Higher Engineering Mathematics

COMPOUND ANGLES 177

B

Hence tan

(

x+

π

4

)

tan

(

x−

π

4

)

=

(

1 +tanx

1 −tanx

)(

tanx− 1

1 +tanx

)

=

tanx− 1

1 −tanx

=

−(1−tanx)

1 −tanx

=− 1

Problem 4. If sinP= 0 .8142 and cosQ=

0 .4432 evaluate, correct to 3 decimal places:

(a) sin(P−Q), (b) cos(P+Q) and

(c) tan(P+Q), using the compound-angle

formulae.

Since sinP= 0 .8142 then
P=sin−^10. 8142 = 54. 51 ◦.
Thus cosP=cos 54. 51 ◦= 0 .5806 and
tanP=tan 54. 51 ◦= 1 .4025.

Since cosQ= 0 .4432,Q=cos−^10. 4432 = 63. 69 ◦.
Thus sinQ=sin 63. 69 ◦= 0 .8964 and
tanQ=tan 63. 69 ◦= 2 .0225.

(a) sin (P−Q)

=sinPcosQ−cosPsinQ

=(0.8142)(0.4432)−(0.5806)(0.8964)

= 0. 3609 − 0. 5204 =− 0. 160

(b) cos (P+Q)

=cosPcosQ−sinPsinQ

=(0.5806)(0.4432)−(0.8142)(0.8964)

= 0. 2573 − 0. 7298 =− 0. 473

(c) tan (P+Q)

=

tanP+tanQ

1 −tanPtanQ

=

(1.4025)+(2.0225)

1 −(1.4025)(2.0225)

=

3. 4250

− 1. 8366

=− 1. 865

Problem 5. Solve the equation

4 sin(x− 20 ◦)=5 cosx

for values ofxbetween 0◦and 90◦.

4 sin(x− 20 ◦)=4[sinxcos 20◦−cosxsin 20◦],

from the formula for sin(A−B)

=4[sinx(0.9397)−cosx(0.3420)]

= 3 .7588 sinx− 1 .3680 cosx

Since 4 sin (x− 20 ◦)=5 cosxthen

3 .7588 sinx− 1 .3680 cosx=5 cosx

Rearranging gives:

3 .7588 sinx=5 cosx+ 1 .3680 cosx

= 6 .3680 cosx

and

sinx

cosx

=

6. 3680

3. 7588

= 1. 6942

i.e. tanx= 1 .6942, andx=tan−^11. 6942 = 59. 449 ◦

or 59 ◦ 27 ′

[Check: LHS=4 sin (59. 449 ◦− 20 ◦)

=4 sin 39. 449 ◦= 2. 542

RHS=5 cosx=5 cos 59. 449 ◦= 2 .542]

Now try the following exercise.

Exercise 80 Further problems on com-

pound angle formulae

1. Reduce the following to the sine of one
   angle:

(a) sin 37◦cos 21◦+cos 37◦sin 21◦

(b) sin 7tcos 3t−cos 7tsin 3t

[(a) sin 58◦ (b) sin 4t]

2. Reduce the following to the cosine of one
   angle:
   (a) cos 71◦cos 33◦−sin 71◦sin 33◦

(b) cos

π

3

cos

π

4

+sin

π

3

sin

π

4

[(a) cos 104◦≡−cos 76◦

(b) cos

π

12

]

3. Show that:
   (a) sin

(

x+

π

3

)

+sin

(

x+

2 π

3

)

=

√

3 cosx

and

(b)−sin

(

3 π

2

−φ

)

=cosφ

4. Prove that:
   (a) sin

(

θ+

π

4

)

−sin

(

θ−

3 π

4

)

=

√

2( sinθ+cosθ)

(b)

cos (270◦+θ)

cos (360◦−θ)

=tanθ