Mathametics 04 – Trignometry

1) The value of sin 12°. sin48°. sin54° =

(a) 1/8               (b) 1/6               (c) 1/4               (d) 1/2

(a) sin 12°. sin48°. sin54° = 1/2[cos36° – cos60°] sin54° = 1/2[cos36°sin54° – 1/2sin54°]

= 1/4[2cos36°sin54°] = 1/4[sin90°+sin18° – sin54°]

= 1/4[1 – (sin54° – sin18°)]

= 1/4[1 – 2sin18°cos36°]

= 1/4[1 – (2sin18°/cos18°)cos18° cos36°]

= 1/4[1 – (sin36°cos36°)/(cos18°)]

= 1/4[1 – (2sin36°cos36°)/2cos18°]

= 1/4[1 – (sin72°/2sin72°)]

= 1/4[1-(1/2)]

= 1/8

Alternative Method Let θ = 12°

sin 12°. sin48°. sin54° = 1/sin72° (sin12° sin48°sin72°sin54°)

= 1/4 (sin3(12°)sin54°)/sin72°

= (sin36°sin54°)/(8sin36°cos36°) = cos36°/8cos36° = 1/8

2) The smallest positive value of x (in degrees) for which tan(x + 100°) = tan(x + 50°) tan x tan(x – 50°) is: 

(a) 30°               (b) 45°              (c) 60°               (d) 90°

(a) The relation may be written as [tan(x+100°)/[tan(x-50°) = tan(x+50º)tan x 

⇒ [sin(x+100º)cos(x-50º)]/[sin(x-50º)cos(x+100º)] = [sin(x+50º)sin x]/[cos(x+50º)cos x]

⇒ [sin(2x+50º)+sin(150º)]/[sin(2x+50º)-sin(150º)] = [cos(50º)-cos(2x+50º)]/[cos(50º)+cos(2x+50º)]

⇒ [sin(2x+50º)]/sin150º = -cos50º/[cos(2x+50º)]

⇒ cos50°+ 2sin(2x + 50°) cos(2x + 50°) = 0

⇒ cos50°+ sin (4x + 100°) = 0

⇒ cos50° + cos(4x + 10°) = 0

⇒ cos(2x + 30°) cos(2x – 20°) = 0

⇒ x = 30°, 55°

⇒ The smallest value of x = 30

3) If A + B + C = 180° then the value of tanA + tanB + tanC is 

(a) ≥ 3√3               (b) ≥ 2√3              (c) > 3√3               (d) > 2√3

(a) tan(A + B) = tan(180º – C)

or, (tanA+tanB)/(1-tanAtanB) = tanC

or, tanA + tanB + tanC = tanA tanB tanC

(tanA+tanB+tanC)/3 ≥ ³√(tanAtanBtanC)                         [since A.M.  G.M.] 

or, tanA tanB tanC ≥ ³√(tanAtanBtanC)

or, tan2A tan2B tan2C ≥ 27                                                    [cubing both sides]

or tanA tanB tanC ≥ 3 ³√3

⇒ tanA + tanB + tanC ≥ 3 √3.

4) Let 0 < A, B < π/2 satisfying the equalities 3 sin2A + 2 sin2B = 1 and 3sin2A – 2sin2B = 0. Then A + 2B =:

(a) π/4               (b) π/3              (c) π/2               (d) None of these

(c) From the second equation, we have

sin2B = 3/2 sin2A …(1)

and from the first equality

3 sin2A = 1 – 2 sin2B = cos2B …(2)

Now cos (A + 2B) = cosA. cos2B – sinA . sin2B

= 3 cosA . sin2A – 3/2 . sinA . sin2A

= 3cosA. sin2 A – 3 sin2A . cosA = 0

⇒ A + 2B = π/2 or 3π/2 

Given that 0 < A < π/2 and 0 < B < π/2 

⇒  0 < A + 2B < π + π/2 

Hence, A + 2B = π/2 .

5) If a cos3 θ + 3a cos θ sin2 θ = x and a sin3 θ + 3a cos2 θ sin θ = y, then (x + y)2/3 + (x – y)2/3 =

(a) 2a2/3               (b) a2/3              (c) 3a2/3               (d) 2a1/3

(a) a cos3 θ + 3a cos θ sin2 θ = x

a sin3 θ + 3a cos2 θ sin θ = y

x + y = a[sin3 θ + cos3 θ + 3 sin θ cos θ(sin θ + cos θ)] = a(sinθ + cosθ)3

[(x+y)/a]1/3 = sin θ + cos θ ……(1)

x – y = a[cos3 θ– sin3 θ + 3 cosθ sin2 θ – 3 cos2 θ sin θ] = a[cosθ – sinθ]3

[(x-y)/a]1/3 = cos  – sin  ……(2)

(sin θ + cos θ)2 + (cos θ – sin θ)2 = [(x+y)2/3+(x-y)2/3]/a2/3

2 (sin2 θ + cos2 θ) = [(x+y)2/3+(x-y)2/3]/a2/3

(x + y)2/3 + (x – y)2/3 = 2a2/3

6) If (1+ √(1+a)tanα = 1+√(1-a) , then sin4α =

(a) a/2               (b) a             (c) a2/3               (d) 2a

(b) Let a = sin 4θ ⇒ √(1+a) = cos 2θ + sin 2θ and √(1-a) = cos 2θ – sin 2θ

[1 + √(1+a)] tan α = [1 + √(1-a)]

⇒ (1 + cos 2θ + sin 2θ) tan α = 1 + cos 2θ – sin 2θ

⇒ [2cosθ(cosθ+sinθ)]/[ 2cosθ(cosθ-sinθ)] = cot α

⇒ (cosθ+sinθ)/(cosθ-sinθ) = cot

⇒ (1+tanθ)/(1-tanθ) = -cotα
⇒ tan (π/4+θ) = tan(π/2+α)

⇒ θ = (π/4 – α)
⇒ a = sin 4θ = sin (π – 4α) = sin 4α .

7) If cos2 θ = 1/3(a2 -1) and tan2 θ/2 = tan2/3 α, then cos2/3 α + sin2/3 α =

(a) 2a2/3              (b) (2/a)2/3             (c) (2/a)1/3              (d) 2a1/3

(b) cos2 θ = (a2 – 1)/3 , tan2 θ/2 = tan2/3 α

tan3 θ/2 = tan α

⇒ (sin3 θ/2)/(cos3 θ/2) = sin α/cos α

(sin3 θ/2)/sin α = (cos3 θ/2)/cos α = k

(sin3 θ/2) = k sin α ……(1)

(cos3 θ/2) = k cos α ……(2)

k2/3 sin2/3 α + k2/3 cos α = 1

sin2/3 α + cos2/3 α = 1/k2/3

Squaring and adding (1) and (2)

k2(sin2 α + cos2 α) = sin6 θ/2 + cos6 θ/2 = (sin2 θ/2+cos2 θ/2)3 – 3sin2 θ/2 (cos2 θ/2)[(sin2 θ/2 (cos2 θ/2)]

k2 = 1 – 3/4 sin2 θ = 1 – 3/4 + 3/4 cos2 θ

k2 = a2/4 ⇒ k = a/2

sin2/3 α + cos2/3 α = (2/a)2/3

8) If 3 sin2 α + 2 sin2 β = 1 and 3sin 2α –2sin 2β = 0, where α, β are positive acute angles, then α + 2β =

(a) π/2              (b) π/3             (c) π/4              (d) π/6

(a) 3 sin2 α + 2 sin2 β = 1 ……(1)

3 sin 2α = 2 sin 2β ……(2)

3 sin2 α = 1 – 2 sin2 β = cos 2β

3 sin α sin α = cos 2β ……(3)

from equation (2)

3 . 2 sin α cos α = 2 sin 2β

3 sin α = sin2 β/cos α

from equation (3)

3 sin α = sin 2β/cosα = cos 2β

cos α cos 2β – sin α sin 2β = 0

cos (α + 2β) = 0

⇒ α + 2β = π/2 .

                           5

9) The value of cos(2r-1) π/11 is:

                          r=1

 

(a) 1/2              (b) 1/3             (c) 1/4              (d) 1/6

       5

(a) cos(2r-1)π/11

        r=1


cos π/11+cos 3π/11+cos 5π/11+cos 7π/11+cos 9π/11


= (2 sin π/11)/(2 sin π/11)[cos π/11+cos 3π/11+cos 5π/11+cos 7π/11+cos 9π/11]


= (sin 2π/11+sin 4π/11-sin 2π/11+sin 6π/11-sin 4π/11+sin 8π/11-sin 6π/11+sin 10π/11-sin 8π/11)/(2 sin π/11)


= (sin 10π/11)/(2 sin π/11)


= sin [π-(π/11)]/(2 sin π/11)


= 1/2

10) If tanx = n. tany, n ∈ R+ , then maximum value of sec2(x – y) is equal to:

(a) (n+1)2/2n              (b) (n+1)2/n             (c) (n+1)2/2              (d) (n+1)2/4n

(d) tanx = n tany,

cos(x – y) = cosx. cosy + sinx.siny.

⇒ cos(x – y) = cosx.cosy(1 + tanx.tany)

= cosx. cosy (1 + n tan2y)

⇒ sec2(x-y) = (sec2x×sec2y)/(1+n tan2y)2

= (1+tan2x)(1+tan2y)/(1+n tan2y)2

= (1+n2tan2y)(1+tan2y)/(1+n tan2y)2

= 1+[(n-1)2 tan2y]/(1+n tan2y)2

Now, [(1+n tan2y)/2]2 ≥ n tan2y

⇒ tan2y/(1+n tan2y) ≤ 1/4n

⇒ sec2(x-y) ≤ 1+(n-1)2/4n = (n+1)2/4n

11) Let fn(θ) = tan θ/2 .(1 + secθ) (1 + sec2θ) (1 + sec4θ) … (1 + sec 2nθ). Then

(a) f2 (π/16) = 4              (b) f3 (π/32) = 3             (c) f4 (π/64) = 2              (d) f5 (π/128) = 1

(d) 1 + sec θ = (1+cos θ)/cos θ = (2cos2θ/2)/cos θ . Similarly for others,

∴ fn(θ) = tan θ/2 × [(2cos2θ/2)/cos θ] ×2cos2θ/cos 2θ ×….×(2 cos22n-1θ)/cos2nθ

= tan θ/2 × 2n+1 × (cos θ×cos 2θ×….×cos 2n-1θ) × (cos2θ/2)/cos 2nθ

= sin θ/cos 2nθ × 2n {(2 sin θ×cos θ×cos 2θ×…×cos 2n-1θ)/2 sin θ}

= sin θ/(cos 2nθ) × 2n × (sin 2nθ)/(2n sin θ)

= tan 2nθ

∴ f5 (π/128) = 1

12) The set of values of λ ∈ R such that tan2θ + secθ = λ holds for some θ is

(a) (-,1]              (b) (-,1]            (c) Φ              (d) [-1,+)

(d) sec 2θ + sec θ – (λ + 1) = 0

∴ sec θ = (-1±√[1+4(λ+1)])/2 = 

For real sec θ, 4λ + 5 ≥ 0, i.e. λ ≥ -5/4 

Also sec θ ≥ 1 or sec θ ≤ -1

∴ (-1±√(4λ+5)/2 ≥ 1 or (-1±√(4λ+5)/2 ≤ -1

⇒ -1+√(4λ+5) ≥ 2 or -1-√(4λ+5) ≤ -2

⇒ 4λ+5 ≥ 9 or 4λ+5 ≥ 1

⇒ λ ≥ 1 or λ ≥ -1

∴ λ ≥ -5/4 and λ ≥ 1 or λ ≥ -5/4 and λ ≥ -1

∴ λ ≥ 1 or λ ≥ -1

∴ λ ∈ [-1, +∞)

13) For what and only what values of α lying between 0 and π is the inequality sin α cos3 α > sin3 α cos α valid ?

(a) a ∈ (0,π/4)              (b) a ∈ (0,π/2)            (c) a ∈ (π/4 , π/4)              (d) none of these

(a) We have sin α cos3α > sin3α cosα

⇒ sin α cos α(cos2α – sin2α) > 0

⇒ cos3α sin α(1 – tan2α) > 0

⇒ cos α (1 – tan2α) > 0         (∴ sin α > 0 for 0 < α < π)

⇒ cos α > 0 and 1 – tan2α > 0

or cos α < 0 and 1 – tan2α < 0

⇒ α ∈ (0, π/4) or α ∈ (3π/4 , π)

14) If an + 1 = √1/2(1+an) , then cos [(√1 – (ao)2)/(a1a2a3 …. to ∞)] is equal to

(a) 1              (b) -1            (c) ao              (d) 1/ao

(c) Let a0 = cos θ

 

Then a1 = √1/2(1+a0) = √1/2(1+cos θ) = cos θ/2

 

a2 = √1/2(1+a1) = √1/2(1+cos θ/2) = cos (θ/22) and so on.

 

Now, √(1-a02)/(a1a2a3…∞) = sin θ/[cos (θ/2)cos (θ/22)cos (θ/23)…∞]

 

= lim sin θ/[cos (θ/2)cos (θ/22)cos (θ/23)…cos (θ/2n)]

    n→∞

 

= lim 2sin (θ/2)cos (θ/2)/[cos (θ/2)cos (θ/22)cos (θ/23)…cos (θ/2n)]

    n→∞

 

= lim 2sin (θ/2)/[cos (θ/22)cos (θ/23)…cos (θ/2n)]

    n→∞

 

= lim 22 sin (θ/22)/[cos (θ/23)…cos (θ/2n)]

    n→∞

 

= lim 2n sin (θ/2n)

    n→∞

 

= lim sin (θ/2n)/(θ/2n

    n→∞

 

θ = cos-1(a0)

 

cos √(1-a02)/(a1a2a3…∞) = a

15) The number of points inside or on the circle x2 + y2 = 4 satisfying tan4x + cot4x + 1 = 3sin2y is

(a) one              (b) two            (c) four              (d) infinite

(c) tan4x + cot4x + 1 = (tan2x – cot2x)2 + 3 ≥ 3 3 sin2y ≤ 3

⇒ tan2 x = cot2 x , sin2y =1

⇒ tanx = ± 1, siny = ±1

⇒ x = ± π/4, ± 3π/4 , . . .

But x2 ≤ 4 ⇒ -2 ≤ x ≤ 2

⇒ x = ± π/4 only

Siny = ± 1 ⇒ y = ± π/2, ± 3π/2, . . ..

But y2 ≤ 4 ⇒ y = ± π/2 only.

So, four solutions are possible.

16) The values of x between 0 and 2π which satisfy the equation sinx √8cos2x = 1 are in A.P. with common difference 

(a) π/4              (b) π/8            (c) 3π/8              (d) 5π/8

(a) We have sinx √(8cos2x) = 1

⇒ sinx |cosx| = 1/2√2

Case – I: when cosx > 0.

In this case sinx cosx = 1/2√2

⇒ sin 2x = 1/√2 ⇒ 2x = π/4, 3π/4 , 9π/4, 13π/4

⇒ x = π/8, 3π/8, 9π/8, 13π/8

As x lies between 0 and 2π and cosx > 0, x =π/8, 3π/8

Case – II: When cos x < 0

In this case sinx | cosx| = 1/2√2

⇒ sin x cos x = – 1/2√2 or sin 2x = – 1/√2

⇒ x = 5π/8 , 7π/8, 13π/8, 15π/8

⇒ x = 5π/8, 7π/8 as cosx < 0

Thus the values of x satisfying the given equation which lie between 0 and 2π are π/8 , 3π/8 , 5π/8 , 7π/8

These are in A.P. with common difference π/4

17) General solution for θ if sin[2θ+(π/6)]+cos[θ+(5π/6)] = 2 , is 

(a) 2nπ+(7π/6)              (b) 2nπ+(π/6)           (c) 2nπ-(7π/6)              (d) None of these

(a) sin (2θ+π/6)+cos (θ+5π/6) = 2 . . . (1)

sin (2θ+π/6) ≤ 1 and cos (θ+5π/6) ≤ 1

∴ (1) may holds true if sin (2θ+π/6) and cos (θ+5π/6) both equal to 1 simultaneously.

First common value of θ is 7π/6 for which sin (2θ+π/6) = sin 5π/2 = sin π/2 = 1

and cos(θ+5π/6) = cos (7π/6+5π/6) = cos 2π = 1

and since periodicity of sin (2θ+π/6) is π

and periodicity of cos (θ+5π/6) is 2π,

therefore, periodicity

sin(2θ+π/6)+cos(θ+5π/6) is 2π

Therefore, general solution is θ = 2nπ+7π/6

18) The general solution of the equation (√3 – 1)sinθ+(√3+1)cosθ = 2 is

(a) nπ+(-1)n(π/4) – (π/12)                             (b) 2nπ±(π/4) – (π/12)           

(c) nπ+(-1)n(π/4)+(π/12)                               (d) 2nπ±(π/4)+(π/12)

(d) Let √3+1 = rcos α, √3-1 = rsin α

∴  r2 = (√3+1)2+(√3-1)2 = 8 or r = 2√2

and tan α = (√3-1)/(√3+1) = 1-(1/√3)/1+(1/√3)

or tan α = tan(45º – 30º) = tan15º

∴ α = 15º = π/12

Using these in the given equation, we get rcos(θ-α) = 2

or cos(θ – π/12) = 2/r = 2/2√2 = 1/√2

cos(π/4)

∴ θ-(π/12) = 2nπ ± π/4 or θ = 2nπ ± π/12 , n ∈ 1

19) The solution of the equation tan θ . tan 2θ = 1 is :

(a) nπ+(5π/12)              (b) nπ – (π/12)           (c) 2nπ ± (π/4)              (d) nπ ± (π/6)

(d) Given tan θ. tan 2θ = 1 ⇒ 2 tan2θ/(1 – tan2θ) = 1

⇒ 2 tan2θ = 1 – tan2θ ⇒ 3 tan2 θ = 1

⇒ tan θ = ±1/√3

⇒ θ = nπ ± π/6

20) Solve for x, the equation sin3x + sin x cos x + cos3x = 1:

(a) 2mπ              (b) (4n+1)(π/2)           (c) Both              (d) None of these

(c) The given equation is sin3 x + cos3 x + sin x cos x = 1

⇒ (sin x + cos x) (sin2 x – sin x cos x + cos2 x ) + sin x cos x – 1 = 0

⇒ (1 – sin x cos x)[sin x + cos x – 1] = 0

Either 1 – sin x cos x = 0 ⇒ sin 2 x = 2 which is not possible

Or, sin x + cos x – 1 = 0 ⇒ cos (x – π/4) = 1/√2 ⇒ x – (π/4) = 2mπ

⇒ x = 2mπ and x = (4n + 1) π/2

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