8/13/2019 Answers Trig 1&2 General

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8/13/2019 Answers Trig 1&2 General

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Answers Trig 1 General

1. Using the rules for angles greater than 90 i. cos150 cos30

3

2

ii. sin 180 30sin210

cos330 cos 360 30

sin30

cos30

tan30

1

3

iii.1

tan60 sec240 tan60

cos(180 60)1

3cos60

3 2

2 3

2.i. tan320 tan 360 40

tan40

0.84

ii. cos 0.6

related angle = 53.13

180 53.13,180 53.13

126.87, 233.13

iii.

2 2

if cos 0, then must be in 2nd and 3rd quadrant. But since tan 0, then is restricted

to be in 3rd quadrant.

Hypotenuse calculation: 4 1 17

4sin

17

iv. if cos 0.4, must be in 1st quadrant or in 4th quadrant.

if cos 0.4, 66.42, 293.58

66.42 90, 293.58 360

x x

x x

x x

3.i.

11 1tan sin sin 2

6 6 63

1 sin63

1 1

23

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ii.2 7

sec cos sec cos 23 4 3 4

1cos

4cos

3

1 11 2

2

12

2

4. 2sin 1 cos 3 0, 0 2x x x 2sin 1, cos 3

for cos 3, trig ratios cannot be >1, therefore not a solution.

1sin

2

5, .

6 6

x x

x

x

x

5.i. 2 2 2 2

4

2

2 2

1 sin sec 1 cos cot

cos

sinor cos cot

2 2 2 2

2 2 2 2

sin cos 1 1 sin cos i.e.

1 cot sec cot sec 1

j

ii. 2 2

2

3 3cot 3 1 cot

3sec

2 21 cot sec

6. Here, we use the complementary angle formula cos 90 sin and note that 180 lies in the3

rdquadrant, and since sine is negative in the 3

rdquadrant, so sin 180 sin

cos 90 sin

1sin 180 sin

7. 24sin 3 0, 0 360x x Factorising (difference of squares), we obtain

2sin 3 2sin 3 0x x 3

sin2

x ,2 4 5

, , ,3 3 3 3

x

8. Related angle6

, 2x belongs in the 2ndquadrant and 3rdquadrant.

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5 72 ,

6 6

5 7,

12 12

x

x

9.i.

2 21 tan cosecx x 2 2

2 2

sin cos 1

tan 1 cosec

j

ii.

2

2

2

2

2

2 2 2

2 2 2

2

2

sec tan 1

1 sin1

cos cos

1 sin cos1

cos

1 sin cos cos

cos sin cos 1 0

cos sin cos sin cos 0

cos sin cos sin cos 0

sin cos sin 0

sin sin cos 0

sin sin cos 0

sin cos

x x

x

x x

x x

x

x x x

x x x

x x x x x

x x x x x

x x x

x x x

x x x

x x

sin 0

tan 1 0, 180, 360

180 45, 360 45

0, 135, 180, 315, 360

x

x x

x

x

10. 0 90x , all ratios are positive.

2 2sin 1 cos

1 cos 1 cos

1 cos 1 cos

1 cos

1 cos

x x

x x

x x

x

x

Least value occurs whencos 0

least value is 1

x

Answers Trig 2 General

1. 3sin 4 2sin 2 0x x 4

sin , sin 13

4But 0 sin 1 so sin not a solution.

3sin 1

3

2

x x

x x

x

x

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2. Ifcos 0 , then must be in 2ndor 3rdquadrant. But since tan 0 , then is restricted to be in 3rdquadrant.

2 2Hypotenuse calculation: 3 7 58

3sin

58

3. First, we combine the fractions and obtain:

2

2

sin 1 cos sin 1 cos sin sin cos sin sin cos

1 cos 1 cos 1 cos 1 cos 1 cos 1 cos

2sin

1 cos

2sin

sin

2

sin

2cosec

x x x x x x x x x x

x x x x x x

x

x

x

x

x

x

4.i.

3 5tan cos tan cos

4 6 4 6

tan cos4 6

31

2

ii. sin 225 cos300 sin 180 45 cos 360 60

sin45 cos60

1 1

22

5. Here, we use the complementary angle formula cos 90 sin and note that 360 lies in the4

thquadrant, and since tangent is negative in the 4

thquadrant, so tan 360 tan

tan 360 tan

cos 90 sin

sin

cos

sin

1

cos

sec

6. Starting from LHS, we have:sin

tan cosLHS1sec 1

1cos

xx x

x

x

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2

2

sin

cosLHS1 cos

cos

sin cos

cos 1 cos

sin 1 cos

1 cos 1 cos

sin 1 cos

1 cos

sin 1 cos

sin

1 cos

sin

x

x

x

x

x x

x x

x x

x x

x x

x

x x

x

x

x

Now, dividing top and bottom by cosx , we obtain:

1 coscosLHSsin

cos

11

cos

tan

sec 1

tan

RHS

x

x

x

x

x

x

x

x

LHS = RHS as required

7. Given that x , and related angle 30 , and since sine is negative, in the 1stand 2ndquadrant when going anticlockwise, we obtain

Factorising (difference of squares), we obtain

1sin2

2

52 ,

6 6

x

x

5,12 12

x

8. This equation is in quadratic form, so letting cosY x and substituting, we have:

2

2

2cos 5cos 3 0

2 5 3 0

2 1 3 0

1, 3

2

x x

Y Y

Y Y

Y Y

Substituting back cosY x gives:1

cos , cos 32

x x

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But -1 cos 1, so cos 3 not a solution.

1cos

2

5,

3 3

x x

x

x

9.i. 2 2 2 21 sec cosec 1 cot tan

1

2 2 2 2

2 2 2 2

1 cot sec 1 sec cot i.e.

tan 1 cosec cosec 1 tan

j

ii. 2 2

2

sec 1 sin sec cos

1cos

cos

cos

A A A A

AA

A

10. If x is acute, then it is a normal right-angled triangle.Given that

2

2sin

1

tx

t

, using Pythagoras and working out the sides of the triangle gives the adjacent

side as:

2 22

2 4 2

2 4

22

2

adjacent = 1 2

1 2 4

1 2

1

1

t t

t t t

t t

t

t

And sinceadjacent

coshypotenuse

x ,

2

2

1cos

1

tx

t