Solving Trig EquationsExample (i): 2cos2 3sin 2 sin(2 ).

2cos2 3sin 2 1 0 360 .

a) Express in the form

b) Hence solve for o

x x k x

x x x

Step 1:2cos2 3sin 2 sin(2 )x x k x Expand sin(a-b)

sin 2 cos cos2 sink x k x Equate coefficients

sin 2

cos 3

eqn 1

eqn 2

k

k

Square & add eqns 1&22 13k

13k

Subst. for k in eqn 213cos 3

1 3cos

13

cas

tnd rd is in 2 or 3 quadrants.

Take 2nd quadrant: 146.3o

02cos2 3sin 2 13sin(2 146.3)b) We now have x x x

2cos2 3sin 2 1

13sin(2 146.3) 1

We solve

by solving

x x

x

1sin(2 146.3)

13x 1 01

sin 16.113

stIn the 1 quadrant

2x-146.3 = …., 16.1o , 180-16.1o, 360+16.1o, 360+180-16.1o, ….

2x-146.3 = …., 16.1o , 163.9o, 376.1o, 523.9o, ….

2x = …., 162.4o , 310.2o, 522.4o, 670.2o, ….

cas

t

x = …., 81.2o , 155.1o, 261.2o, 335.1o, ….

Return to Main

Example (ii): (From a past paper)

A builder has obtained a large supply of 4 metre rafters. He wishes to use them to build some holiday chalets. The planning department insists that the gable end of each chalet should be in the form of an isosceles triangle surmounting two squares, as shown in the diagram.

a) If o is the angle shown in the diagram and A is the area (in square metres) of the gable end, show that

8 2 sin 2coso oA

8 2 sin 2cos sin .b) Express in the form oo o k

c) Find algebraically the value of for which the area of the gable end is 30 square metres.

44

44

ss

Let the side of the square frames be s.

Part (a)

Use the cosine rule in the isosceles triangle:

2 2 2 2 cosa b c bc A

2 2 22 4 4 2.4.4coss

24 32 1 coss

2 8 1 coss This is the area of one of the squares.

Use the formula for the area of an isosceles triangle.21

2 sina 21

2 .4 sinAreaof Triangle 8sin

Total area = Triangle + 2 x square:

8sin 2.8 1 cosA 8 2 sin 2cosA

Part (b) 8sin 16cos sin .Express in the form k

8 sin 2cos 8 sinConsider t

sin 2cos sin 8 .Solve and remember t k t

Expand sin(a+b)

sin 2cos sin cos cos sint t Equate coefficients.

cos 1

sin 2

eqn 1

eqn 2

t

t

Square & add eqns 1 & 22 5t

5t Only interested in +ve root.

Subst. for t in eqn 1.1 1cos

5

63.4o Only interested in 1st quadrant.

Finally:

8sin 16cos 8 5 sin 63.4o

Part (c) Find algebraically the value of o for which the area is the 30m2

16 8 5 sin 63.4o

A

30 16 8 5 sin 63.4o

14 8 5 sin 63.4o

7sin 63.4

4 5

1 7 563.4 sin

20

51.5o

114.9o

Return to Main