Trigonometric Equations Involving Addition Formula

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 Ө^o - 16 sin⁴ Ө^o + 2 cos 4Ө^o = 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 Ө^o - 16 sin⁴ Ө^o + 2 cos 4Ө^o = 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 1^st Qd or 4^th Qd

                                    2Ө = cos^-1 (0.25/3)      …1^st 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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