RBSE Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

RBSE Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Rajasthan Board RBSE Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1

Question 1.
Find the radian measures corresponding to the following degree measures
(i) 25°
(ii) -47°30′
(iii) 520°
Solution :

Question 2.
Find the degree measures corresponding to the following radian measures (useπ=22/7)
(i) 11/16
(ii) -4
(iii) 5π/3
Solution:

Question 3.
A wheel makes 360 revolutions in 1 minute than how many radians does it turn in one second ?
Solution:
∵ The wheel revolves 360 times in 1 minute, So, in 60 second there are 360 revolution, then
In 1 Second = 360° / 60 = 6 revolution
Angle make in 1 revolution = 2π radian
then angle make in 6 revolution
= 6  x 2π radian
= 12π radian
Hence, angle make in 1 second by the wheel
= 12π radian

Question 4.
Find the degree measure of the angle sub¬tended at the centre of a circle of radius 100 cm by an are of length 22 cm (useπ=22/7)
Solution :
Radius of circle ( r ) = 100 cm
length of arc ( l ) – 22 cm

Hence, the angle subtended at the centre of the circle is = 12°36′

Question 5.
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of corresponding chord.
Solution:
Diameter of circle = 40 cm
Radius of circle (r) = 20 cm
Let AB is a chord of the circle, whose length is 20 cm.

After joining A and B to centre of circle O.
We get an equilateral triangle ΔOAB
Hence, angle at the centre

Hence, the length of minor arc of corresponding chord is
20π/3 cm or 20.95 cm

Question 6.
If in two circles, arcs of the same length subtend angles of 60° and 75° at the centre, find the ratio of their radii.
Solution:
Let radii of circle r₁ and r₂.

then angle substened by an arc at the centre of first circle is
θ = 60° = π/3 radian
Angle subtended by an arc at the centre of second circle is
= 75° = 75π/180 = 5π/12
From fomula : Length of arc ( l )
= radius (r) x angle (θ)
∴ Length of arc of first circle = π/3 x r₁
Length of arc of second circle = 5π/12 x r₂
Given that: Arcs of two circles are of same length

Question 7.
Find the angle in radian through which a pendulum swings, if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 21 cm
Solution:
(i) Length of pendulum ( r ) = 75 cm
Length of arc ( l ) = 10 cm
Let pendulum makes an angle θ

Hence, angle make by swings pendulum = 2/15 radian
(ii) Length of pendulum ( r ) = 75 cm
length of arc ( l ) = 21 cm
Let pendulum makes an angle θ
θ = l/r = 21/5 = 7/25 radian
Hence, angle makes by swings pendulum
= 7/25

Rajasthan Board RBSE Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2

Question 1.
cos x = – 1/2, x lies in third quadrant.
Solution :
According to question, cos x = 1/2 and x lies in IIIrd quadrant

If x lies in IIIrd quadrant 180° ≤ x ≤ 270°, (π ≤ x ≤ 3/2 π)
Hence, in it tan and cot are only in positive and other trigonometric functions are in negative.

Question 2.
cot x = 3/4, x lies in third quadrant.
Solution:

According to rule ASTC all the trigonometric functions are in negative in IIIrd quadrant except tan and cot.

Question 3.
sec x = 13/5, x lies in fourth quadrant.
Solution:

Question 4.
Find the value of sin 765°
Solution :
we know that values of sin x repeats after an interval of 2π
So, sin (765°) = sin (720° + 45°)
= sin (2 x 360° + 45°)
= sin 45°
(∵ After repetition it wil comes in 1st quardrant)
= 1/√2
Hence, sin 765° = 1/√2

Question 5.
Find the value of tan 19π/3
Solution:
We know that values of tan x repeats after an interval of Ilnd quardrant so,

(After repetition it will comes in 1st quardrant and all trigonometric ratios are positive in 1st quadrant.)
= √3
Hence, tan = (−19π/3) = √3

Question 6.
Find the value of sin (−11π/3)
Solution:
We know that values of sin x repeats after an interval of 2π.

(So, this angle will be in IV quadrant in which sin is negative.)

Question 7.
Find the value of cot (−15π/4)
Solution:

(So, this angle will be in IV quadrant. In which cot is negative)

Rajasthan Board RBSE Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3

Question 1.
Prove that

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
Find the value of :
(i) sin 75°
(ii) tan 15°
Solution :
(i) sin 75° = sin (45° + 30°)
= sin 45°.cos 30° + cos 45°.sin 30°
(∵ sin (A + B) = sin A cos B + cos A sin B)

Question 6.
Prove that:
tan 225° cot 405° + tan 765° cot 675° = 0.
Solution :
L.H.S.
= tan 225° cot 405° + tan 765° cot 675°
= tan (360° – 135°) cot (360° + 45°) + tan (2 × 360° + 45°) cot (2 × 360° – 45°)
= (- tan 135°) (cot 45°) + (tan 45°) (- cot 45°)
= – tan 135° cot 45° – tan 45° cot 45°
= – tan (180° – 45°) cot 45° – tan 45° cot 45°
= – (- tan 45°) cot 45° – tan 45° cot 45°
= tan 45° cot 45° – tan 45° cot 45° = 0
= R.H.S.
Hence Proved.

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.
sin (n + 1) x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Solution :
L.H.S.
= sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x
= cos (n + 1)x cos (n + 2)x + sin (n + 1)x × sin (n + 2)x
[Formula cos (x – y) – cos x cos y + sin x sin y]
= cos [(n + 1)x – (n + 2)x]
= cos [nx + x – nx – 2x]
= cos (- x) cos x = R.H.S.
Hence Proved.

Question 12.
sin² 6x – sin² 4x = sin 2x sin 10x.
Solution :
L.H.S. = sin² 6x – sin² 4x
= 1/2 [2 sin² 6x – 2sin² 4x] [Multiply and divide by 2]
= 1/2 [(1 – cos 12x) – (1 – cos 8x)]

= sin 10x sin 2x =sin 2x sin 10x = R.H.S.
Second Method: R.H.S.
= sin² 6x – sin² 4x
= (sin 6x + sin 4x) (sin 6x – sin 4x)

= (2 sin 5x cos 5x).(2 sin x cos x)
= sin 10x.sin 2x [∵ sin 2θ = 2 sin θ cos θ]
= sin 2x sin 10x = R.H.S. Hence Proved.

Question 13.
sin 2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x.
Solution :
L.H.S. = sin 2x + 2 sin 4x + sin 6x
= (sin 2x + sin 6x) + 2 sin 4x

Question 14.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution :
L.H.S. = cot 4x (sin 5x + sin 3x)


From (i) and (ii), we get
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x).
L.H.S. = R.H.S.
Hence Proved.

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.
cos 4x = 1 – 8 sin² x cos² x.
Solution :
L.H.S. = cos 4x = cos 2(2x)
(cos θ = 1 – 2 sin² θ)
= 1 – 2 sin² 2x
= 1 – 2 (2 sin x cos x)² [∵ sin 2x = 2 sin x cos x]
= 1 – 2 x 4 sin² x cos² x
= 1 – 8 sin² x cos² x = R.H.S.
Hence Proved.

Question 21.
cos 6x = 32 cos⁶ x- 48 cos⁴ x + 18 cos² x- 1.
Solution :
L.H.S. = cos 6x = cos 3(2x) [∵ cos 3x = 4 cos³ x – 3 cos x]
= 4 cos³ 2x – 3 cos 2x
= cos 2x (4 cos² 2x – 3)
= cos 2x [4(cos 2x)² – 3] [∵ cos 2x = 2 cos² x – 1]
= cos 2x [4(2 cos² x – l)² – 3]
= cos 2x [4(4 cos⁴ x + 1 – 4 cos² x) – 3]
= cos 2x (16 cos⁴ x + 4 – 16 cos² x – 3)
= (2 cos² x – 1) (16 cos⁴ x + 4 – 16 cos² x – 3)
= 32 cos⁶ x + 8 cos² x – 32 cos⁴ x – 6 cos² x – 16 cos⁴ x – 4 + 16 cos² x + 3
= 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1
= R.H.S.  Hence Proved.

Question 22.
[1 + cot θ – sec (θ + π/(2)]
[1 + cot θ + sec (θ + π/(2)] = 2 cot θ
Solution :
L.H.S.
[1 + cot θ – sec (θ + π/(2)]
[1 + cot θ + sec (θ + π/(2)]
= [1 + cot θ + cosec θ] [1+ cot θ – cosec θ]
= (1 + cot)² – cosec²θ
= 1 + cot²θ + 2 cotθ – cosec²θ
= 1 + 2 cot θ – (cosec2θ – cot2θ)
= 1+2 cot θ – 1
= 2 cot θ = R.H.S.
Hence Proved.

Rajasthan Board RBSE Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise

Question 1.
A right angle is:
(A) equal to a radian
(B) equal to 90 degree
(C) equal to 18°
(D) equal to 90 radian
Solution :
(B) equal to 90 degree

Question 2.
Which trigonometric function is positive in third quadrant:
(A) sin θ
(B) tan θ
(C) cos θ
(D) sec θ
Solution:
(B) tan θ

Question 3.
cosec (- θ) is equal to:
(A) sin θ
(B) tan θ
(C) cos θ
(D) -cosec θ
Solution:
(D) -cosec θ

Question 4.
tan (90° – θ) is equal to:
(A) -tan θ
(B) cot θ
(C) tan θ
(D) -cot θ
Solution :
(B) cot θ

Question 5.
If cos θ = 1/2 then value of θ will be:
(A) 2π/3
(B) π/3
(C) –2π/3
(D) 3π/4
Solution:
(A) 2π/3

Question 6.
If n is an even integer, then the value of sin (2nπ ± θ) will be:
(A) ± cos θ
(B) ± tan θ
(C) ± sin θ
(D) ± cot θ
Solution:
(C) ± sin θ

Question 7.
The value of cot 15° will be:
(A) 2 + √3
(B) – 2 + √3
(C) 2 – √3
(D) – 2 – √3
Solution:
(A) 2 + √3

Question 8.

Question 9.
The value of 2 sin 5π12 cos π12 will be:
(A) 1
(B)√3/2
(C) √3/2 -1
(D) √3/2 +1
Solution:
(D) √3/2 +1
2 sin 5π/12 cos π/12
= 2 sin 75° cos 15°
= 2 sin (90° – 15°) cos 15°
= 2 cos 15° cos 15°
= 2 cos² 15°
= cos² × 15° + 1
= cos 30° + 1
= √3/2 +1

Question 10.

Question 11.
If sin A = 3/5 than value sin 2A will be:
(A) 3/5
(B) 5/25
(C) 24/25
(D) 4/5
solution:

Hence,Option (C) is correct.

Question 12.
If sin A = 3/4 than value of sin 3A will be:
(A) 9/16
(B) – 9/16
(C) 9/32
(D) 7/16
Solution:
sin 3A
3 sin A – 4 sin³ A

Hence,Option (A) is correct.

Question 13.
If tan A = 1/5 then value of tan 3A will be:
(A) 47/25
(B) 37/55
(C) 37/11
(D) 47/55
Solution:

Hence,Option (B) is correct.

Question 14.
If A + B = π/4 then value (1 + tan A) (1 + tan B) will be:
(A) 3
(B) 2
(C) 4
(D) 1
Solution:
(1 + tan A) (1 + tan B)
= 1 + tan A + tan B + tan A tan B ….. (i)

From equation (1) and (2)
(1 + tan A) (1 + tan B) = 2.

Question 15.
General value of θ in equation sec² θ = 2 will be:

Solution:

Hence,Option (A) is correct.

Question 16.
Prove that:
(i) cos θ + sin (27θ° + θ) – sin (27θ° – θ)+ cos (18θ° + θ) = θ

Solution:
(i) sin (27θ° + θ) → IV quardant, -ive = -cosθ … (i)
sin (27θ° – θ) → III quardant, -ive = – cos θ …. (ii)
cos (18θ° + θ) → III quardrant, -ive = – cos θ … (iii)
L.H.S. = cos θ + sin (27θ° + θ) – sin (27θ° – θ) + cos (18θ° + 9)
= cos θ – cos θ + cos θ – cos θ [from (i), (ii) and (iii)]
= θ
= R.H.S.
Hence Proved.

= sec (27θ° – θ) sec (θ – 45θ°)+ tan (45θ° + θ) tan (θ – 27θ°)
= – cosec θ sec (θ – 45θ°)+ tan (36θ° + 9θ° + θ) tan (θ – 27θ°)
= – cosec θ sec (45θ° – θ) – tan (9θ° + θ) tan (27θ° – θ)
= – cosec θ sec (36θ° + 9θ° – θ) – (- cot θ) cot θ
= – cosec θ sec (9θ° – θ) + cot² θ
= – cosec θ cosec θ + cot² θ
= – cosec² θ + cot2 θ
= – 1 (∵ 1 + cot² θ = cosec² θ cot2 θ – cosec² θ = – 1)
R.H.S.
Hence Proved.

Question 17.

Solution:

Question 18.

Solution:

Question 19.
If A + B + C – 180°, then prove that:
(i) cos 2A + cos 2B – cos 2C = 1 – 4 sin A sin 5 cos C
(ii) sin A – sin B + sin C = 4 sin A/2 cos B/2 sin C/2
Solution:
(i) L.H.S.
= cos 2A + cos 2B – cos 2C
= (cos 2A + cos 2B) – cos 2C
= 2 cos (A + B) cos (A – B) – cos 2C
= 2 cos (180° – C) cos (A – B) – (2 cos² C – 1)
= -2 cos C cos (A – B) -2 cos² C + 1
= 1 – 2 cos C cos (A – B) -2 cos² C
= 1 – 2 cos C [cos (A – B) + cos C]

Question 20.
If A + B + C = 2π, then prove that cos² B + cos² C – sin² A = 2 cos A cos B cos C.
Solution:
We know that:

= [cos²B + cos (2π + B) cos (A – C)]
= cos²B + cos B cos (A – C)
= cos B (cos B + cos (A – C)
= cos B (cos (2π – (A – C) + cos (A – C) – cos B [cos (A – B) + cos (A + C)]
= cos B [(cos (A + B) + cos (A – C)]
= cos B [2 cos A cos C]
= 2 cos A cos B cos C
Hence Proved

Question 21.
Find the following equation 2 tan θ – cot θ +1 = 0.
Solution:

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