin 7𝑥
⇒ (sin 𝑥 + sin 5𝑥) + (sin 3𝑥 + sin 7𝑥)
⇒ 2sin (𝑥 + 5𝑥
2) cos (
𝑥 − 5𝑥
2) + 2sin (
3𝑥 + 7𝑥
2) cos (
3𝑥 − 7𝑥
2)
[∵ sin𝐴 + sin𝐵 = 2sin (𝐴 + 𝐵
2) cos (
𝐴 − 𝐵
2)]
⇒ 2 sin 3𝑥 cos(−2𝑥) + 2 sin 5𝑥 cos(−2𝑥)
Step2:
= 2 sin 3𝑥 cos2𝑥 + 2 sin 5𝑥 cos 2𝑥
= 2 cos 2𝑥 (sin 3𝑥 + sin 5𝑥)
= 2 cos 2𝑥 [2sin (3𝑥 + 5𝑥
2) cos (
3 𝑥 − 5𝑥
2)]
= 4 cos 2𝑥 sin 4𝑥 cos(−𝑥)
= 4 cos 2𝑥 sin 4𝑥 cos 𝑥
=R.H.S.
Hence RHS = LHS
Overall Hint: Use sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵
2) cos (
𝐴−𝐵
2) formula to show that the
LHS is equal to RHS
6. Prove that (sin7𝑥+sin5𝑥)+(sin9𝑥+sin3𝑥)
(cos7𝑥+cos5𝑥)+(cos9𝑥+cos3𝑥)= tan6x
Solution:
Given, L.H.S.=(sin7𝑥+sin5𝑥)+(sin9𝑥+sin3𝑥)
(cos7𝑥+cos5𝑥)+(cos9𝑥+cos3𝑥)
We know that,
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sin 𝐴 + sin 𝐵 = 2sin (𝐴 + 𝐵
2) cos (
𝐴 − 𝐵
2) & cos 𝐴 + cos 𝐵
= 2cos (𝐴 + 𝐵
2) cos (
𝐴 − 𝐵
2)
L.H.S. ={2sin(
7𝑥+5𝑥
2)cos(
7𝑥−5𝑥
2)}+{2sin(
9𝑥+3𝑥
2)cos(
9𝑥−3𝑥
2)}
{2cos(7𝑥+5𝑥
2)cos(
7𝑥−5𝑥
2)}+{2cos(
9𝑥+3𝑥
2)cos(
9𝑥−3𝑥
2)}
Step2:
=2 sin 6𝑥 cos 𝑥 + 2 sin 6𝑥 cos 3𝑥
2 cos 𝑥 cos 𝑥 + 2 cos 6𝑥 cos 3𝑥
=2 sin 6𝑥 (cos 𝑥 + cos 3𝑥)
2 cos 6𝑥 (cos 𝑥 + cos 3𝑥)
=sin 6𝑥
cos 6𝑥
= tan 6𝑥
=R.H.S.
Hence LHS = RHS.
Overall Hint: Change the LHS( (sin7𝑥+sin5𝑥)+(sin9𝑥+sin3𝑥)
(cos7𝑥+cos5𝑥)+(cos9𝑥+cos3𝑥) to
{2sin(7𝑥+5𝑥
2)cos(
7𝑥−5𝑥
2)}+{2sin(
9𝑥+3𝑥
2)cos(
9𝑥−3𝑥
2)}
{2cos(7𝑥+5𝑥
2)cos(
7𝑥−5𝑥
2)}+{2cos(
9𝑥+3𝑥
2)cos(
9𝑥−3𝑥
2)}
using formula
sin 𝐴 + sin 𝐵 = 2sin (𝐴 + 𝐵
2) cos (
𝐴 − 𝐵
2) & cos 𝐴 + cos 𝐵
= 2cos (𝐴 + 𝐵
2) cos (
𝐴 − 𝐵
2)
And show it is equal to RHS
7. Prove that: sin 3𝑥 + sin 2𝑥 − sin 𝑥 = 4 sin 𝑥 cos𝑥
2cos
3𝑥
2
Solution:
Step1:
Given, L.H.S.sin 3𝑥 + sin 2𝑥 − sin 𝑥 = sin 3𝑥 + sin 2𝑥 − sin 𝑥
= sin 3𝑥 + 2cos (2𝑥 + 𝑥
2) sin (
2𝑥 − 𝑥
2)
[ ∵ sin𝐴 − sin𝐵 = 2cos (𝐴 + 𝐵
2) sin (
𝐴 − 𝐵
2)]
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= sin 3𝑥 + 2cos3𝑥
2sin
𝑥
2
= 2sin3𝑥
2cos
3𝑥
2+ 2cos
3𝑥
2sin
𝑥
2
= 2cos3𝑥
2(sin
3𝑥
2+ sin
𝑥
2)
= 2cos3𝑥
2(2sin (
3𝑥
2+
𝑥
2
2) cos (
3𝑥
2−
𝑥
2
2))
Step2:
[∵ sin𝐴 + sin𝐵 = 2sin (𝐴 + 𝐵
2) cos (
𝐴 − 𝐵
2)]
= 2cos3𝑥
2(2sin𝑥 cos (
𝑥
2))
= 4 sin 𝑥 cos𝑥
2cos
3𝑥
2
=R.H.S.
Hence LHS = RHS.
Overall Hint: Change LHS sin 3𝑥 + sin 2𝑥 − sin 𝑥 to sin 3𝑥 + 2cos (2𝑥+𝑥
2) sin (
2𝑥−𝑥
2)
by using the formula
Sin𝐴 − sin𝐵 = 2cos (𝐴+𝐵
2) sin (
𝐴−𝐵
2) use sin𝐴 + sin𝐵 = 2sin (
𝐴+𝐵
2) cos (
𝐴−𝐵
2) for
similarly changing the RHS and showing they both are equal
8. Find sin𝑥
2, cos
𝑥
2andtan
𝑥
2intan 𝑥 = −
4
3, 𝑥 in quadrant II
Solution:
Step1:
Here, 𝑥 lies in second quadrant.
So,𝜋
2< 𝑥 < 𝜋
⇒𝜋
4<
𝑥
2<
𝜋
2
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Therefore, sin𝑥
2, cos
𝑥
2, tan
𝑥
2liesinfirstquadrant so they all are positive.
Given,tan 𝑥 = −4
3
We know that ,tan 2𝐴 =2tan𝐴
1−tan2𝐴
So,tan𝑥 = 2tan
𝑥
2
1−tan2𝑥
2
⇒ −4
3=
2tan𝑥
2
1 − tan2 𝑥
2
⇒ −4 (1 − tan2𝑥
2) = 6tan
𝑥
2
Step2:
⇒ 4tan2𝑥
2− 6tan
𝑥
2− 4 = 0
⇒ 2tan2𝑥
2− 3tan
𝑥
2− 2 = 0
⇒ 2tan2𝑥
2− 4tan
𝑥
2+ tan
𝑥
2− 2 = 0
⇒ 2tan𝑥
2(tan
𝑥
2− 2) + 1 (tan
𝑥
2− 2) = 0
⇒ (tan𝑥
2− 2) (2tan
𝑥
2+ 1) = 0
tan𝑥
2= 2ortan
𝑥
2= −
1
2
tan𝑥
2= 2 [tan
𝑥
2 Cannot be negative as
𝑥
2 is in first quadrant]
Step3:
sec2𝑥
2= 1 + tan2
𝑥
2
sec2𝑥
2= 1 + (2)2
⇒ sec2𝑥
2= 5
⇒ sec𝑥
2= √5 [sec
𝑥
2 Cannot be negative as
𝑥
2 is in first quadrant]
⇒ cos𝑥
2=
1
√5
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Therefore, tan𝑥
2= 2
sin𝑥
2
cos𝑥
2
= 2
⇒ sin𝑥
2= 2cos
𝑥
2
Therefore, sin𝑥
2=
2
√5
Overall Hint: Find tan𝑥
2 using tan2x =
2tan𝑥
2
1−tan2𝑥
2
then find similarly values of
sin𝑥
2, cos
𝑥
2 using sin
𝑥
2= tan
𝑥
2cos
𝑥
2 and cos
𝑥
2 = sin
𝑥
2/ tan
𝑥
2
9. Find sin𝑥
2, cos
𝑥
2 and 𝑡𝑎𝑛
𝑥
2 for cos 𝑥 = −
1
3, 𝑥 in quadrant III
Solution:
Step1:
Here,𝑥liesinthirdquadrant.
So,𝜋 < 𝑥 <3𝜋
2
⇒𝜋
2<
𝑥
2<
3𝜋
4
It is known that sin𝑥
2is positive whereas, cos
𝑥
2andtan
𝑥
2are negative, lies in third
quadrant.
Given,cos 𝑥 = −1
3
We know that,cos 2𝐴 = 2cos2𝐴 − 1
So,cos 𝑥 = 2cos2 𝑥
2− 1
⇒ −1
3= 2cos2
𝑥
2− 1
⇒ cos2𝑥
2=
1 + cos 𝑥
2=
1 + (−1
3)
2=
(2
3)
2=
1
3
⇒ cos𝑥
2= ±
1
√3
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Step2:
Whereas, cos𝑥
2 is negative as
𝑥
2 lies in second quadrant.
So,cos𝑥
2= −
1
√3
We know that, sin2𝐴 + cos2𝐴 = 1
sin2𝑥
2+ (
1
√3)
2
= 1
sin2𝑥
2= 1 −
1
3
⇒ sin2𝑥
2=
2
3
sin𝑥
2= ±√
2
3
Step3:
But sin𝑥
2 is positive as
𝑥
2 lies in second quadrant.
So,sin𝑥
2= √
2
3
Also,tan𝑥
2=
sin𝑥
2
cos𝑥
2
tan𝑥
2=
√2
3
−1
√3
Therefore, tan𝑥
2= −√2
Overall Hint: Find
cos𝑥
2 using
𝑐𝑜𝑠 2𝐴= 2cos2𝐴 − 1 and then the values of
sin𝑥
2= √1– cos2 𝑥
2
𝑡𝑎𝑛𝑥
2 using tan𝑥 =
2tan𝑥
2
1−tan2𝑥
2
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10. Find sin𝑥
2, cos
𝑥
2 and tan
𝑥
2 for sin 𝑥 =
1
4, 𝑥 in quadrant II.
Solution:
Step1:
Here,𝑥 lies in second quadrant.
So,𝜋
2< 𝑥 < 𝜋
⇒𝜋
4<
𝑥
2<
𝜋
2
That means sin𝑥
2, cos
𝑥
2, tan
𝑥
2 lies in first quadrant.
Given, sin 𝑥 =1
4
We know that, sin2𝐴 + cos2𝐴 = 1
sin2𝑥 + cos2𝑥 = 1
= cos2𝑥 = 1 − (1
4)
2
⇒ cos2𝑥 = 1 −1
16=
15
16
cos 𝑥 = ±√15
4
Step2:
But cos 𝑥 is negative as 𝑥 lies in second quadrant.
So,cos𝑥 = −√15
4
sin2𝑥
2=
1 − cos𝑥
2[∵ cos2𝐴 = 1 − 2sin2𝐴]
sin2𝑥
2=
1 +√15
4
2
sin2𝑥
2=
4 + √15
8
sin𝑥
2= √(
4+√15
8) [sin
𝑥
2is positive as
𝑥
2lies in first quadrant]
cos2 𝑥
2=
1+cos𝑥
2 [∵ cos2𝐴 = 1 − 2sin2𝐴]
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cos2𝑥
2=
1 −√15
4
2
cos2𝑥
2=
4 − √15
8
cos𝑥
2= √(
4−√15
8) [cos
𝑥
2 Is positive as
𝑥
2 lies in first quadrant]
Step3:
Also,tan𝑥
2=
sin𝑥
2
cos𝑥
2
tan𝑥
2=
√(4+√15
8)
√(4−√15
8)
⇒ tan𝑥
2=
√4 + √15
√4 − √15
⇒ √4 + √15
4 − √15×
4 + √15
4 + √15
⇒ √(4 + √15)2
16 − 15
⇒ 4 + √15
Therefore, the values of sin𝑥
2 , 𝑐𝑜𝑠
𝑥
2, tan
𝑥
2 are √(
4+√15
8), √(
4−√15
8) and 4 + √15.
Overall Hint: First find cos 𝑥 using the formula sin2𝐴 + cos2𝐴 = 1 then proceed accordingly