170 Higher Engineering Mathematics
LHS=
1 −cos2θ
sin2θ
=
1 −( 1 −2sin^2 θ)
2sinθcosθ
=
2sin^2 θ
2sinθcosθ
=
sinθ
cosθ
=tanθ=RHS
Problem 13. Prove that
cot 2x+cosec 2x=cotx.
LHS=cot2x+cosec 2x=
cos2x
sin2x
+
1
sin2x
=
cos2x+ 1
sin2x
=
(2cos^2 x− 1 )+ 1
sin2x
=
2cos^2 x
sin2x
=
2cos^2 x
2sinxcosx
=
cosx
sinx
=cotx=RHS
Problem 14. Solve the equation
cos2θ+3sinθ=2forθin the range 0◦≤θ≤ 360 ◦.
Replacing the double angle term with the relationship
cos2θ= 1 −2sin^2 θgives:
1 −2sin^2 θ+3sinθ= 2
Rearranging gives: −2sin^2 θ+3sinθ− 1 = 0
or 2sin^2 θ−3sinθ+ 1 = 0
which is a quadratic in sinθ
Using the quadratic formula or by factorising gives:
(2sinθ− 1 )(sinθ− 1 )= 0
from which, 2sinθ− 1 =0orsinθ− 1 = 0
and sinθ=^12 or sinθ= 1
from which, θ= 30 ◦or 150◦or 90◦
Now try the following exercise
Exercise 74 Further problems on double
angles
- The powerpin an electrical circuit is given by
p=
v^2
R
. Determine the power in terms ofV,
Rand cos2twhenv=Vcos[t.
V^2
2 R
( 1 +cos2t)
]
- Prove the following identities:
(a) 1−
cos2φ
cos^2 φ
=tan^2 φ
(b)
1 +cos2t
sin^2 t
=2cot^2 t
(c)
(tan2x)( 1 +tanx)
tanx
=
2
1 −tanx
(d) 2cosec2θcos2θ=cotθ−tanθ
- If the third harmonic of a waveform is given by
V 3 cos3θ, express the third harmonic in terms
of the first harmonic cosθ,whenV 3 =1.
[cos3θ=4cos^3 θ−3cosθ]
In Problems 4 to 8, solve for θ in the range
− 180 ◦≤θ≤ 180 ◦ - cos2θ=sinθ [− 90 ◦, 30 ◦, 150 ◦]
- 3sin2θ+2cosθ= 0
[− 160. 47 ◦,− 90 ◦,− 19. 47 ◦, 90 ◦] - sin2θ+cosθ= 0
[− 150 ◦,− 90 ◦,− 30 ◦, 90 ◦] - cos2θ+2sinθ=−3[− 90 ◦]
- tanθ+cotθ=2[ 45 ◦,− 135 ◦]
17.4 Changing products of sines and
cosines into sums or differences
(i) sin(A+B)+sin(A−B)=2sinAcosB(fromthe
formulae in Section 17.1)
i.e. sinAcosB
=^12 [sin(A+B)+sin(A−B)] (1)
(ii) sin(A+B)−sin(A−B)=2cosAsinB
i.e. cosAsinB
=^12 [sin(A+B)−sin(A−B)] (2)
(iii) cos(A+B)+cos(A−B)=2cosAcosB
i.e. cosAcosB
=^12 [cos(A+B)+cos(A−B)] (3)