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3.1 Overview
3.1.1 The word ‘trigonometry’ is derived from the Greek words ‘trigon’ and ‘metron’ which means measuring the sides of a triangle. An angle is the amount of rotation of a revolving line with respect to a fixed line. If the rotation is in clockwise direction the angle is negative and it is positive if the rotation is in the anti-clockwise direction. Usually we follow two types of conventions for measuring angles, i.e., (i) Sexagesimal system (ii) Circular system. In sexagesimal system, the unit of measurement is degree. If the rotation from the
initial to terminal side is
th
360 of a revolution, the angle is said to have a measure of
1°. The classifications in this system are as follows: 1° = 60′ 1 ′ = 60″ In circular system of measurement, the unit of measurement is radian. One radian is the angle subtended, at the centre of a circle, by an arc equal in length to the radius of the circle. The length s of an arc PQ of a circle of radius r is given by s = r θ, where θ is the angle subtended by the arc PQ at the centre of the circle measured in terms of radians.
3.1.2 Relation between degree and radian The circumference of a circle always bears a constant ratio to its diameter. This constant
ratio is a number denoted by π which is taken approximately as
for all practical
purpose. The relationship between degree and radian measurements is as follows: 2 right angle = 180° = π radians
1 radian =
= 57°16′ (approx)
radian = 0.01746 radians (approx)
Chapter 3
TRIGONOMETRIC FUNCTIONS
3.1.3 Trigonometric functions Trigonometric ratios are defined for acute angles as the ratio of the sides of a right angled triangle. The extension of trigonometric ratios to any angle in terms of radian measure (real numbers) are called trigonometric functions. The signs of trigonometric functions in different quadrants have been given in the following table:
I II III IV sin x + + – – cos x + – – + tan x + – + – cosec x + + – – sec x + – – + cot x + – + –
3.1.4 Domain and range of trigonometric functions
Functions Domain Range sine R [–1, 1] cosine R [–1, 1]
tan R – {(2 n + 1)
: n ∈ Z } R
cot R – { n π : n ∈ Z } R
sec R – {(2 n + 1)
: n ∈ Z } R – (–1, 1)
cosec R – { n π : n ∈ Z } R – (–1, 1)
3.1.5 Sine, cosine and tangent of some angles less than 90°
0° 15° 18° 30° 36° 45° 60° 90°
sine 0 6 2
TRIGONOMETRIC FUNCTIONS 35
TRIGONOMETRIC FUNCTIONS 37
(vii) cot (A + B) =
cot A cot B 1 cot A cot B
(viii) cot (A – B) = cot A cot B 1 cotB cot A
(ix) sin 2A = 2 sin A cos A = (^2) 2 tan A 1 tan A
(x) cos 2A = cos^2 A – sin 2 A = 1 – 2 sin 2 A = 2 cos 2 A – 1 =
2 2
1 – tan A 1+ tan A
(xi) tan 2A = (^2)
2 tan A 1 – tan A (xii) sin 3A = 3sin A – 4sin 3 A (xiii) cos 3A = 4cos 3 A – 3cos A
(xiv) tan 3A =
3 2
3 tanA – tan A 1 – 3tan A
(xv) cos A + cos B =
A + B A – B
2 cos cos 2 2
(xvi) cos A – cos B =
A + B B – A
2sin sin 2 2
(xvii) sin A + sin B = 2sin A^ B^ cosA^ B 2 2
(xviii) sin A – sin B = 2cos A^ B^ sinA^ B 2 2
(xix) 2sin A cos B = sin (A + B) + sin (A – B) (xx) 2cos A sin B = sin (A + B) – sin (A – B) (xxi) 2cos A cos B = cos (A + B) + cos (A – B) (xxii) 2sin A sin B = cos (A – B) – cos (A + B)
(xxiii)
if Alies in quadrants I or II sin A^1 cos A^2 (^2 2) – if Alies in III or IV quadrants 2
38 EXEMPLAR PROBLEMS – MATHEMATICS
(xxiv)
if Alies in I or IV quadrants cos A^1 cos A^2 (^2 2) – if Alies in II or III quadrants 2
(xxv)
if Alies in I or III quadrants tan A^1 cosA^2 2 1 cosA (^) – if Alies in II or IV quadrants 2
Trigonometric functions of an angle of 18° Let θ = 18°. Then 2θ = 90° – 3θ Therefore, sin 2θ = sin (90° – 3θ) = cos 3θ or sin 2θ = 4cos^3 θ – 3cos θ Since, cos θ ≠ 0, we get 2sin θ = 4cos^2 θ – 3 = 1 – 4sin 2 θ or 4sin 2 θ + 2sin θ – 1 = 0.
Hence, sin θ =
Since, θ = 18°, sin θ > 0, therefore, sin 18° =
Also, cos18° = 2 1 sin 18 1 6 2 5 10 2 5 16 4
^
Now, we can easily find cos 36° and sin 36° as follows:
cos 36° = 1 – 2sin 2 18° =
Hence, cos 36° = 5 1 4
Also, sin 36° = 1 cos 2 36 1 6 2 5 16
3.1.9 Trigonometric equations Equations involving trigonometric functions of a variables are called trigonometric equations. Equations are called identities, if they are satisfied by all values of the
40 EXEMPLAR PROBLEMS – MATHEMATICS
Solution Given that circular wire is of radius 3 cm, so when it is cut then its length = 2π × 3 = 6π cm. Again, it is being placed along a circular hoop of radius 48 cm. Here, s = 6π cm is the length of arc and r = 48 cm is the radius of the circle. Therefore, the angle θ, in radian, subtended by the arc at the centre of the circle is given by
θ = Arc^6 22. Radius 48 8
^ .
Example 2 If A = cos 2 θ + sin^4 θ for all values of θ, then prove that
≤ A ≤ 1.
Solution We have A =cos^2 θ + sin^4 θ = cos^2 θ + sin^2 θ sin 2 θ ≤ cos 2 θ + sin^2 θ Therefore, A ≤ 1 Also, A = cos 2 θ + sin 4 θ = (1 – sin^2 θ) + sin^4 θ
2 sin 2 1 1 1 2 4
2 sin 2 1 3 3 2 4 4
Hence,
A 1
Example 3 Find the value of 3 cosec 20° – sec 20°
Solution We have
3 cosec 20° – sec 20° =^
sin 20 cos 20
3 cos 20 – sin 20 sin 20 cos 20
(^3) cos 20 – 1 sin 20 4 2 2 2 sin 20 cos 20
^
^
4 sin 60 cos 20 – cos 60 sin 20 sin 40
^ ^ ^
(Why?)
= 4 sin (60^ – 20 ) sin 40
^
(^) = 4^ (Why?)
TRIGONOMETRIC FUNCTIONS 41
Example 4 If θ lies in the second quadrant, then show that
1 sin 1 sin 2sec 1 sin 1 sin
Solution We have
1 sin 1 sin 1 sin 1 sin
1 sin 1 sin 1 sin 1 sin
cos
=
| cos | (Since^ ^2 = |α| for every real number^ α) Given that θ lies in the second quadrant so |cos θ| = – cos θ (since cos θ < 0).
Hence, the required value of the expression is
cos = –2 secθ Example 5 Find the value of tan 9° – tan 27° – tan 63° + tan 81° Solution We have tan 9° – tan 27° – tan 63° + tan 81° = tan 9° + tan 81° – tan 27° – tan 63° = tan 9° + tan (90° – 9°) – tan 27° – tan (90° – 27°) = tan 9° + cot 9° – (tan 27° + cot 27°) (1)
Also tan 9° + cot 9° =
sin 9 cos9 sin
(Why?)^ (2)
Similarly, tan 27° + cot 27° =
sin 27 cos 27
sin 54 cos
(Why?) (3)
Using (2) and (3) in (1), we get
tan 9° – tan 27° – tan 63° + tan 81° =
sin 18 cos 36 (^5 1 5 )
Example 6 Prove that sec8 1 tan sec 4 1 tan 2
Solution We have
sec8 1 sec4 1
=^
(1 cos8 ) cos cos8 (1 – cos 4 )
2 sin 4^2 cos 4 cos8 2sin^2
(Why?)
TRIGONOMETRIC FUNCTIONS 43
If we take tan x =
3 , then^ x^ =^
or^7 6 6
(Why?)
Again, if we take tan x =
(^1) , then 5 or 11 3 6 6
(^) x (Why?)
Therefore, the possible solutions of above equations are
x = 6
and
where 0 ≤ x ≤ 2 π
Long Answer Type
Example 9 Find the value of
1 cos 1 cos 1 cos 1 cos 8 8 8 8
Solution Write
1 cos 1 cos 1 cos 1 cos 8 8 8 8
1 cos 1 cos 1 cos 1 cos 8 8 8 8
^ ^
1 cos 1 cos 8 8
(Why?)
sin sin 8 8
1 cos 1 cos 4 4 4
(Why?)
1 cos 1 cos 4 4 4
(Why?)
1 cos 4 4
Example 10 If x cos θ = y cos (θ +
) = z cos ( θ +
), then find the value of
xy + yz + zx.
44 EXEMPLAR PROBLEMS – MATHEMATICS
Solution Note that xy + yz + zx = xyz^1 1 x y z
If we put x cos θ = y cos (θ +
) = z cos
= k (say).
Then x = cos
k ,^ y^ =^2 cos 3
k
^
and z = cos 4 3
k
^
so that
x y z
cos cos cos k 3 3
^ ^ ^
[cos cos cos sin sin k 3 3
cos cos sin sin 3 3
]
[cos cos ( ) k 2 2
sin cos sin ] 2 2
(^) (Why?)
k
Hence, xy + yz + zx = 0 Example 11 If α and β are the solutions of the equation a tan θ + b sec θ = c ,
then show that tan (α + β) = (^2 )
2 ac a c
Solution Given that a tanθ + b secθ = c or a sinθ + b = c cos θ Using the identities,
sin θ =
2
2 2
2 tan 1 tan (^2) and cos 2 1 tan 1 tan 2 2
^
46 EXEMPLAR PROBLEMS – MATHEMATICS
⇒ ( a tanθ – c ) 2 = b^2 (1 + tan^2 θ) ⇒ a^2 tan^2 θ – 2 ac tanθ + c^2 = b^2 + b^2 tan^2 θ ⇒ ( a^2 – b^2 ) tan^2 θ – 2 ac tanθ + c^2 – b^2 = 0 ... (1) Since α and β are the roots of the equation (1), so
tanα + tanβ = (^2 )
2 ac a b
and tanα tanβ =
2 2 2 2
c b a b
Therefore, tan (α + β) = tan tan 1 tan tan
2 2 2 2 2 2
2 ac a b c b a b
2 ac a c
Example 12 Show that 2 sin^2 β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α
Solution LHS = 2 sin 2 β + 4 cos (α + β) sin α sin β + cos 2(α + β) = 2 sin^2 β + 4 (cos α cos β – sin α sin β) sin α sin β
- (cos 2α cos 2β – sin 2α sin 2β) = 2 sin^2 β + 4 sin α cos α sin β cos β – 4 sin^2 α sin^2 β
- cos 2α cos 2β – sin 2α sin 2β = 2 sin^2 β + sin 2α sin 2β – 4 sin^2 α sin^2 β + cos 2α cos 2β – sin 2 α sin 2β = (1 – cos 2β) – (2 sin 2 α) (2 sin^2 β) + cos 2α cos 2β (Why?) = (1 – cos 2β) – (1 – cos 2α) (1 – cos 2β) + cos 2α cos 2β = cos 2α (Why?)
Example 13 If angle θ is divided into two parts such that the tangent of one part is k times the tangent of other, and φ is their difference, then show that
sin θ =
k k
sin φ
Solution Let θ = α + β. Then tan α = k tan β
TRIGONOMETRIC FUNCTIONS 47
or
tan tan
k
Applying componendo and dividendo, we have
tan tan tan tan
=^
k k
or sin cos cos sin sin cos cos sin
k k
(Why?)
i.e., sin ( ) sin ( )
k k
(Why?)
Given that α – β = φ and α + β = θ. Therefore,
sin sin
=^
k + k – or sin θ =
k k
sin φ
Example 14 Solve 3 cos θ + sin θ = (^2)
Solution Divide the given equation by 2 to get
(^3) cos 1 sin 1 (^2 2 )
(^) or cos cos sin sin cos 6 6 4
or cos^6 cos^4 or cos^6 cos 4
(Why?)
Thus, the solution are given by, i.e., θ = 2 m π ±
Hence, the solution are
θ = 2 m π + 4 6
(^) and 2 m π – 4 6
(^) , i.e., θ = 2 m π +^5 12
and θ = 2 m π – 12
Objective Type Questions
Choose the correct answer from the given four options against each of the Examples 15 to 19
Example 15 If tan θ =
, then sinθ is
TRIGONOMETRIC FUNCTIONS 49
Solution Correct choice is (C). Indeed sin 20° sin 40° sin 60° sin 80°.
3 2
(^) sin 20° sin (60° – 20°) sin (60° + 20°) (since sin 60° = 3 2
(^) sin 20° [sin^2 60° – sin 2 20°] (Why?)
3 2
(^) sin 20° [
4 – sin
2 20°]
(^) [3sin 20° – 4sin^3 20°]
3 1 2 4
(sin 60°) (Why?)
3 1 3 2 4 2
Example 19 The value of cos 5
cos
cos
cos
is
(A)
16 (B)^0 (C)^
(D)
Solution (D) is the correct answer. We have
cos 5
cos
cos
cos
2 sin cos cos cos cos 2sin^5 5 5 5 5
sin cos cos cos 2sin^5 5 5 5
(^) (Why?)
sin cos cos 4sin^5 5 5
(^) (Why?)
50 EXEMPLAR PROBLEMS – MATHEMATICS
sin cos 8 sin^5 5
(Why?)
sin 16 5 16 sin 5
sin 3 5 16 sin 5
^
sin 5 16 sin 5
^
(Why?)
Fill in the blank :
Example 20 If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = _________
Solution Given that 3 tan (θ – 15°) = tan (θ + 15°) which can be rewritten as
tan( 15°) 3 tan( 15°) 1
Applying componendo and Dividendo; we get
tan ( 15°) + tan ( – 15°) (^2) tan ( 15°) tan ( –15°)
sin ( 15°) cos ( 15°) + sin ( 15°) cos ( 15°) 2 sin ( 15°) cos ( 15°) sin ( 15°) cos( 15°)
sin 2 2 sin 30
i.e., sin 2θ = 1 (Why?)
giving 4
State whether the following statement is True or False. Justify your answer
Example 21 “The inequality 2 sinθ^ + 2cosθ^ ≥ 1 1 2 ^2 holds for all real values of^ θ”
52 EXEMPLAR PROBLEMS – MATHEMATICS
(b)
1 cos 1 cos
x x
2 2 2
2sin (^2) cot (^2) 2sin (^2)
x x x ^.^ Hence (b) matches with (i) i.e., (b)^ ↔^ (i)
(c)
1 cos sin
x x
2 cos 2 (^2) cot 2sin cos^2 2 2
x x x x
Hence (c) matches with (ii) i.e., (c) ↔ (ii)
(d) 1 sin 2 x = sin 2 x cos 2 x 2 sin x cos x
= (^) (sin x cos x )^2
= (^) sin x cos x . Hence (d) matches with (iii), i.e., (d) ↔ (iii)
3.3 EXERCISE
Short Answer Type
1. Prove that
tan A secA – 1 1 sin A tan A secA 1 cos A
2. If 2 sin 1 cos sin
y
, then prove that^1 cos^ sin 1 sin
is also equal to y.
1 cos sin 1 cos sin 1 cos sin :Express. 1 sin 1 sin 1 cos sin
Hint
3. If m sin θ = n sin (θ + 2α), then prove that tan (θ + α) cot α =
m n m n
[ Hint: Express sin ( 2 ) sin
m n
and apply componendo and dividendo]
4. If cos (α + β) =
5 and sin (α^ –^ β) =^
13 , where^ α^ lie between 0 and 4
, find the value of tan2α [ Hint: Express tan 2 α as tan (α + β + α – β]
TRIGONOMETRIC FUNCTIONS 53
5. If tan x = b a , then find the value of
a b a b a b a b
6. Prove that cosθ cos (^2)
(^) = sin 7θ sin 8θ.
[ Hint: Express L.H.S. =
2 [2cosθ^ cos 2
]
7. If a cos θ + b sin θ = m and a sin θ – b cos θ = n , then show that a^2 + b^2 = m^2 + n^2
8. Find the value of tan 22°30 ^.
[ Hint: Let θ = 45°, use
sin 2 sin cos (^) sin tan^2 2 (^2) cos 2 cos 2 1 cos 2 2
]
9. Prove that sin 4A = 4sinA cos 3 A – 4 cosA sin^3 A. 10. If tanθ + sinθ = m and tanθ – sinθ = n , then prove that m^2 – n^2 = 4sinθ tanθ [ Hint: m + n = 2tanθ, m – n = 2 sinθ, then use m^2 – n^2 = ( m + n ) ( m – n )] 11. If tan (A + B) = p , tan (A – B) = q , then show that tan 2 A = (^1)
p q pq
[ Hint: Use 2A = (A + B) + (A – B)]
12. If cosα + cosβ = 0 = sinα + sinβ, then prove that cos 2α + cos 2β = – 2cos (α + β). [ Hint: (cosα + cosβ) 2 – (sinα + sinβ) 2 = 0] 13. If
sin ( ) sin ( )
x y a b x y a b
, then show that
tan tan
x a y b
(^) [ Hint: Use Componendo and
Dividendo].
14. If tanθ =
sin cos sin cos
, then show that sinα^ + cosα^ =^2 cosθ.
[ Hint: Express tanθ = tan (α – (^4)
]
15. If sinθ + cosθ = 1, then find the general value of θ. 16. Find the most general value of θ satisfying the equation tanθ = –1 and
cosθ =
