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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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