Understanding Engineering Mathematics

(xx) cos

(

−

π

3

)

=cos

(π

3

)

=

1

2

(xxi) sin( 585 °)=sin( 225 °)=−sin( 45 °)=−

1

√

2

(xxii) cos( 225 °)=−cos( 45 °)=−

1

√

2

(xxiii) tan(− 135 °)=−tan( 135 °)=−(−tan 45°)=−(− 1 )= 1

(xxiv) sec 30°= 1 /cos 30°= 2 /

√

3

(xxv) cosecπ/ 4 = 1 /sin(π/ 4 )=

√

2

(xxvi) cot 60°= 1 /tan 60°= 1 /

√

3

(xxvii) sec 120°= 1 /cos 120°=− 1 /cos 60°=− 2

(xxviii) cot

(

−

π

2

)

=−cot(π/ 2 )= 0

(xxix) cosec(− 60 °)=− 1 /sin 60°=− 2 /

√

3

6.2.3 Sine and cosine rules and solutions of triangles

➤

172 194➤

For general triangles two important rules apply. We use the standard notation – capital
A,B,Cfor the angles with corresponding lower case letters labelling the opposite sides
(Figure 6.6).

C B

A

b

a

c

Figure 6.6Standard labelling of angles and sides.

Thesine rulestates that

a

sinA

=

b

sinB

=

c

sinC

i.e. the sides of a triangle are proportional to the sines of the opposite angles. We can see
this by using the result for the area of a triangle−^12 base×height. We can calculate this
area in three different ways, giving:

1

2 bcsinA=

1

2 acsinB=

1

2 absinC

The sine rule then follows by dividing by^12 abcand taking the reciprocals of the results.
Incidentally, there is another useful formula (Heron’s formula) for calculating the area
of a triangle with sidesa,b,c:

area=

√

s(s−a)(s−b)(s−c)