Solving
Trigonometric Equations Involving Addition Formula
Q1) Using cos(A + B)
≡ cos A cos B – sin A sin B,
(a) show
that
(i) sin2 Ө =
(1/2)(1 – cos 2Ө),
(ii) cos2 Ө = (1/2)(cos 2Ө +
1)
(4)
f(Ө) = 1 + 10 sin2 Ө - 16 sin4 Ө .
(b) Show
that f(Ө) = 3 cos 2Ө – 2 cos
4Ө.
(4)
(c) Solve
the equation
1 + 10 sin2 Өo - 16 sin4 Өo + 2 cos 4Өo = 0.25, for 0 ≤ Ө ≤ 180
giving your solutions to 1
decimal
place. (4)
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Possible Solutions:
Given: cos (A + B) ≡ cos A cos B – sin A sin B …Eqn
(1)
a)
To show:
(i)
sin2 Ө = (1/2)(1 – cos 2Ө) …Eqn (2)
RHS = (1/2)(1 – cos 2Ө) = (1/2)[1 – cos (Ө + Ө)]
= (1/2)[1 – (cos2 Ө - sin2
Ө)] = (1/2)[1 – (cos2 Ө) + sin2 Ө]
Use identity: cos2 Ө = 1 - sin2 Ө
RHS = (1/2)[1 – (1 - sin2 Ө) + sin2 Ө]
= (1/2)[1 – 1 + sin2
Ө + sin2 Ө]
= (1/2)[2 sin2 Ө]
= sin2 Ө = LHS
(ii)
cos2 Ө = (1/2)(cos 2Ө + 1) …Eqn (3)
RHS = (1/2)(cos 2Ө + 1) = (1/2)[cos (Ө + Ө) + 1]
= (1/2)[(cos2 Ө - sin2
Ө) + 1] = (1/2)[1 + cos2 Ө – (sin2 Ө)]
Use identity: sin2 Ө = 1 - cos2 Ө
RHS = (1/2)[1 + cos2 Ө – (1 - cos2 Ө)]
= (1/2)[1 + cos2
Ө – 1 + cos2 Ө]
= (1/2)[2 cos2 Ө]
= cos2 Ө = LHS
Given:
f(Ө) = 1 + 10 sin2 Ө – 16 sin4 Ө …Eqn (4)
b)
To show: f(Ө) = 3 cos 2Ө – 2 cos 4Ө …Eqn (5)
Fr. (4): f(Ө)
= 1 + 10 sin2 Ө – 16 sin4 Ө
f(Ө) = 1 + 2 sin2 Ө (5 – 8 sin2
Ө)
Use (2): f(Ө)
= 1 + 2 [(1/2)(1 – cos 2Ө)][5 – 8 ((1/2)(1 – cos 2Ө))
f(Ө)
= 1 + (1 – cos 2Ө)[5 – 4(1 – cos 2Ө)]
f(Ө)
= 1 + (1 – cos 2Ө)[5 – 4 + 4 cos 2Ө)]
f(Ө)
= 1 + (1 – cos 2Ө)(1 + 4 cos 2Ө)
f(Ө)
= 1 + (1 + 3 cos 2Ө - 4 cos2
2Ө)
Use (3): f(Ө)
= 1 + [1 + 3 cos 2Ө - 4 ((1/2)(cos 4Ө
+ 1))]
f(Ө)
= 1 + [1 + 3 cos 2Ө - 2(cos 4Ө + 1)]
f(Ө)
= 1 + [1 + 3 cos 2Ө - 2cos 4Ө - 2]
f(Ө)
= 1 + 1 + 3 cos 2Ө - 2cos 4Ө - 2
f(Ө)
= 3 cos 2Ө – 2 cos 4Ө (Shown Eqn 5 from
Eqn 4)
c)
To solve, for 0 ≤ Ө ≤ 180
1 + 10 sin2 Өo - 16 sin4 Өo + 2 cos 4Өo = 0.25 …Eqn
(6)
Fr. (5): 2 cos 4Ө = 3 cos 2Ө – f(Ө) …Eqn
(7)
Put (7) and (4) into (6): (by “double
substitution”)
f(Ө)
+ (3 cos 2Ө – f(Ө)) = 0.25
3
cos 2Ө = 0.25
cos 2Ө = (0.25/3)
Domain for 2Ө: 2 x 0 ≤ 2Ө ≤ 2 x 180 or 0 ≤ 2Ө ≤ 360
Since cos 2Ө is
+ve: 2Ө is in 1st Qd or 4th
Qd
2Ө = cos-1
(0.25/3) …1st Qd
Ө = (1/2) cos-1 (0.25/3)
Ө = 42.6099…
Ө =
42.6 (1 dp)
Or, 2Ө = 360 - cos-1
(0.25/3) …4th Qd
Ө = (1/2) [360 - cos-1
(0.25/3)]
Ө = (1/2) (274.78019..)
Ө = 137.390…
Ө = 137.4 (1 dp)
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