Index1. Theory2. Short Revision3. Exercise (1 To 5)4. Assertion & Reason5. Que. from Compt. Exams

Subject : MathematicsTopic: Trigonometric Ratios & Identities

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Trigonometric Ratios& Identities

1. Basic Trigonometric Identities:

(a) sin² θ + cos² θ = 1; −1 ≤ sin θ ≤ 1; −1 ≤ cos θ ≤ 1 ∀ θ ∈ R

(b) sec² θ − tan² θ = 1 ; sec θ ≥ 1 ∀ θ ∈ R – ( )

Ι∈π+ n,

21n2

(c) cosec² θ − cot² θ = 1 ; cosec θ ≥ 1 ∀ θ ∈ R – { }Ι∈π n,n

Solved Example # 1

Prove that(i) cos4A – sin4A + 1 = 2 cos2A

(ii) 1AsecAtan1AsecAtan

+−−+

= AcosAsin1+

Solution(i) cos4A – sin4A + 1

= (cos2A – sin2A) (cos2A + sin2A) + 1= cos2A – sin2A + 1 [∴ cos2A + sin2A = 1]= 2 cos2A

(ii) 1AsecAtan1AsecAtan

+−−+

= 1AsecAtan

)AtanA(secAsecAtan 22

+−−−+

= 1AsecAtan)AtanAsec1)(AsecA(tan

+−+−+

= tan A + sec A = AcosAsin1+

Solved Example # 2

If sin x + sin2x = 1, then find the value ofcos12x + 3 cos10x + 3 cos8x + cos6x – 1

Solution

cos12x + 3 cos10x + 3 cos8x + cos6x – 1= (cos4x + cos2x)3 – 1= (sin2x + sinx)3 – 1 [∵ cos2x = sin x]= 1 – 1 = 0

Solved Example # 3

If tan θ = m – m41

, then show that sec θ – tan θ = – 2m or m21

Solution

Depending on quadrant in which θ falls, sec θ can be ± m4

1m4 2 +

So, if sec θ = m4

1m4 2 + = m + m41

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3 of 2

4

⇒ sec θ – tan θ = m21

and if sec θ = –

+

m41m

⇒ sec θ – tan θ = – 2m

Self Practice Problem

1. Prove the followings :(i) cos6A + sin6A + 3 sin2A cos2A = 1(ii) sec2A + cosec2A = (tan A + cot A)2

(iii) sec2A cosec2A = tan2A + cot2A + 2(iv) (tan α + cosec β)2 – (cot β – sec α)2 = 2 tan α cot β (cosec α + sec β)

(v)

α−α+

α−α 2222 sineccos1

cossec1

cos2α sin2α = αα+αα−

22

22

cossin2cossin1

2. If sin θ = 22

2

n2mn2mmn2m++

+, then prove that tan θ = 2

2

n2mn2mn2m

++

2. Circular Definition Of Trigonometric Functions:

sin θ = OPPM

cos θ = OPOM

tan θ = θθ

cossin

, cos θ ≠ 0

cot θ = θθ

sincos

, sin θ ≠ 0

sec θ = θcos1

, cos θ ≠ 0 cosec θ = θsin1

, sin θ ≠ 0

3. Trigonometric Functions Of Allied Angles:

If θ is any angle, then − θ, 90 ± θ, 180 ± θ, 270 ± θ, 360 ± θ etc. are called ALLIED ANGLES.(a) sin (− θ) = − sin θ ; cos (− θ) = cos θ(b) sin (90° − θ) = cos θ ; cos (90° − θ) = sin θ(c) sin (90° + θ) = cos θ ; cos (90° + θ) = − sin θ(d) sin (180° − θ) = sin θ ; cos (180° − θ) = − cos θ(e) sin (180° + θ) = − sin θ ; cos (180° + θ) = − cos θ(f) sin (270° − θ) = − cos θ ; cos (270° − θ) = − sin θ(g) sin (270° + θ) = − cos θ ; cos (270° + θ) = sin θ(h) tan (90° − θ) = cot θ ; cot (90° − θ) = tan θ

Solved Example # 4

Prove that(i) cot A + tan (180º + A) + tan (90º + A) + tan (360º – A) = 0(ii) sec (270º – A) sec (90º – A) – tan (270º – A) tan (90º + A) + 1 = 0

Solution(i) cot A + tan (180º + A) + tan (90º + A) + tan (360º – A)

= cot A + tan A – cot A – tan A = 0(ii) sec (270º – A) sec (90º – A) – tan (270º – A) tan (90º + A) + 1

= – cosec2A + cot2A + 1 = 0Self Practice Problem3. Prove that

(i) sin 420º cos 390º + cos (–300º) sin (–330º) = 1(ii) tan 225º cot 405º + tan 765º cot 675º = 0

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4 of 2

44. Graphs of Trigonometric functions:

(a) y = sin x x ∈ R; y ∈ [–1, 1]

(b) y = cos x x ∈ R; y ∈ [ – 1, 1]

(c) y = tan x x ∈ R – (2n + 1) π/2, n ∈ Ι ; y ∈ R

(d) y = cot x x ∈ R – nπ , n ∈ Ι ; y ∈ R

(e) y = cosec x x ∈ R – nπ , n ∈ Ι ; y ∈ (− − − − − ∞, −−−−− 1] ∪∪∪∪∪ [1, ∞∞∞∞∞)

(f) y = sec x x ∈ R – (2n + 1) π/2, n ∈ Ι ; y ∈∈∈∈∈ (− ∞, − 1] ∪ [1, ∞)

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5 of 2

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Solved Example # 5

Find number of solutions of the equation cos x = |x|

Solution

Clearly graph of cos x & |x| intersect at two points. Hence no. of solutions is 2

Solved Example # 6

Find range of y = sin2x + 2 sin x + 3 ∀ x ∈ R

SolutionWe know – 1 ≤ sin x ≤ 1⇒ 0 ≤ sin x +1 ≤ 2⇒ 2 ≤ (sin x +1)2 + 2 ≤ 6Hence range is y ∈ [2, 6]

Self Practice Problem

4. Show that the equation sec2θ = 2)yx(xy4

+ is only possible when x = y ≠ 0

5. Find range of the followings.(i) y = 2 sin2x + 5 sin x +1∀ x ∈ R Answer [–2, 8]

(ii) y = cos2x – cos x + 1 ∀ x ∈ R Answer

3,43

6. Find range of y = sin x, x ∈

ππ 2

32

Answer

−

23,1

5. Trigonometric Functions of Sum or Difference of Two Angles:(a) sin (A ± B) = sinA cosB ± cosA sinB(b) cos (A ± B) = cosA cosB ∓ sinA sinB(c) sin²A − sin²B = cos²B − cos²A = sin (A+B). sin (A− B)(d) cos²A − sin²B = cos²B − sin²A = cos (A+B). cos (A − B)

(e) tan (A ± B) = BtanAtan1BtanAtan

∓±

(f) cot (A ± B) = AcotBcot1BcotAcot

±∓

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6 of 2

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(g) tan (A + B + C) = AtanCtanCtanBtanBtanAtan1CtanBtanAtanCtanBtanAtan

−−−−++

.

Solved Example # 7

Prove that(i) sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = cos (A – B)

(ii) tan

θ+π

4 tan

θ+π

43

= –1

Solution

(i) Clearly sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B)= sin (45º + A + 45º – B)= sin (90º + A – B)= cos (A – B)

(ii) tan

θ+π

4 × tan

θ+π

43

= θ−θ+

tan1tan1

× θ+θ+−

tan1tan1

= – 1

Self Practice Problem

7. If sin α = 53

, cos β = 135

, then find sin (α + β) Answer – 6533 ,

6563

8. Find the value of sin 105º Answer22

13 +

9. Prove that 1 + tan A tan 2A

= tan A cot 2A

– 1 = sec A

6. Factorisation of the Sum or Difference of Two Sines or Cosines:

(a) sinC + sinD = 2 sin2DC+

cos2DC−

(b) sinC − sinD = 2 cos2DC+

sin2DC−

(c) cosC + cosD = 2 cos2DC+

cos2DC−

(d) cosC − cosD = − 2 sin2DC+

sin2DC−

Solved Example # 8

Prove that sin 5A + sin 3A = 2sin 4A cos A

SolutionL.H.S. sin 5A + sin 3A = 2sin 4A cos A = R.H.S.

[∵ sin C + sin D = 2 sin 2

DC + cos

2DC −

]

Solved Example # 9

Find the value of 2 sin 3θ cos θ – sin 4θ – sin 2θSolution

2 sin 3θ cos θ – sin 4θ – sin 2θ = 2 sin 3θ cos θ – [2 sin 3θ cos θ ] = 0

Self Practice Problem

10. Proved that

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7 of 2

4

(i) cos 8x – cos 5x = – 2 sin 2x13

sin 2x3

(ii) A2cosAcosA2sinAsin

−+

= cot 2A

(iii) A7cosA5cosA3cosAcosA7sinA5sinA3sinAsin

++++++

= tan 4A

(iv) A7sinA5sin2A3sinA5sinA3sin2Asin

++++

= A5sinA3sin

(v) A13cosA9cosA5cosAcosA13sinA9sinA5sinAsin

+−−−+−

= cot 4A

7. Transformation of Products into Sum or Difference of Sines &Cosines:

(a) 2 sinA cosB = sin(A+B) + sin(A−B) (b) 2 cosA sinB = sin(A+B) − sin(A−B)

(c) 2 cosA cosB = cos(A+B) + cos(A−B) (d) 2 sinA sinB = cos(A−B) − cos(A+B)

Solved Example # 10

Prove that

(i) θθ−θθθθ−θθ

4sin3sincos2cos3cos6sincos8sin

= tan 2θ

(ii) θ−θθ+θ

3tan5tan3tan5tan

= 4 cos 2θ cos 4θ

Solution

(i) θθ−θθθθ−θθ

4sin3sin2cos2cos23cos6sin2cos8sin2

= θ+θ−θ+θθ−θ−θ+θ

7coscoscos3cos3sin9sin7sin9sin

= θθθθ

2cos5cos25cos2sin2

= tan 2θ

(ii) θ−θθ+θ

3tan5tan3tan5tan

= θθ−θθθθ+θθ

5cos3sin3cos5sin5cos3sin3cos5sin

= θθ

2sin8sin

= 4 cos2θ cos 4θ

Self Practice Problem

11. Prove that sin 2θ

sin 27θ

+ sin 23θ

sin 2

11θ = sin 2θ sin 5θ

12. Prove that cos A sin (B – C) + cos B sin (C – A) + cos C sin (A – B) = 0

13. Prove that 2 cos 13π

cos 139π

+ cos 133π

+ cos 135π

= 0

8. Multiple and Sub-multiple Angles :

(a) sin 2A = 2 sinA cosA ; sin θ = 2 sinθ2

cosθ2

(b) cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2 sin²A; 2 cos²2θ

= 1 + cos θ, 2 sin²2θ

= 1 − cos θ.

(c) tan 2A = Atan1Atan22−

; tan θ =2

22

tan1tan2

θ

θ

−

(d) sin 2A = Atan1Atan22+

, cos 2A =Atan1Atan1

2

2

+−

(e) sin 3A = 3 sinA − 4 sin3A (f) cos 3A = 4 cos3A − 3 cosA

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8 of 2

4

(g) tan 3A = Atan31

AtanAtan32

3

−−

Solved Example # 11

Prove that

(i) A2cos1A2sin

+ = tan A

(ii) tan A + cot A = 2 cosec 2 A

(iii) )BAcos(BcosAcos1)BAcos(BcosAcos1

+−−++−+−

= tan 2A

cot 2B

Solution

(i) L.H.S. A2cos1A2sin

+ = Acos2AcosAsin2

2 = tan A

(ii) L.H.S. tan A + cot A = Atan

Atan1 2+ = 2

+Atan2

Atan1 2

= A2sin2

= 2 cosec 2 A

(iii) L.H.S. )BAcos(BcosAcos1)BAcos(BcosAcos1

+−−++−+−

=

+−

++

B2Acos

2Acos2

2Acos2

B2Asin

2Asin2

2Asin2

2

2

= tan 2A

+−

++

B2Acos

2Acos

B2Asin

2Asin

= tan 2A

+

+

2Bsin

2BAsin2

2Bcos

2BAsin2

= tan 2A

cot 2B

Self Practice Problem

14. Prove that θ+θ+θ+θ2coscos1

2sinsin= tan θ

15. Prove that sin 20º sin 40º sin 60º sin 80º = 163

16. Prove that tan 3A tan 2A tan A = tan 3A – tan 2A – tan A

17. Prove that tan

+

2Aº45 = sec A + tan A

9. Important Trigonometric Ratios:

(a) sin n π = 0 ; cos n π = (−1)n ; tan n π = 0, where n ∈ Ι

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9 of 2

4

(b) sin 15° or sin12π

= 2213−

= cos 75° or cos 125π

;

cos 15° or cos12π

= 2213+

= sin 75° or sin 125π

;

tan 15° = 1313

+−

= 32− = cot 75° ; tan 75° = 1313

−+

= 32+ = cot 15°

(c) sin 10π

or sin 18° =4

15− & cos 36° or cos 5π

= 4

15+

10. Conditional Identities:

If A + B + C = π then :

(i) sin2A + sin2B + sin2C = 4 sinA sinB sinC

(ii) sinA + sinB + sinC = 4 cos2A

cos2B

cos2C

(iii) cos 2 A + cos 2 B + cos 2 C = − 1 − 4 cos A cos B cos C

(iv) cos A + cos B + cos C = 1 + 4 sin2A

sin2B

sin2C

(v) tanA + tanB + tanC = tanA tanB tanC

(vi) tan2A

tan2B

+ tan2B

tan2C

+ tan2C

tan2A

= 1

(vii) cot2A

+ cot2B

+ cot2C

= cot2A

. cot2B

. cot2C

(viii) cot A cot B + cot B cot C + cot C cot A = 1

(ix) A + B + C =π2

then tan A tan B + tan B tan C + tan C tan A = 1

Solved Example # 12

If A + B + C = 180°, Prove that, sin2A + sin2B + sin2C = 2 + 2cosA cosB cosC.

Solution.Let S = sin2A + sin2B + sin2Cso that 2S = 2sin2A + 1 – cos2B +1 – cos2C

= 2 sin2A + 2 – 2cos(B + C) cos(B – C)= 2 – 2 cos2A + 2 – 2cos(B + C) cos(B – C)

∴ S = 2 + cosA [cos(B – C) + cos(B+ C)]since cosA = – cos(B+C)∴ S = 2 + 2 cos A cos B cos C

Solved Example # 13

If x + y + z = xyz, Prove that 2x1x2

− + 2y1

y2−

+ 2z1z2

− =

2x1x2

−. 2y1

y2−

. 2z1z2

−.

Solution.Put x = tanA, y = tanB and z = tanC,so that we have

tanA + tanB + tanC = tanA tanB tanC ⇒ A + B + C = nπ, where n ∈ ΙHenceL.H.S.

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10 of

24

∴ 2x1x2

− + 2y1

y2− + 2z1

z2−

= Atan1Atan22−

+ Btan1Btan22−

+ Ctan1Ctan22−

.

= tan2A + tan2B + tan2C [∵ A + B + C = nπ ]= tan2A tan2B tan2C

= 2x1x2

− . 2y1

y2− . 2z1

z2−

Self Practice Problem

18. If A + B + C = 180°, prove that

(i) sin(B + 2C) + sin(C + 2A) + sin(A + 2B) = 4sin 2

CB − sin

2AC −

sin2

BA −

(ii) CsinBsinAsinC2sinB2sinA2sin

++++

= 8 sin2A

sin2B

sin2C

.

19. If A + B + C = 2S, prove that(i) sin(S – A) sin(S – B) + sinS sin (S – C) = sinA sinB.

(ii) sin(S – A) + sin (S – B) + sin(S – C) – sin S = 4sin 2A

sin2B

sin2C

.

11. Range of Trigonometric Expression:

E = a sin θ + b cos θ

E = 22 ba + sin (θ + α), where tan α = ab

= 22 ba + cos (θ − β), where tan β = ba

Hence for any real value of θ, 2222 baEba +≤≤+−

Solved Example # 14

Find maximum and minimum values of following :(i) 3sinx + 4cosx(ii) 1 + 2sinx + 3cos2x

Solution.(i) We know

– 22 43 + ≤ 3sinx + 4cosx ≤ 22 43 +– 5 ≤ 3sinx + 4cosx ≤ 5

(ii) 1+ 2sinx + 3cos2x= – 3sin2x + 2sinx + 4

= – 3

−

3xsin2xsin2

+ 4

= – 3 2

31xsin

− + 3

13

Now 0 ≤ 2

31xsin

− ≤ 9

16

⇒ – 316

≤ – 3 2

31xsin

− ≤ 0

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11 of

24

– 1 ≤ – 3 2

31xsin

− + 3

13 ≤ 3

13

Self Practice Problem20. Find maximum and minimum values of following

(i) 3 + (sinx – 2)2 Answer max = 12, min = 4.(ii) 10cos2x – 6sinx cosx + 2sin2x Answer max = 11, min = 1.

(iii) cosθ + 3 2 sin

π+θ

4 + 6 Answer max = 11, min = 1

12. Sine and Cosine Series:

sin α + sin (α + β) + sin (α + 2β ) +...... + sin ( )β−+α 1n =2

2n

sin

sinβ

β

sin

β−+α

21n

cos α + cos (α + β) + cos (α + 2β ) +...... + cos ( )β−+α 1n =2

2n

sin

sinβ

β

cos

β−+α

21n

Solved Example # 15

Find the summation of the following

(i) cos72π

+ cos74π

+ cos76π

(ii) cos7π

+ cos72π

+ cos7

3π + cos

74π

+ cos7

5π + cos

76π

(iii) cos11π

+ cos113π

+ cos115π

+ cos117π

+ cos119π

Solution.

(i) cos72π

+ cos74π

+ cos76π

=

7sin

73sin

276

72

cos

π

π

π+π

=

7sin

73sin

74cos

π

ππ

=

7sin

73sin

73cos

π

ππ− = –

7sin2

76sin

π

π

= – 21

(ii) cos7π

+ cos72π

+ cos7

3π + cos

74π

+ cos7

5π + cos

76π

=

14sin

146sin

276

7cos

π

π

π+π

=

14sin

146sin

2cos

π

ππ

= 0

(iii) cos11π

+ cos113π

+ cos115π

+ cos117π

+ cos119π

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07

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2 0

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2 0

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12 of

24

=

11sin

115sin

2210cos

π

ππ

=

11sin2

1110sin

π

π

= 21

Self Practice Problem

Find sum of the following series :

21. cos 1n2 +

π + cos

1n23

+π

+ cos 1n2

5+π

+ ...... + to n terms. Answer21

22. sin2α + sin3α + sin4α + ..... + sin nα, where (n + 2)α = 2π Answer 0.

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13 of

24SHORT REVISION1 . BASIC TRIGONOMETRIC IDENTITIES :

(a)sin2θ + cos2θ = 1 ; −1 ≤ sin θ ≤ 1 ;−1 ≤ cos θ ≤ 1 ∀ θ ∈ R

(b)sec2θ − tan2θ = 1 ; sec θ ≥ 1 ∀ θ ∈ R(c)cosec2θ − cot2θ = 1 ; cosec θ ≥ 1 ∀ θ ∈ R

2. IMPORTANT T′ RATIOS:(a)sin n π = 0 ; cos n π = (-1)n ; tan n π = 0 where n ∈ I

(b)sin2

)1n2( π+ = (−1)n &cos 2

)1n2( π+= 0 where n ∈ I

(c)sin 15° or sin 12π

= 2213− = cos 75° or cos 12

5π;

cos 15° or cos 12π

= 2213+

= sin 75° or sin 125π

;

tan 15° = 1313

+−

= 2 3− = cot 75° ; tan 75° = 1313

−+

= 2 3+ = cot 15°

(d)sin 8π

= 2

22− ; cos 8π

= 2

22+ ; tan 8π

= 12− ; tan 83π

= 12+

(e) sin10π

or sin 18° = 4

15− & cos 36° or cos 5π

= 4

15+

3. TRIGONOMETRIC FUNCTIONS OF ALLIED ANGLES :If θ is any angle, then − θ, 90 ± θ, 180 ± θ, 270 ± θ, 360 ± θ etc. are called ALLIED ANGLES.(a) sin (− θ) = − sin θ ; cos (− θ) = cos θ(b) sin (90°- θ) = cos θ ; cos (90° − θ) = sin θ(c) sin (90°+ θ) = cos θ ; cos (90°+ θ) = − sin θ (d)sin (180°− θ) = sin θ; cos (180°− θ) = − cos θ(e) sin (180°+ θ) = − sin θ ; cos (180°+ θ) = − cos θ(f) sin (270°− θ) = − cos θ ; cos (270°− θ) = − sin θ (g) sin (270°+ θ) = − cos θ ; cos (270°+ θ) = sin θ

4. TRIGONOMETRIC FUNCTIONS OF SUM OR DIFFERENCE OF TWO ANGLES :(a) sin (A ± B) = sinA cosB ± cosA sinB (b) cos (A ± B) = cosA cosB ∓ sinA sinB(c) sin²A − sin²B = cos²B − cos²A = sin (A+B) . sin (A− B)(d) cos²A − sin²B = cos²B − sin²A = cos (A+B) . cos (A − B)

(e) tan (A ± B) = tan tantan tanA B

A B±

1 ∓ (f) cot (A ± B) = cot cot

cot cotA BB A

∓ 1±

5. FACTORISATION OF THE SUM OR DIFFERENCE OF TWO SINES OR COSINES :

(a) sinC + sinD = 2 sin2

DC+ cos

2DC−

(b) sinC − sinD = 2 cos2

DC+ sin

2DC−

(c) cosC + cosD = 2 cos2DC+

cos2DC− (d) cosC − cosD = − 2 sin

2DC+

sin2DC−

6. TRANSFORMATION OF PRODUCTS INTO SUM OR DIFFERENCE OF SINES & COSINES :(a) 2 sinA cosB = sin(A+B) + sin(A−B) (b) 2 cosA sinB = sin(A+B) − sin(A−B)(c) 2 cosA cosB = cos(A+B) + cos(A−B) (d) 2 sinA sinB = cos(A−B) − cos(A+B)

7. MULTIPLE ANGLES AND HALF ANGLES :(a) sin 2A = 2 sinA cosA ; sin θ = 2 sin θ

2cos θ

2(b) cos2A = cos2A − sin2A = 2cos2A − 1 = 1 − 2 sin2A ;

cos θ = cos2 θ2

− sin² θ2

= 2cos2 θ2

− 1 = 1 − 2sin2 θ2

.

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14 of

242 cos2A = 1 + cos 2A , 2sin2A = 1 − cos 2A ; tan2A = A2cos1A2cos1

+−

2 cos22θ

= 1 + cos θ , 2 sin22θ

= 1 − cos θ.

(c) tan 2A = Atan1

Atan22−

; tan θ = )2(tan1

)2(tan22 θ−θ

(d) sin 2A = Atan1

Atan22+

, cos 2A = Atan1Atan1

2

2

+− (e) sin 3A = 3 sinA − 4 sin3A

(f) cos 3A = 4 cos3A − 3 cosA (g) tan 3A = Atan31

AtanAtan32

3

−−

8. THREE ANGLES :

(a) tan (A+B+C) = AtanCtanCtanBtanBtanAtan1CtanBtanAtanCtanBtanAtan

−−−−++

NOTE IF : (i) A+B+C = π then tanA + tanB + tanC = tanA tanB tanC(ii) A+B+C =

2π

then tanA tanB + tanB tanC + tanC tanA = 1(b) If A + B + C = π then : (i) sin2A + sin2B + sin2C = 4 sinA sinB sinC

(ii) sinA + sinB + sinC = 4 cos2A

cos2B

cos2C

9. MAXIMUM & MINIMUM VALUES OF TRIGONOMETRIC FUNCTIONS:(a) Min. value of a2tan2θ + b2cot2θ = 2ab where θ ∈ R(b) Max. and Min. value of acosθ + bsinθ are 22 ba + and – 22 ba +(c) If f(θ) = acos(α + θ) + bcos(β + θ) where a, b, α and β are known quantities then

– )cos(ab2ba 22 β−α++ < f(θ) < )cos(ab2ba 22 β−α++

(d) If α,β ∈ 02

, π

and α + β = σ (constant) then the maximum values of the expression

cosα cosβ, cosα + cosβ, sinα + sinβ and sinα sinβoccurs when α = β = σ/2.

(e) If α,β ∈ 02

, π

and α + β = σ(constant) then the minimum values of the expression

secα + secβ, tanα + tanβ, cosecα + cosecβ occurs when α = β = σ/2.(f) If A, B, C are the angles of a triangle then maximum value of

sinA + sinB + sinC and sinA sinB sinC occurs when A = B = C = 600

(g) In case a quadratic in sinθ or cosθ is given then the maximum or minimum values can be interpretedby making a perfect square.

10. Sum of sines or cosines of n angles,

sin α + sin (α + β) + sin (α + 2β ) + ...... + sin ( )α β+ −n 1 = 2

2n

sinsin

β

β

sin

β−+α

21n

cos α + cos (α + β) + cos (α + 2β ) + ...... + cos ( )α β+ −n 1 = 2

2n

sin

sinβ

β

cos

β−+α

21n

EXEREXEREXEREXEREXERCISE–ICISE–ICISE–ICISE–ICISE–IQ.1 Prove that cos²α + cos² (α + β) − 2cos α cos β cos (α + β) = sin²βQ.2 Prove that cos 2α = 2 sin²β + 4cos (α + β) sin α sin β + cos 2(α + β)Q.3 Prove that , tan α + 2 tan 2α + 4 tan 4α + 8 cot 8 α = cot α .Q.4 Prove that : (a) tan 20° . tan 40° . tan 60° . tan 80° = 3

(b) tan 9° − tan 27° − tan 63° + tan 81° = 4 . (c) 23

167sin

165sin

163sin

16sin 4444 =π+π+π+π

Q.5 Calculate without using trigonometric tables :

(a) cosec 10° − 3 sec 10° (b) 4 cos 20° − 3 cot 20° (c) °°−°

20sin20cos40cos2

(d)

°−

°°+°° 35sin2

5sin40cos

25sec10sin22 (e) cos6

16π

+ cos6163π

+ cos6165π

+ cos6167π

(f) tan 10° − tan 50° + tan 70°

Q.6(a) If X = sin

π+θ

127

+ sin

π−θ

12 + sin

π+θ

123 , Y = cos

π+θ

127

+ cos

π−θ

12 + cos

π+θ

123

then prove that XY

YX − = 2 tan2θ.

(b) Prove that sin²12° + sin² 21° + sin² 39° + sin² 48° = 1+ sin² 9° + sin² 18° .

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15 of

24Q.7 Show that : (a) cot 7 12° or tan 82 1

2° = ( )( )1223 ++ or 6432 +++

(b) tan 142 12° = 2 + 2 3 6− − .

Q.8 If m tan (θ - 30°) = n tan (θ + 120°), show that cos 2 θ = )nm(2nm

−+

.

Q.9 If tan

+π

2y

4 = tan3

+π

2x

4, prove that

xsinysin =

xsin31xsin3

2

2

++ .

Q.10 If cos (α + β) = 45

; sin (α - β) = 513

& α , β lie between 0 & 4π

, then find the value of tan 2α.

Q.11 Prove that if the angles α & β satisfy the relation ( ) ( )nmmn

2sinsin >=

β+αβ

then nmtantan1

nm

1 tantan

−βα−

+

+=α

β

.

Q.12 (a) If y = 10 cos²x − 6 sin x cos x + 2 sin²x , then find the greatest & least value of y .(b) If y = 1 + 2 sin x + 3 cos2 x , find the maximum & minimum values of y ∀ x ∈ R .(c) If y = 9 sec2x + 16 cosec2x, find the minimum value of y ∀ x ∈ R.

(d) Prove that 3 cos θ π+

3

+ 5 cos θ + 3 lies from - 4 & 10 .

(e) Prove that ( )2 3 4+ sin θ + 4 cos θ lies between − ( )522 + & ( )522 + .

Q.13 If A + B + C = π, prove that ∑

Ctan.Btan

Atan = ∑ (tan A) − 2∑ (cot A).

Q.14 If α + β = c where α, β > 0 each lying between 0 and π/2 and c is a constant, find the maximum orminimum value of(a) sin α + sin β (b) sin α sin β(c) tan α + tan β (d) cosec α + cosec β

Q.15 Let A1 , A2 , ...... , An be the vertices of an n-sided regular polygon such that ;

413121 AA1

AA1

AA1 += . Find the value of n.

Q.16 Prove that : cosec θ + cosec 2 θ + cosec 22 θ + ...... + cosec 2 n − 1 θ = cot (θ/2) − cot 2 n - 1θQ.17 For all values of α , β , γ prove that;

cos α + cos β + cos γ + cos (α + β + γ) = 4 cos2

β+α .cos2

γ+β. cos

2α+γ .

Q.18 Show that BcosAcos)BAsin(Bsin2Asin2

Bsin1Bcos

AcosAsin1

−+−−=

−++

.

Q.19 If tan β = tan tantan . tanα γ

α γ+

+1 , prove that sin 2β = sin sin

sin . sin2 2

1 2 2α γ

α γ+

+ .

Q.20 If α + β = γ , prove that cos² α + cos² β + cos² γ = 1 + 2 cos α cos β cos γ .

Q.21 If α + β + γ = π2

, show that ( )( )( )( )( )( )222

222

tan1tan1tan1tan1tan1tan1

γβα

γβα

+++

−−− = sin sin sin

cos cos cosα β γ

α β γ+ + −

+ +1 .

Q.22 If A + B + C = π and cot θ = cot A + cot B + cot C, show that ,sin (A − θ) . sin (B − θ) . sin (C − θ) = sin3 θ .

Q.23 If P = 1917cos.........

195cos

193cos

19cos π++π+π+π

and

Q = 2120cos.........

216cos

214cos

212cos π++π+π+π

, then find P – Q.Q.24 If A, B, C denote the angles of a triangle ABC then prove that the triangle is right angled if and only if

sin4A + sin4B + sin4C = 0.Q.25 Given that (1 + tan 1°)(1 + tan 2°)......(1 + tan 45°) = 2n, find n.

EXEREXEREXEREXEREXERCISE–IICISE–IICISE–IICISE–IICISE–IIQ.1 If tan α = p/q where α = 6β, α being an acute angle, prove that;

21

(p cosec 2 β − q sec 2 β) = 22 qp + .Q.2 Let A1 , A2 , A3 ............ An are the vertices of a regular n sided polygon inscribed in a circle of radius R.

If (A1 A2)2 + (A1 A3)

2 + ......... + (A1 An)2 = 14 R2 , find the number of sides in the polygon.

Q.3 Prove that: 1)cos(2

3cos3cos−φ−θφ+θ

= (cosθ + cosφ) cos(θ + φ) – (sinθ + sinφ) sin(θ + φ)

Q.4 Without using the surd value for sin 180 or cos 360 , prove that 4 sin 360 cos 180 = 5

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16 of

24Q.5 Show that , 21

x27cosx9sin

x9cosx3sin

x3cosxsin =++ (tan27x − tanx)

Q.6 Let x1 = ∏=

π5

1r 11rcos and x2 = ∑

=

π5

1r 11rcos , then show that

x1 · x2 = 641

−π 1

22eccos , where ΠΠΠΠΠ denotes the continued product.

Q.7 If θ = 72π

, prove that tan θ . tan 2 θ + tan 2 θ . tan 4 θ + tan 4 θ . tan θ = − 7.

Q.8 For 0 < x < π4

prove that , )xsinx(cosxsinxcos

2 − > 8.

Q.9 (a) If α = 72π

prove that, sin α + sin 2α + sin 4α = 27

(b) sin7π . sin

72π . sin

73π =

87

Q.10 Let k = 1°, then prove that ∑= +

88

0n k)1ncos(·nkcos1

= ksinkcos

2

Q.11 Prove that the value of cos A + cos B + cos C lies between 1 & 23

where A + B + C = π.Q.12 If cosA = tanB, cosB = tanC and cosC = tanA , then prove that sinA = sinB = sinC = 2 sin18°.

Q.13 Show that xsin

xcos3+ Rx∈∀ can not have any value between 22− and 22 . What inference

can you draw about the values of xcos3

xsin+

?

Q.14 If (1 + sin t)(1 + cos t) = 45

. Find the value of (1 – sin t)(1 – cos t).

Q.15 Prove that from the equality ba1

bcos

asin 44

+=α+α

follows the relation ; ( )33

8

3

8

ba1

bcos

asin

+=α+α

.

Q.16 Prove that the triangle ABC is equilateral iff , cot A + cot B + cot C = 3 .Q.17 Prove that the average of the numbers n sin n°, n = 2, 4, 6, ......., 180, is cot 1°.Q.18 Prove that : 4 sin 27° = ( ) ( ) 2/12/1

5355 −−+ .

Q.19 If A+B+C = π; prove that tan22A + tan2

2B + tan2

2C

≥ 1.

Q.20 If A+B+C = π (A , B , C > 0) , prove that sin 2A

. sin 2B

. sin 2C

≤ 81

.Q.21 Show that elliminating x & y from the equations , sin x + sin y = a ;

cos x + cos y = b & tan x + tan y = c gives ( ) 2222 a4ba

ba8

−+ = c.

Q.22 Determine the smallest positive value of x (in degrees) for whichtan(x + 100°) = tan(x + 50°) tan x tan (x – 50°).

Q.23 Evaluate : ∑=

−−

n

1n1n

1n

n

2xcos2

2xtan

Q.24 If α + β + γ = π &

γ−β+α

β−α+γ

α−γ+β

4tan·

4tan·

4tan = 1, then prove that;

1 + cos α + cos β + cos γ = 0.Q.25 ∀ x ∈ R, find the range of the function, f (x) = cos x (sin x + sin sin2 2x + α ) ; α ∈ [0, π]

EXEREXEREXEREXEREXERCISE–IIICISE–IIICISE–IIICISE–IIICISE–IIIQ.1 sec2θ = 2)yx(

xy4+

is true if and only if : [JEE ’96, 1]

(A) x + y ≠ 0 (B) x = y , x ≠ 0 (C) x = y (D) x ≠ 0 , y ≠ 0

Q.2 (a) Let n be an odd integer. If sin nθ = r

n

=∑

0br sinr θ, for every value of θ, then :

(A) b0 = 1, b1 = 3 (B) b0 = 0, b1 = n(C) b0 = − 1, b1 = n (D) b0 = 0, b1 = n2 − 3n + 3

(b) Let A0 A1 A2 A3 A4 A5 be a regular hexagon inscribed in a circle of unit radius .Then the product of the lengths of the line segments A0 A1, A0 A2 & A0 A4 is :

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17 of

24(A) 34 (B) 3 3 (C) 3 (D) 3 3

2(c) Which of the following number(s) is/are rational ? [ JEE '98, 2 + 2 + 2 = 6 out of 200 ]

(A) sin 15º (B) cos 15º (C) sin 15º cos 15º (D) sin 15º cos 75º

Q.3 For a positive integer n, let fn (θ) =

θ

2tan (1+ sec θ) (1+ sec 2θ) (1+ sec 4θ) .... (1 + sec2nθ) Then

(A) f2 π

16

= 1 (B) f3

π32

= 1 (C) f4

π64

= 1 (D) f5

π128

= 1 [JEE '99,3]

Q.4(a) Let f (θ) = sin θ (sin θ + sin 3 θ) . Then f (θ) : [ JEE 2000 Screening. 1 out of 35 ](A) ≥ 0 only when θ ≥ 0 (B) ≤ 0 for all real θ(C) ≥ 0 for all real θ (D) ≤ 0 only when θ ≤ 0 .

(b) In any triangle ABC, prove that, cot 2A

+ cot 2B

+ cot 2C

= cot 2A

cot 2B

cot 2C

. [JEE 2000]

Q.5(a) Find the maximum and minimum values of 27cos 2x · 81sin 2x. (b) Find the smallest positive values of x & y satisfying, x − y =

4π

, cot x + cot y = 2. [REE 2000, 3]

Q.6 If α + β = π2

and β + γ = α then tanα equals [ JEE 2001 (Screening), 1 out of 35 ](A) 2(tanβ + tanγ) (B) tanβ + tanγ (C) tanβ + 2tanγ (D) 2tanβ + tanγ

Q.7 If θ and φ are acute angles satisfying sinθ = 21

, cos φ = 31

, then θ + φ ∈ [JEE 2004 (Screening)]

(A)

ππ

2,

3 (B)

ππ

32,

2 (C)

ππ

65,

32

(D)

ππ,

65

Q.8 In an equilateral triangle, 3 coins of radii 1 unit each are kept so that theytouch each other and also the sides of the triangle. Area of the triangle is(A) 4 + 2 3 (B) 6 + 4 3

(C) 12 + 4

37(D) 3 +

437

[JEE 2005 (Screening)]

Q.9 Let θ ∈

π

4,0 and t1 = (tanθ)tanθ, t2 = (tanθ)cotθ, t3 = (cotθ)tanθ , t4 = (cotθ)cotθ, then

(A) t1 > t2 > t3 > t4 (B) t4 > t3 > t1 > t2 (C) t3 > t1 > t2 > t4 (D) t2 > t3 > t1 > t4[JEE 2006, 3]

EXEREXEREXEREXEREXERCISE–IV (CISE–IV (CISE–IV (CISE–IV (CISE–IV (ObjectiObjectiObjectiObjectiObjectivvvvve)e)e)e)e)Part : (A) Only one correct option

1.( ) ( ) ( )

( ) ( )xtan.xcos

xsinxcos.xtan

23

2

27

23

23

+−

−−+−ππ

πππ when simplified reduces to:

(A) sin x cos x (B) − sin2 x (C) − sin x cos x (D) sin2x

2. The expression 3

α+π+

α−π )3(sin

23sin 44

– 2

α+π+

α+π )5(sin

2sin 66

is equal to

(A) 0 (B) 1 (C) 3 (D) sin 4α + sin 6α3. If tan A & tan B are the roots of the quadratic equation x2 − ax + b = 0, then the value of sin2 (A + B).

(A) 22

2

)b1(aa

−+ (B) 22

2

baa+ (C) 2

2

)cb(a+ (D) 22

2

)a1(ba−

4. The value of log2 [cos2 (α + β) + cos2 (α − β) − cos 2α. cos 2β] :(A) depends on α & β both (B) depends on α but not on β(C) depends on β but not on α (D) independent of both α & β.

5. °°°°+°

80sin10sin50sin70sin820cos

2 is equal to:(A) 1 (B) 2 (C) 3/4 (D) none

6. If cos A = 3/4, then the value of 16cos2 (A/2) – 32 sin (A/2) sin (5A/2) is(A) – 4 (B) – 3 (C) 3 (D) 4

7. If y = cos2 (45º + x) + (sin x − cos x)2 then the maximum & minimum values of y are:(A) 2 & 0 (B) 3 & 0 (C) 3 & 1 (D) none

8. The value of cos 19π

+ cos 193π

+ cos 195π

+...... + cos 1917π

is equal to:(A) 1/2 (B) 0 (C) 1 (D) none

9. The greatest and least value of ( )23xcosxsinlog 2 +− are respectively:(A) 2 & 1 (B) 5 & 3 (C) 7 & 5 (D) 9 & 7

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18 of

2410. In a right angled triangle the hypotenuse is 2 2 times the perpendicular drawn from the oppositevertex. Then the other acute angles of the triangle are

(A) 3π

& 6π

(B) 8π

& 83π

(C)4π

& 4π

(D) 5π

& 103π

11. °290cos1

+ °250sin31

=

(A) 3

32(B)

334

(C) 3 (D) none

12. If 4

3π < α < π, then

α+α 2sin

1cot2 is equal to

(A) 1 + cot α (B) – 1 – cot α (C) 1 – cot α (D) – 1 + cot α

13. If x ∈

ππ

23, then 4 cos2

−π

2x

4 + x2sinxsin4 24 + is always equal to

(A) 1 (B) 2 (C) – 2 (D) none of these14. If 2 cos x + sin x = 1, then value of 7 cos x + 6 sin x is equal to

(A) 2 or 6 (B) 1 or 3 (C) 2 or 3 (D) none of these

15. If cosec A + cot A = 211

, then tan A is

(A) 2221

(B) 1615

(C) 11744

(D) 43117

16. If cot α + tan α = m and αcos

1 – cos α = n, then

(A) m (mn2)1/3 – n(nm2)1/3 = 1 (B) m(m2n)1/3 – n(nm2)1/3 = 1(C) n(mn2)1/3 – m(nm2)1/3 = 1 (D) n(m2n)1/3 m(mn2)1/3 = 1

17. The expression xcos10x3cos5x5cos10x2cos15x4cos6x6cos

+++++

is equal to(A) cos 2x (B) 2 cos x (C) cos2 x (D) 1 + cos x

18. If BsinAsin

= 23

and BcosAcos

= 25

, 0 < A, B < π/2, then tan A + tan B is equal to

(A) 5/3 (B) 3/5 (C) 1 (D) 5/)35( +

19. If sin 2θ = k, then the value of θ+

θ2

3

tan1tan

+ θ+

θ2

3

cot1cot

is equal to

(A) kk1 2−

(B) kk2 2−

(C) k2 + 1 (D) 2 – k2

Part : (B) May have more than one options correct20. Which of the following is correct ?

(A) sin 1° > sin 1 (B) sin 1° < sin 1 (C) cos 1° > cos 1 (D) cos 1° < cos 121. If 3 sin β = sin (2α + β), then tan (α + β) – 2 tan α is

(A) independent of α (B) independent of β(C) dependent of both α and β (D) independent of α but dependent of β

22. It is known that sin β = 54

& 0 < β < π then the value of α

β+α−β+α π

sin

)(cos)(sin36cos

2

is:

(A) independent of α for all β in (0, π) (B) 35

for tan β > 0

(C)15

)cot247(3 α+ for tan β < 0 (D) none

23. If the sides of a right angled triangle are {cos2α + cos2β + 2cos(α + β)} and{sin2α + sin2β + 2sin(α + β)}, then the length of the hypotenuse is:

(A) 2[1+cos(α − β)] (B) 2[1 − cos(α + β)] (C) 4 cos2

2β−α

(D) 4sin2

2β+α

24. If x = sec φ − tan φ & y = cosec φ + cot φ then:

(A) x = 1y1y

−+

(B) y = x1x1

−+

(C) x = 1y1y

+−

(D) xy + x − y + 1 = 025. (a + 2) sin α + (2a – 1) cos α = (2a + 1) if tan α =

(A) 43

(B) 34

(C) 1aa2

2 +(D) 1a

a22 −

26. If tan x = cab2− , (a ≠ c)

y = a cos2x + 2b sin x cos x + c sin2xz = a sin2x – 2b sin x cos x + c cos2x, then

(A) y = z (B) y + z = a + c (C) y – z = a – c (D) y – z = (a – c)2 + 4b2

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19 of

2427.n

BsinAsinBcosAcos

−+

+ n

BcosAcosBsinAsin

−+

(A) 2 tann 2

BA −(B) 2 cotn

2BA −

: n is even (C) 0 : n is odd (D) none28. The equation sin6x + cos6x = a2 has real solution if

(A) a ∈ (–1, 1) (B) a ∈

−−

21,1 (C) a ∈

−

21

21

(D) a ∈

1,

21

EXEREXEREXEREXEREXERCISE–V (CISE–V (CISE–V (CISE–V (CISE–V (SubjectiSubjectiSubjectiSubjectiSubjectivvvvve)e)e)e)e)1. The minute hand of a watch is 1.5 cm long. How far does its tip move in 50 minutes?

(Use π = 3.14).2. If the arcs of the same length in two circles subtend angles 75° and 120° at the centre, find the ratio of

their radii.3. Sketch the following graphs :

(i) y = 3 sin 2x (ii) y = 2 tan x (iii) y = sin 2x

4. Prove that cos

θ+π

23

cos (2π + θ)

θ+π+

θ−π )2(cot

23cot = 1.

5. Prove that cos 2 θ cos 2θ

– cos 3 θ cos 2

9θ = sin 5 θ sin

25θ

.

6. If tan x = 43

, π < x < 2

3π, find the value of sin

2x

and cos 2x

.

7. prove that

αα+

π−α+

π−α−

4cot2

cos

4cot1

4cot1

2

2

sec2

9α = cosec 4α.

8. Prove that, sin 3 x. sin3 x + cos 3 x. cos3 x = cos3 2 x.

9. If tan α = qp

where α = 6 β, α being an acute angle, prove that; 21

(p cosec 2 β − q sec 2 β) = 22 qp + .

10. If tan β = γα+γ+α

tan.tan1tantan

, prove that sin 2β = γα+γ+α2sin.2sin1

2sin2sin.

11. Show that: (i) cot 721°

or tan 8221°

= ( ) ( )1223 ++ or 6432 +++

(ii) tan 14221°

= 2 + 632 −− . (iii) 4 sin 27° = ( ) ( ) 2/12/15355 −−+

12. Prove that, tan α + 2 tan 2α + 4 tan 4α + 8 cot 8 α = cot α.

13. If cos (β − γ) + cos (γ − α) + cos (α − β) = 23−

, prove that cos α + cos β + cos γ = 0, sin α + sin β + sin γ = 0.

14. Prove that from the equality ba

1b

cosa

sin 44

+=α+α follows the relation ( )33

8

3

8

ba1

bcos

asin

+=α+α

15. Prove that: cosec θ + cosec 2 θ + cosec 22 θ +... + cosec 2 n − 1θ = cot (θ/2) − cot 2n − 1 θ. Hence or

otherwise prove that cosec 154π

+ cosec 158π

+ cosec 1516π

+ cosec 1532π

= 0

16. Let A1, A2,......, An be the vertices of an n−sided regular polygon such that; 413121 AA

1AA1

AA1 += .

Find the value of n.

17. If A + B + C = π, then prove that: (i) tan²2A + tan²

2B + tan²

2C

≥ 1

(ii) sin2A

. sin2B

. sin2C

≤ 81

. (iii) cos A + cos B + cos C ≤ 23

18. If θcosax

+ θsinby

= a2 – b2, θθ

2cossinax

– θθ

2sincosby

= 0. Show that (ax)2/3 + (by)2/3 = (a2 – b2)2/3

19. If Pn = cosnθ + sinnθ and Qn = cosnθ – sinnθ, then show thatPn – Pn – 2 = – sin2θ cos2θ Pn – 4Qn – Qn – 2

= – sin2θ cos2θ Qn – 4 and hence show thatP4 = 1 – 2 sin2θ cos2θ , Q4 = cos2θ – sin2θ

20. If sin (θ + α) = a & sin (θ + β) = b (0 < α, β, θ < π/2) then find the value ofcos2 (α − β) − 4 ab cos(α − β)

21. If A + B + C = π, prove thattan B tan C + tan C tan A + tan A tan B = 1 + sec A. sec B. sec C.

22. If tan2α + 2tanα. tan2β = tan2β + 2tanβ. tan2α, then prove that each side is equal to 1 ortan α = ± tan β.

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20 of

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ANSWER SHEETEXERCISE–I

Q 5. (a) 4 (b) −1 (c) 3 (d) 4 (e) 45

(f) 3 Q 10. 5633

Q 12. (a) ymax = 11 ; ymin = 1 (b) ymax = 133

; ymin = − 1, (c) 49Q14. (a) max = 2 sin (c/2), (b) max. = sin2 (c/2), (c) min = 2 tan (c/2), (d) min = 2 cosec (c/2)Q 15. n = 7 Q23. 1 Q.25 n = 23

EXERCISE –II

Q.2 n = 7 Q.13

−

221,

221

Q.14 104

13 − Q.22 x = 30°

Q 23.1n

1n

2xsin2

1x2sin

2

−−

− Q.25 – 1 2+ sin α ≤ y ≤ 1 2+ sin α

EXERCISE–IIIQ.1 B Q.2 (a) B, (b) C, (c) C Q.3 A, B, C, D Q.4 (a) CQ.5 (a) max. = 35 & min. = 3–5 ; (b) x =

125π

; y = 6π

Q.6 C Q.7 BQ.8 B Q.9 B

EXERCISE–IV

1. D 2. B 3. A 4. D 5. B 6. C 7. B

8. A 9. B 10. B 11. B 12. B 13. B 14. A

15. C 16. A 17. B 18. D 19. B 20. BC

21. AB 22. BC 23. AC 24. BCD 25. BD 26. BC

27. BC 28. BD

EXERCISE–V

1. 7.85 cm 2. r1 : r2 = 8 : 5

6. sin 2x

= 103

and cos 2x

= – 101

16. n = 7 20. 1 − 2a2 − 2b2