Understanding Engineering Mathematics

=

−(tan 45°+tan 30°)

1 −tan 45°tan 30°

=−

(

1 +

1

√

3

)

(

1 −

1

√

3

)=

1 +

√

3

1 −

√

3

=− 2 +

√

3 ( 21

➤

)

D. From cos(A+B)=cosAcosB−sinAsinB, withA=Bwe get

cos 2A=cos^2 A−sin^2 A

Using sin^2 A+cos^2 A=1 this can be expressed in two alternative

forms:

cos 2A=2cos^2 A− 1

= 1 −2sin^2 A

E. (i) Using cos 2θ=^1 −2sin^2 θ

cos 30°= 1 −2sin^215 °

so sin^215 °=

1

2

( 1 −cos 30°)

=

1

2

(

1 −

√

3

2

)

=

2 −

√

3

4

so sin 15°=

√

2 −

√

3

2

(ii) tan 2A=

2tanA

1 −tan^2 A

withA= 15 °gives

tan 30°=

1

√

3

=

2tan15°

1 −tan^215 °

=

2 x

1 −x^2

wherex=tan 15°

So 1−x^2 = 2

√

3 x

orx^2 + 2

√

3 x− 1 =0 giving

x=

− 2

√

3 ±

√

4 × 3 + 4

2

=

− 2

√

3 ± 4

2

= 2 −

√

3 (tan 15°is positive)