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 - 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 + b³ 

  =  (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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