Understanding Engineering Mathematics

Such a conversion simplifies the solution of equations such as

acosθ+bsinθ=c

by conversion to the form

cos(θ+α)=

c

r

=

c

√

a^2 +b^2

for example.

Solution to review question 6.1.9

(i) We assume that we can write

cosθ+sinθ=rsin(θ+α)

=rsinθcosα+rcosθsinα

then rcosα= 1

rsinα= 1

so r^2 = 12 + 12 = 2

and r=

√

2

Then cosα=

1

√

2

sinα=

1

√

2

so tanα=1, in the first quadrant, and hence

α= 45 °

and cosθ+sinθ=

√

2sin(θ+ 45 °)

(ii) In this case

cosθ+sinθ=rcos(θ+α)

=rcosθcosα−rsinθsinα

so rcosα= 1

rsinα=− 1

As before r=

√

2 ,so

cosα=

1

√

2

sinα=−

1

√

2

The appropriate solution in this case is

α=− 45 °

so

cosθ+sinθ=

√

2cos(θ− 45 °)