# PROBLEMS IN TRIGONOMETRY FOR GRADE 11

Problem 1 :

If cos θ  =  (1/2) (a + 1/a), show that

cos3θ  =  (1/2)(a3 + 1/a3)

Solution :

cos 3θ  =  3 cos θ - 4 cos3θ

cos 3θ  =  4 [(1/2) (a + 1/a)]3 (1/2) (a + 1/a)

=  4 [(1/8) (a + 1/a)3] - (3/2) (a + 1/a)

=  (a + 1/a) [(1/2) (a + 1/a)2 - (3/2)]

=  (a + 1/a) [(1/2) (a2 + 1/a2 + 2 - (3/2)]

=  (a + 1/a) [(1/2) (a2 + 1/a2 - (1/2)]

=  (1/2) (a + 1/a)(a2 + 1/a2 - (1/2))

(a + b)(a2 - ab + b2)  =  a3 + b

=  (1/2) (a3 + 1/a3)

Problem 2 :

Prove that

cos 5θ = 16cos5θ - 20 cos3 θ + 5cos θ

Solution :

cos 5θ  =  cos (2θ + 3θ)

=  cos 2θ cos 3θ - sin 2θ sin 3θ

(1)  ==> cos 2θ cos 3θ

By applying the formula for cos 2θ and cos 3θ, we get

cos 2θ cos 3θ  =  (2cos2θ - 1)(4cos3θ - 3cosθ)

=  8cos5θ - 6cos3θ - 4cos3θ + 3cosθ

=  8cos5θ - 10cos3θ + 3cosθ    --------(1)

(2)  ==>  sin 2θ sin 3θ

By applying the formula for sin 2θ and sin 3θ, we get

sin 2θ sin 3θ  =  2sinθcosθ(3sinθ-4sin3θ)

=  6sin2θcos θ - 8sin4θ cosθ

=  6(1-cos2θ)cos θ - 8(1-cos2θ)2 cosθ

=  6 cos θ - 6cos3θ - 8 cosθ (1 - 2cos2θ + cos4θ)

=  6 cos θ - 6cos3θ - 8 cosθ + 16cos3θ - 8 cos5θ

=  10cos3θ - 2 cosθ - 8 cos5θ    --------(2)

(1) - (2)

=  (8cos5θ-10cos3θ+3cosθ) - (10cos3θ-2 cosθ-8 cos5θ)

=  8cos5θ-10cos3θ+3cosθ - 10cos3θ+2 cosθ+8 cos5θ

=  16cos5 θ − 20 cos3 θ + 5cos θ.

Hence proved.

Problem 3 :

Prove that

sin 4α = 4 tan α [(1 − tan2α)/(1 + tan2α)2]

Solution :

R.H.S

4 tan α [(1 − tan2α)/(1 + tan2α)2]

=  2 (2 tan α) [(1 − tan2α)/(1 + tan2α)(1 + tan2α)]

=  2 (2 tan α/(1 + tan2α)) [(1 − tan2α)/(1 + tan2α)]

=  2 (sin 2α) (cos2α)

=  sin 2(2α)

=  sin 4α

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