NCERT Solutions for Class 11 Science Math Chapter 3 Trigonometric Functions are provided here with simple step-by-step explanations. These solutions for Trigonometric Functions are extremely popular among Class 11 Science students for Math Trigonometric Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 11 Science Math are prepared by experts and are 100% accurate.
Page No 54:
Question 1:
Find the radian measures corresponding to the following degree measures:
(i) 25° (ii) – 47° 30' (iii) 240° (iv) 520°
Answer:
(i) 25°
We know that 180° = π radian
(ii) –47° 30'
–47° 30' =
degree [1° = 60']
degree
Since 180° = π radian
(iii) 240°
We know that 180° = π radian
(iv) 520°
We know that 180° = π radian
Page No 55:
Question 2:
Find the degree measures corresponding to the following radian measures
.
(i) (ii) –
4 (iii) (iv)
Answer:
(i)
We know that π radian = 180°
(ii) – 4
We know that π radian = 180°
(iii)
We know that π radian = 180°
(iv)
We know that π radian = 180°
Page No 55:
Question 3:
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Answer:
Number of revolutions made by the wheel in 1 minute = 360
∴Number
of revolutions made by the wheel in 1 second =
In one complete revolution, the wheel turns an angle of 2π radian.
Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e.,
12 π radian
Thus, in one second, the wheel turns an angle of 12π radian.
Page No 55:
Question 4:
Find the
degree measure of the angle subtended at the centre of a circle of
radius 100 cm by an arc of length 22 cm.
Answer:
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then
Therefore, forr = 100 cm, l = 22 cm, we have
Thus, the required angle is 12°36′.
Page No 55:
Question 5:
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Answer:
Diameter of the circle = 40 cm
∴Radius (r) of the circle =
Let AB be a chord (length = 20 cm) of the circle.
In ΔOAB, OA = OB = Radius of circle = 20 cm
Also, AB = 20 cm
Thus, ΔOAB is an equilateral triangle.
∴θ = 60° =
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then.
Thus, the length of the minor arc of the chord is.
Page No 55:
Question 6:
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Answer:
Let
the radii of the two circles be
and.
Let an arc of length l
subtend an angle of 60° at the centre of the circle of radius r1,
while let an arc of length l
subtend
an angle of 75° at the centre of the circle of radius r2.
Now,
60° =and
75° =
We
know that in a circle of radius r
unit, if an arc of length l
unit subtends an angle θ
radian at the centre, then.
Thus, the ratio of the radii is 5:4.
Page No 55:
Question 7:
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) 10 cm (ii) 15 cm (iii) 21 cm
Answer:
We
know that in a circle of radius r
unit, if an arc of length l
unit subtends an angle θ
radian at the centre, then.
It is given that r = 75 cm
(i) Here, l = 10 cm
(ii) Here, l = 15 cm
(iii) Here, l = 21 cm
Page No 63:
Question 1:
Find the values of other five trigonometric functions if , x lies in third quadrant.
Answer:
Since x lies in the 3rd quadrant, the value of sin x will be negative.
Page No 63:
Question 2:
Find the values of other five trigonometric functions if , x lies in second quadrant.
Answer:
Since x lies in the 2nd quadrant, the value of cos x will be negative
Page No 63:
Question 3:
Find the values of other five trigonometric functions if , x lies in third quadrant.
Answer:
Since x lies in the 3rd quadrant, the value of sec x will be negative.
Page No 63:
Question 4:
Find the values of other five trigonometric functions if , x lies in fourth quadrant.
Answer:
Since x lies in the 4th quadrant, the value of sin x will be negative.
Page No 63:
Question 5:
Find the values of other five trigonometric functions if , x lies in second quadrant.
Answer:
Since x lies in the 2nd quadrant, the value of sec x will be negative.
∴sec
x
=
Page No 63:
Question 6:
Find the value of the trigonometric function sin 765°
Answer:
It is known that the values of sin x repeat after an interval of 2π or 360°.
Page No 63:
Question 7:
Find the value of the trigonometric function cosec (–1410°)
Answer:
It is known that the values of cosec x repeat after an interval of 2π or 360°.
Page No 63:
Question 8:
Find the
value of the trigonometric function
Answer:
It is known that the values of tan x repeat after an interval of π or 180°.
Page No 63:
Question 9:
Find the value of the trigonometric function
Answer:
It is known that the values of sin x repeat after an interval of 2π or 360°.
Page No 63:
Question 10:
Find the value of the trigonometric function
Answer:
It is known that the values of cot x repeat after an interval of π or 180°.
Page No 73:
Question 1:
Answer:
L.H.S.
=
Page No 73:
Question 2:
Prove that
Answer:
L.H.S. =
Page No 73:
Question 3:
Prove that
Answer:
L.H.S. =
Page No 73:
Question 4:
Prove that
Answer:
L.H.S =
Page No 73:
Question 5:
Find the value of:
(i) sin 75°
(ii) tan 15°
Answer:
(i) sin 75° = sin (45° + 30°)
= sin 45° cos 30° + cos 45° sin 30°
[sin (x + y) = sin x cos y + cos x sin y]
(ii) tan 15° = tan (45° – 30°)
Page No 73:
Question 6:
Prove
that:
Answer:
Page No 73:
Question 7:
Prove
that:
Answer:
It is
known that
∴L.H.S.
=
Page No 73:
Question 8:
Prove
that
Answer:
Page No 73:
Question 9:
Answer:
L.H.S.
=
Page No 73:
Question 10:
Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Answer:
L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x
Page No 73:
Question 11:
Prove
that
Answer:
It is
known that.
∴L.H.S.
=
Page No 73:
Question 12:
Prove that sin2 6x – sin2 4x = sin 2x sin 10x
Answer:
It is
known that
∴L.H.S. = sin26x – sin24x
= (sin 6x + sin 4x) (sin 6x – sin 4x)
= (2 sin 5x cos x) (2 cos 5x sin x)
= (2 sin 5x cos 5x) (2 sin x cos x)
= sin 10x sin 2x
= R.H.S.
Page No 73:
Question 13:
Prove that cos2 2x – cos2 6x = sin 4x sin 8x
Answer:
It is
known that
∴L.H.S. = cos2 2x – cos2 6x
= (cos 2x + cos 6x) (cos 2x – 6x)
= [2 cos 4x cos 2x] [–2 sin 4x (–sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.
Page No 73:
Question 14:
Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
Answer:
L.H.S. = sin 2x + 2 sin 4x + sin 6x
= [sin 2x + sin 6x] + 2 sin 4x
= 2 sin 4x cos (– 2x) + 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
= 2 sin 4x (2 cos2 x – 1 + 1)
= 2 sin 4x (2 cos2 x)
= 4cos2 x sin 4x
= R.H.S.
Page No 73:
Question 15:
Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Answer:
L.H.S = cot 4x (sin 5x + sin 3x)
= 2 cos 4x cos x
R.H.S. = cot x (sin 5x – sin 3x)
= 2 cos 4x. cos x
L.H.S. = R.H.S.
Page No 73:
Question 16:
Prove that
Answer:
It is known that
∴L.H.S
=
Page No 73:
Question 17:
Prove that
Answer:
It is known that
∴L.H.S.
=
Page No 73:
Question 18:
Prove that
Answer:
It is known that
∴L.H.S.
=
Page No 73:
Question 19:
Prove that
Answer:
It is known that
∴L.H.S.
=
Page No 73:
Question 20:
Prove that
Answer:
It is known that
∴L.H.S.
=
Page No 73:
Question 21:
Prove that
Answer:
L.H.S. =
Page No 74:
Question 22:
Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Answer:
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 x cot 2x – (cot 2x cot x – 1)
= 1 = R.H.S.
Page No 74:
Question 23:
Prove that
Answer:
It is
known that.
∴L.H.S. = tan 4x = tan 2(2x)
Page No 74:
Question 24:
Prove that cos 4x = 1 – 8sin2 x cos2 x
Answer:
L.H.S. = cos 4x
= cos 2(2x)
= 1 – 2 sin2 2x [cos 2A = 1 – 2 sin2 A]
= 1 – 2(2 sin x cos x)2 [sin2A = 2sin A cosA]
= 1 – 8 sin2x cos2x
= R.H.S.
Page No 74:
Question 25:
Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Answer:
L.H.S. = cos 6x
= cos 3(2x)
= 4 cos3 2x – 3 cos 2x [cos 3A = 4 cos3 A – 3 cos A]
= 4 [(2 cos2 x – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 x – 1]
= 4 [(2 cos2 x)3 – (1)3 – 3 (2 cos2 x)2 + 3 (2 cos2 x)] – 6cos2 x + 3
= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3
= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3
= 32 cos6x – 48 cos4x + 18 cos2x – 1
= R.H.S.
Page No 78:
Question 1:
Find the
principal and general solutions of the equation
Answer:
Therefore,
the principal solutions are x =and.
Therefore,
the general solution is
Page No 78:
Question 2:
Find the
principal and general solutions of the equation
Answer:
Therefore,
the principal solutions are x =and.
Therefore,
the general solution is,
where n ∈ Z
Page No 78:
Question 3:
Find the
principal and general solutions of the equation
Answer:
Therefore,
the principal solutions are x =
and.
Therefore,
the general solution is
Page No 78:
Question 4:
Find the general solution of cosec x = –2
Answer:
cosec x = –2
Therefore,
the principal solutions are x
=.
Therefore,
the general solution is
Page No 78:
Question 5:
Find the
general solution of the equation
Answer:
Page No 78:
Question 6:
Find the
general solution of the equation
Answer:
Page No 78:
Question 7:
Find the
general solution of the equation
Answer:
Therefore,
the general solution is.
Page No 78:
Question 8:
Find the
general solution of the equation
Answer:
Therefore,
the general solution is.
Page No 78:
Question 9:
Find the
general solution of the equation
Answer:
Therefore,
the general solution is
Page No 81:
Question 1:
Prove
that:
Answer:
L.H.S.
= 0 = R.H.S
Page No 81:
Question 2:
Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
Answer:
L.H.S.
= (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= RH.S.
Page No 82:
Question 3:
Prove
that:
Answer:
L.H.S. =
Page No 82:
Question 4:
Prove
that:
Answer:
L.H.S. =
Page No 82:
Question 5:
Prove
that:
Answer:
It is
known that.
∴L.H.S.
=
Page No 82:
Question 6:
Prove
that:
Answer:
It is known that
.
L.H.S. =
= tan 6x
= R.H.S.
Page No 82:
Question 7:
Prove
that:
Answer:
L.H.S. =
Page No 82:
Question 8:
,
x
in quadrant II
Answer:
Here, x is in quadrant II.
i.e.,
Therefore, are all positive.
As x is in quadrant II, cosx is negative.
∴
Thus, the respective values of are.
Page No 82:
Question 9:
Find
for
,
x in quadrant III
Answer:
Here, x is in quadrant III.
Therefore,
and
are negative, whereasis
positive.
Now,
Thus, the
respective values of
are.
Page No 82:
Question 10:
Find
for
,
x in quadrant II
Answer:
Here, x is in quadrant II.
Therefore,,
and
are all positive.
[cosx
is negative in quadrant II]
Thus, the
respective values of
are
.
View NCERT Solutions for all chapters of Class 11