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6232E Part 5

1. I like Trigonometry as it compels me to think seriously while solving problems. Analyzing capability increases.

2. Plot of y = cosx between 0 and 360 degrees will be as given below.

[pic]

3. sin2θ = √3/2

= sin600, sin1200, sin4200, sin4800

Hence, 2θ= 600 or θ = 300

Or, 2θ= 1200 or θ = 600

Also θ = 420/2 = 2100, θ=480/2=2400

Hence, θ = 300, 600, 2100 and 2400. Ans.

4. sin2x – cosx = 0

Or, sin2x = cos x = sin(π/2-x)

Or, 2x = π/2-x

Or, x = π/6 = 300

Also sin2x = sin(π – 2x)

Hence, sin2x = sin(π – 2x) = cos x = sin(π/2-x)

Or, π - 2x = π/2-x

Or, x = π/2 = 900

At x = 2700, the equation will also be satisfied.

Also cos x = cos (2π-x)

Hence, sin2x = cos x = cos (2π-x) = sin(π/2+2π-x)

Or, 2x = π/2+2π-x

Or, x = (2 π+ π/2)/3 =1500 Ans.

5. sin(2x -100) = ½

= sin300, sin1500, sin3900, sin5100

When sin(2x -100) = sin300, we have

30 = 2x -10

Or, x = 200 Ans.

When sin(2x -100) = sin1500, we have

150 = 2x -10

Or, x = 800

When sin(2x -100) = sin3900, we have

390 = 2x -10

Or, x =2000

When sin(2x -100) = sin5100, we have

510 = 2x -10

Or, x =2600

Hence, x = 200,800, 2000 and 2600. Ans

6. cos2x –sin2(x/2)+3/4 = 0

We know that cos2x = 1-2sin2x

= 1-2(2 sinx/2 coxx/2)2

= 1- 8sin2x/2cos2x/2

= 1-8sin2x/2(1-sin2x/2)

= 1-8sin2x/2+8sin4x/2

Putting this in the given equation, we get

1-8sin2x/2+8sin4x/2+3/4 = 0

Or, 8sin4x/2-9sin2x/2+3/4 = 0

If p = sin2x/2, we have 8p2- 9p+7/4 = 0

Or, 32p2- 36p+7 = 0

Or, 32p2- 8p- 28p+7 = 0

Or, 8p(4p-1)-7(4p-1) = 0

Or, (8p-7)(4p-1) = 0

Hence, p = ¼, 7/8

Hence, when p = 1/4, we have sin2x/2 = 1/4.Therefore, sinx/2 = ±1/2

Neglecting –ve sign, we get sinx/2 = ½ = sin300, sin1500

Or, x = 600, 3000

Hence, when p =7/8, we have sin2x/2 = 7/8.Therefore, sinx/2 = ±√7/8

Neglecting –ve sign, we get sinx/2 =√7/8 = 0.9354

Or, x/2 = sin-10.9354 = 69.2920, 110.7080

Or, x = 138.5840, 221.4160

Hence, x = 600, 138.5840, 3000, 221.4160 Ans.

7. sin2θ = cos2θ+1/2

Or, cos2θ- sin2θ = - 1/2

Or, cos 2θ = -1/2

= cos 1200, cos2400

Hence, θ = 600, 1200

At θ = 2400, 3000 also, equation will be satisfied.

Hence, θ = 600, 1200, 2400, 3000 Ans.

8. cos4x = sin2x

Or, cos4x = cos [π/2-2x]

Hence, we have x = π/12= 150.

We can also write cos4x = 1-2sin22x

Putting in the given equation, we get

Or, 2sni22x +sin2x -1 = 0

Or, sin2x = [- 1±√(1+8)]/4) = ½,-1

Hence, if sin2x = ½

Or, 2x = 300, 1500, 3900, 5100

Or, x = 150, 750, 1950, 2550

If sin2x = -1, we will have

2x = 2700, 630, or, x = 1350, 3150.

Hence, x = 150,750, 1350, 1950, 2550, 3150 Ans

9. 3sinθ – 4cosθ = 2

We divide both sides by √ (32+42) =5 and get

3/5sinθ – 4/5cosθ = 2/5

If cos α = 3/5, we have sin α = 4/5 and we can write

Sinθ cosα – cosθ sinα= 0.4

Or, sin(θ - α) = 0.4

Or, (θ - α) = sin-10.4 = 23.57810, 156.42190

We have cos α = 3/5 and sinα = 4/5, hence, α = sin-14/5 = 53.13010

Hence, θ = (23.5781+43.1301)0, (156.4219+53.1301)0

= 76.70820, 209.5520 Ans.

10. tan(x+150) = 3 tanx

We have tan(α + β) = (tanα + tanβ)/(1-tanα tanβ)

Hence, we can write the given equation as

(tanx + tan150)/(1- tanxtan150) = 3tanx

Or, tan x + tan150 = 3tanx – 3tan150tan2x

Or, 3tan150tan2x -2tanx +tan150 = 0

We know tan150 = tan (60 - 45) = (tan60 - tan45)/(1 + tan60tan45)

= (√3-1) / (√3+1)

= 0.2679

Putting this value, we get

0.8037tan2x – 2tanx +0.2679 = 0

Or, tanx = (2±√(4- 4x0.8037x0.2679)/(2x0.8037)

= (2±1.7716)/1.6074

= 2.3463, 0.1420

Hence, x = tan-12.3463, tan-10.1420

= 66.9160, 8.08190

Also x = 180+66.9169, 180+8.08190

= 246.9160, 188.08190 will satisfy the equation Ans.

11. (a) x = y + cosθ

Or, cos θ = x - y

Or, θ = cos-1(x-y) Ans.

(b) cos θ = y2

Or, θ = cos-1y2 Ans.

12. (a) sin-1(cos x) = sin-1[sin (π/2-x)]

= π/2-x Ans.

(b) tan ( sin-1x) = ? If sin-1x = θ, we have, sin θ = x.

Hence, tanθ = x/√ (1-x2)

Or, tan(sin-1x) = x/√ (1-x2) Ans.

(c) tan [tan-1(x+1)/(x-1) +tan-1(x-1)/x] = ?

We have tan-1α + tan-1β= tan-1(α + β) (1- α β).

Hence, tan [tan-1(x+1)/(x-1) +tan-1(x-1)/x]

= tan [tan-1{(x+1)/(x-1) + (x-1)/x}/{1-(x+1)/(x-1)x(x-1)/x}]

= (2x2-x+1)/(1-x) Ans.

13. cos-1x+cos-12x = ½

Ifcos-1x = α, we have cos α = x and sin α = √ (1- x2).

Similarly, if we have cos -12x = β, we have cos β = 2x and sin β = √ (1- 4x2)

We have cos (α + β) = cosα cosβ - sinα sinβ

Hence, cos (α + β) = x.2x - √ (1- x2) (1- 4x2)

Or, α + β = cos-1[2x2 - √ (1- x2) (1- 4x2)]

Or, cos-1x+cos-12 x = cos-1[2x2 - √ (1- x2) (1- 4x2)] = 1/2

Hence, 2x2 - √ (1- x2) (1- 4x2) = 1/2

Or, 2x2 – ½ = √ (1- x2) (1- 4x2)

Squaring both sides, we get

4x4+1/4 -2x2 = 4x4-5x2+1

Or, 1/4 -2x2 = -5x2+1

Or, x2 =1/4

Or, x = ±1/2

Neglecting –ve sign, we have x = ½ Ans.

14. tan (2tan-1x) = 2tan(tan-1+tan-1x3)

If tan-1x = α, we have tan α = x and

LHS = tan (2tan-1x) = tan (2α) = 2tanα (1- tan2α)

= 2x/(1-x2)

RHS = 2tan (tan-1+tan-1x3)

= 2 tan [tan-1 (x+x3)/(1-x4)]

= 2(x+x3)/(1-x4)

=2 x (1+x2)/[(1-x2)(1+x2)]

= 2x/(1-x2) = LHS Proved.

6232C-2 Part3

1. The answers are given in the following table.

| |1220 |315012’ |275013’37’’ |193041’51’’ |

|sin |0.8480 |-0.7046 |-0.9958 |-0.2368 |

|cos |-0.5299 |0.7095 |0.0911 |-0.9716 |

|tan |-1.6 |-0.9958 |-10.9311 |0.2437 |

|cot |-0.6248 |-1.0042 |-0.09148 |4.1029 |

2.(a) sin30cos240+sin210sin300[pic]

(b) tan225+tan(-45) = 1-1=0.

3.(a) cos112033’= cos 112.550 = cos (1800-112.550) =-cos 67.450

(b)tan3100=tan(3600-500) = –tan500

(c) cot 138013’10’’ = -cot 41046’50’’

(d) sin(-1400)=-sin1400= -sin(1800-400)=-sin400

4.(a) sin(270+θ)= -cosθ

(b) cos(π+θ)=-cosθ

(c) tan(810+θ)=tan(90+θ)=-cotθ

(d)sin(θ-180) =-sin(180-θ)=-sinθ

5(a) log sin62022’33’’=log(0.8860)-0.0526

(b)logcot 28013’17’’= 0.27028

(c) log cos 125015’23’’= log (-0.5772) is not defined.

(d) log tan78045’50’’ = 0.7019

6.log cosA = 9.12575

I hope we are given LcosA = 10 + logcosA.

Hence, 9.12575 = 10 +logcosA

Or, log cosA = -0.87425

Or, cosA =0.13358

Or, A = 82.3230

(b)LsinA = 10+log sinA = 9.91655

Or,log sinA= -0.08345

Or, sinA=0.82518

Or, A = 55.610

(c)logcotA = 0.11975

Or, cot A = 1.3174

Or, tanA = 0.7590

Or, A = 37.200

(d)log tanA = 0.06323

Or, tanA =1.15672

Or, A = 49.160

7. sin(90+x)sin(180+x)+cos(90+x)cos(180-x)

=cosx*-sinx+(-sinx)*(-cosx)

=0

8.A = 28.5, B = 90-28.5 = 61.5, B = 18.3.

Putting in sin rule, we get sin28.5/a = sin61.5/18.3 = sin90/c or a = 9.936 and c = 20.823. Ans.

9.Applying cos rule, we get cosA = (15^2+17^2-31^2)/(2*15*17)=-0.87647

Or, A = 151.219

Similarly, cosB = (31^2+17^2-15^2)/(2*31*17)=0.972486

Or, B = 13.382

Hence, C = 180-13.382-151.219 = 15.399

10. B+C = 115030’+20029’ = 135059’

Hence, A = 180-135059’=4401’

Applying sin rule, we get sin4401’/23.47 = sin115.5/b = sin 20029/c

Or, b =23.47sin 115.5/sin 4401’= 30.486 and similarly c = 23.47sin20029/sin 4401’=11.819

11. Applying sin rule, we get sinA/134.2=sin5209’11’’/84.54=sin C/c. This gives sinA>1, which is not possible; this triangle is not possible.

12. Applying sin rule, we get sin 66047’/627.7 = sinB/412.2=SINc/C.

Hence, sinB = 412.2sin 66047’’/627.7 = 0.6035 or B = 37.120; C = 180-37.12-4401’ = 76.10.

Hence, c = 627.7sin 76.1/sin66047’ = 663

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