164 Higher Engineering Mathematics
tan
(
x+
π
4
)
=
tanx+tanπ 4
1 −tanxtanπ 4
from the formula fortan(A+B)
=
tanx+ 1
1 −(tanx)( 1 )
=
(
1 +tanx
1 −tanx
)
since tan
π
4
= 1
tan
(
x−
π
4
)
=
tanx−tanπ 4
1 +tanxtanπ 4
=
(
tanx− 1
1 +tanx
)
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=cos54. 51 ◦= 0 .5806 and
tanP=tan54. 51 ◦= 1. 4025
Since cosQ= 0 .4432,Q=cos−^10. 4432 = 63. 69 ◦.
Thus sinQ=sin63. 69 ◦= 0 .8964 and
tanQ=tan63. 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
4sin(x− 20 ◦)=5cosx
for values ofxbetween 0◦and 90◦.
4sin(x− 20 ◦)=4[sinxcos20◦−cosxsin20◦],
from the formula forsin(A−B)
=4[sinx( 0. 9397 )−cosx( 0. 3420 )]
= 3 .7588sinx− 1 .3680cosx
Since 4sin(x− 20 ◦)=5cosxthen
3 .7588sinx− 1 .3680cosx=5cosx
Rearranging gives:
3 .7588sinx=5cosx+ 1 .3680cosx
= 6 .3680cosx
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=4sin( 59. 449 ◦− 20 ◦)
=4sin39. 449 ◦= 2. 542
RHS=5cosx=5cos59. 449 ◦= 2 .542]
Now try the following exercise
Exercise 72 Further problems on
compound angle formulae
- Reduce the following to the sine of one
angle:
(a) sin37◦cos21◦+cos37◦sin21◦
(b) sin7tcos3t−cos 7tsin3t
[(a) sin58◦(b) sin4t] - Reduce the following to the cosine of one
angle:
(a) cos71◦cos33◦−sin71◦sin33◦
(b) cos
π
3
cos
π
4
+sin
π
3
sin
π
4
⎡
⎣
(a)cos104◦≡−cos76◦
(b)cos
π
12
⎤
⎦