Cambridge International AS and A Level Mathematics Pure Mathematics 1

Trigonometry

218

P1^

7

From triangle ABD,

sin; 60 23 cos; 60 1 tan;

2

°= °= 60 °= 3

sin; 60 °= 23 cos; 60 °= 21 tan; 60 °= 3

sin; 60 °= 23 cos; 60 °= 21 tan; 60 °= 3

sin; 30 12 cos; 30 23 tan. 30 1

3

°= °= °=

sin; 30 cos; tan.

1

2 30

3

2 30

1

3

sin;°= 30 12 cos;°= 30 23 tan.°= 30 1

3

°= °= °=

ExAmPlE 7.1 Without using a calculator, find the value of cos 60°sin 30° + cos^2 30°.

(Note that cos^2 30° means (cos 30°)^2 .)

SOlUTION

cos 60°sin 30° + cos^2 30°

(ii) The angle 45°

In figure 7.6, triangle PQR is a right-angled isosceles triangle with equal sides of

length 1 unit.

Using Pythagoras’ theorem, PQ = 2.

This gives

sin; 45 1 cos; tan.

2

45 1

2

°= °= 45 °= 1

(iii) The angles 0° and 90°

Although you cannot have an angle of 0° in a triangle (because one side would be

lying on top of another), you can still imagine what it might look like. In figure

7.7, the hypotenuse has length 1 unit and the angle at X is very small.

=× +







=+

=

1

2

1

2

3

2

1

4

3

4

1

2

.

45°

P R

Q

1

1

Figure 7.6

opposite

X Y

Z

adjacent

hypotenuse

Figure 7.7