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EXERCISE 3.
1. Find the radian measures corresponding to the following degree measures: ( i ) 25° ( ii ) – 47°30 ′′′′′ ( iii ) 240° ( iv ) 520°.
Sol. Since 180° = π radians ∴ 1° =
π radians
( i ) 25° = 25 ×
π radians =
π radians
∴ Radian measure of 25° is
π .
( ii ) – 47°30′ = – 47
×
π radians
^
^ °^ °
′ =^ =
π (^) radians
∴ Radian measure of – 47°30′ is –
π .
( iii ) 240° = 240 ×
π radians =
π radians
∴ Radian measure of 240° is
π .
( iv ) 520° = 520 ×
π radians =
π radians
∴ Radian measure of 520° is
π .
2. Find the degree measures corresponding to the
following radian measure: ^
π
( i )
( ii ) – 4 ( iii ) π ( iv )
π .
Sol. Since π radians = 180° ∴ 1 radian =
π
( i )
radians =
×
π
^ × ^
×
D
Ans. 39°, 22', 30''
( ii ) – 4 radians = – 4 × ° π
− ×
D
θ =
radians
radians =
×
π
= ^ ^
× ×
D
Ans. 12°, 36'
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.
Sol. Radius of circle =
× 40 cm = 20 cm Chord AB = 20 cm ∴ In triangle OAB, OA = OB = AB ⇒ Triangle is equilateral ⇒ Each angle = 60° Let arc AB = l cm.
Now, θ = ∠AOB = 60° = 60 ×
π
π radians r = 20 cm
∴ l = r θ = 20 × π cm [ l and r have same units]
π cm.
6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
20 c
m
(^20) (^) cm
(^20) c m
O
A
B
Sol. Let the radii of the two circles be r 1 and r 2 respectively. Also, let the length of arc in each case be l. [∵Arcs are of same length (given)]
For the first circle, θ = 60° = 60 ×
π
π radians We know that l = r θ
∴ r = (^) θ
⇒ r 1 = π
π
(^) ...( i )
For the second circle, θ = 75° = 75 ×
π (^) =
π (^) radians.
∴ r 2 = θ
π
π
(^) ...( ii )
Dividing ( i ) by ( ii ), we get
π
(^) × π
∴ r 1 : r 2 = 5 : 4.
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 ) 15 cm ( iii ) 21 cm. Sol. [If one end of an inelastic string is attached to a fixed point and to the other end is attached a heavy particle (called bob ), then this system is called a pendulum. Length of the string is called the length of the pendulum.] Here, r = 75 cm. ( i ) l = 10 cm
∴ θ =
radians. ( ii ) l = 15 cm
∴ θ =
radians. ( iii ) l = 21 cm
∴ θ =
radians.
⇒ cos x = ± (^) Since x lies in second quadrant, cos x will be negative.
∴ cos x = – ...( ii )
and sec x =
...( iii )
Also, tan x =
= – ...( iv )
and cot x =
= –. ...( v )
3. cot x = , x lies in third quadrant.
Sol. Given, cot x = , x lies in third quadrant.
∴ tan x =
= ...( i )
Now, cosec 2 x = 1 + cot 2 x = 1 +
⇒ cosec x = ±
Since x lies in third quadrant, cosec x will be negative.
∴ cosec x = –
...( ii )
and sin x =
...( iii )
Also, cos x = cos sin
x x
. sin x = cot x sin x
= –^ ...( iv )
and sec x =
. ...( v )
4. sec x = ^ , x lies in fourth quadrant.
Sol. Given, sec x =
, x lies in fourth quadrant.
∴ cos x =
=^
...( i )
Now, sin 2 x = 1 – cos 2 x = 1 –
⇒ sin x = ±
Since x lies in fourth quadrant, sin x will be negative. ∴ sin x = –
...( ii )
and cosec x =
...( iii )
Also, tan x =
...( iv )
and cot x =
. ...( v ) 5. tan x = –
, x lies in second quadrant.
Sol. Given, tan x = –
, x lies in second quadrant.
∴ cot x =
...( i )
Now, sec 2 x = 1 + tan 2 x = 1 +
=^
⇒ sec x = ±
Since x lies in second quadrant, sec x will be negative.
∴ sec x = –
...( ii )
and cos x =
= –^
...( iii )
EXERCISE 3.
1. Prove that: sin 2
π (^) + cos 2
π (^) – tan 2
π (^) = –
Sol. We know that sin
π =
, cos
π =
, tan
π (^) = 1
∴ L.H.S. =
^
^
– 1^
= R.H.S.
2. Prove that: 2 sin 2
π (^) + cosec 2
π (^) cos 2
π (^) =
Sol. We know that sin
π
cosec
π = cosec
π + π = cosec^
(^) π + π
= – cosec
π = – 2, cos
π
∴ L.H.S. = 2
^
+ (– 2)^
2
^
= R.H.S.
3. Prove that: cot 2
π (^) + cosec
π (^) + 3 tan 2
π (^) = 6.
Sol. We know that cot
π = (^)
cosec
π = cosec
π π = cosec^
(^) π − π
= cosec
π = 2, tan
π
∴ L.H.S. = ( ) 2 + 2 + 3 ^
= 6 = R.H.S.
4. Prove that: 2 sin 2 π^ + 2 cos 2 π^ + 2 sec 2 π^ = 10.
Sol. We know that sin π = sin 4 4
π − π = sin^
π π −
= sin π =
cos π =
, sec
π = 2
∴ L.H.S. = 2
= 1 + 1 + 8 = 10 = R.H.S.
5. Find the value of: ( i ) sin 75° ( ii ) tan 15°. Sol. ( 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°) =
tan – tan^ tan 1 tan tan
∵ x y x^ y x y
^
+ ×
Rationalising
=
×
III IV
Sol. L.H.S. = ^ ^
π + − π − ^ π+
II II
=
= cot 2 x = R.H.S.
9. Prove that:
cos ^
π (^) cos (2 πππππ + x )
π (^) π = 1.
Sol. L.H.S. = cos
π +
cos (2π + x )
IV I
π− + π + ^ III I = sin x cos x [tan x + cot x ]
= sin x cos x
= sin x cos x
^
= 1 = R.H.S.
10. Prove that: sin ( n + 1) x sin ( n + 2) x + cos ( n + 1) x cos ( n + 2) x = cos x****. Sol. L.H.S. = sin ( n + 1) x sin ( n + 2) x + cos ( n + 1) x cos ( n + 2) x = Put ( n + 1) x = A and ( n + 2) x = B ∴ L.H.S. = sin A sin B + cos A cos B = cos A cos B + sin A sin B = cos (A–B) = cos [( n + 1) x – ( n + 2) x ] = cos ( nx + x – nx – 2 x ) = cos (– x ) = cos x = R.H.S. IV 11. Prove that:
cos ^
π (^) – cos
π (^) = – sin x****.
Sol. L.H.S. = cos
π + – cos^
π −
= cos
π cos x – sin
π sin x –
cos cos sin sin 4 4 x x π^ + π [... cos (A + B) = cos A cos B – sin A sin B and cos (A – B) = cos A cos B + sin A sin B]
= cos
π cos x – sin
π sin x – cos
π cos x – sin
π sin x
= –2 sin
π sin x = –2 sin
π^ π sin^ x
= –2 sin –^4
(^) π^ π sin x = –2 sin (^4)
π sin x = –2.
2 sin^ x = – 2 sin x
12. Prove that: sin 2 6 x – sin 2 4 x = sin 2 x sin 10 x****.
Sol. L.H.S. = sin 2 6 x – sin 2 4 x
= sin (6 x + 4 x ) sin (6 x – 4 x ) [...^ sin 2 A – sin 2 B = sin (A + B) sin (A – B)] = sin 10 x sin 2 x = R.H.S.
13. Prove that: cos 2 2 x – cos 2 6 x = sin 4 x sin 8 x****.
Sol. L.H.S. = cos 2 2 x – cos 2 6 x [ Remember: cos 2 A – cos 2 B ≠ cos (A + B) cos (A – B)] = (1 – sin 2 2 x ) – (1 – sin 2 6 x ) [... cos 2 θ = 1–sin 2 θ] = sin 2 6 x – sin 2 2 x = sin (6 x + 2 x ) sin (6 x – 2 x ) [...^ sin^2 A – sin^2 B = sin (A + B) sin (A – B)] = sin 8 x sin 4 x = R.H.S.
14. Prove that: sin 2 x + 2 sin 4 x + sin 6 x = 4 cos^2 x sin 4 x****.
Sol. L.H.S. = (sin 6 x + sin 2 x ) + 2 sin 4 x
= 2 sin ^
+ cos ^
^ ^
^ ∵
= 2 sin 4 x cos 2 x + 2 sin 4 x
17. Prove that: ^
= tan 4 x****.
Sol. L.H.S. =
^ ^ ^
^ ^ ^
[Using C – D formulae]
sin 4 cos 4
x x = tan 4 x^ = R.H.S.
18. Prove that: ^
= tan ^ ^.
Sol. L.H.S. =
[Using C – D formulae]
Cancelling (^) 2 cos 2
x + y ,
− = tan^
− = R.H.S.
19. Prove that: ^
= tan 2 x****.
Sol. L.H.S. = ^
^
[Using C – D formulae]
= tan 2 x = R.H.S.
20. Prove that:^ ^
= 2 sin x****.
Sol. L.H.S. = (^)
^
[Using C – D formulae]
[... cos 2 x – sin 2 x = cos 2 x ]
=
= 2 sin^ x^ = R.H.S.
21. Prove that:
= cot 3 x****.
Sol. L.H.S. =
(... 4 x + 2 x = 2 × 3 x )
^
[Using C – D formulae]
=
Taking cos 3 x common both from numerator and denominator. = ^
= cot 3 x = R.H.S.
22. Prove that: cot x cot 2 x – cot 2 x cot 3 x – cot 3 x cot x = 1. Sol. We know that 3 x = 2 x + x ∴ cot 3 x = cot (2 x + x )
⇒ cot 3 x =
^ ^
25. Prove that: cos 6 x = 32 cos^6 x – 48 cos^4 x + 18 cos^2 x – 1. Sol. Since cos 2A = 2 cos 2 A – 1 Put A = 3 x , ∴ cos 6 x = 2 cos 2 3 x – 1 = 2(4 cos 3 x – 3 cos x ) 2 – 1 = 2(16 cos 6 x – 24 cos 4 x + 9 cos 2 x ) – 1 = 32 cos 6 x – 48 cos 4 x + 18 cos 2 x – 1.
EXERCISE 3.
Note : For general solutions, it is customary to use the following relations:
- sin α =α =α =α =α = sin ( π +π +π +π +π + ααααα ), – cos α =α = α =α =α = cos ( πππππ – ααααα **) and
- tan** αα ααα ===== tan ( πππππ – ααα αα **) Find the principal and general solutions of the following equations:
- tan** x = 2. sec x **= 2
- cot** x = – 4. cosec x = – 2 Sol. 1. tan x = ( i ) Since tan x is positive, principal solutions ( i.e. values of x between 0° and 360° = 2 πππππ ) are in first and third quadrants. Now, tan x = III
= tan π = tan π π +
, i.e., tan
π
∴ Principal solutions are π , π .
( ii ) Since tan x = = tan π ,
∴ x = n π + π , n ∈ Z is the general solution.
2. sec x = 2 ⇒ cos x = ( i ) Since cos x is positive, principal solutions are in first and fourth quadrants.
